I will solve the exam chemistry 33 task. The structure of the examination work consists of two blocks

Date of: 16.12.2021

For 2-3 months it is impossible to learn (repeat, pull up) such a complex discipline as chemistry.

There are no changes in KIM USE 2020 in chemistry.

Don't delay your preparation.

Before starting the analysis of tasks, first study theory. The theory on the site is presented for each task in the form of recommendations that you need to know when completing the task. guides in the study of the main topics and determines what knowledge and skills will be required when completing the USE tasks in chemistry. For the successful passing of the exam in chemistry, theory is the most important thing.
Theory needs to be backed up practice constantly solving problems. Since most of the errors are due to the fact that I read the exercise incorrectly, I did not understand what is required in the task. The more often you solve thematic tests, the faster you will understand the structure of the exam. Training tasks developed on the basis of demos from FIPI give them the opportunity to decide and find out the answers. But do not rush to peek. First, decide for yourself and see how many points you have scored.

Points for each task in chemistry

1 point - for 1-6, 11-15, 19-21, 26-28 tasks.
2 points - 7-10, 16-18, 22-25, 30, 31.
3 points - 35.
4 points - 32, 34.
5 points - 33.

Total: 60 points.

The structure of the examination paper consists of two blocks:

Questions that require a short answer (in the form of a number or word) - tasks 1-29.
Tasks with detailed answers - tasks 30-35.

3.5 hours (210 minutes) are allotted to complete the examination paper in chemistry.

There will be three cheat sheets on the exam. And they need to be dealt with.

This is 70% of the information that will help you successfully pass the exam in chemistry. The remaining 30% is the ability to use the provided cheat sheets.

If you want to get more than 90 points, you need to spend a lot of time on chemistry.
To successfully pass the exam in chemistry, you need to solve a lot: training tasks, even if they seem easy and of the same type.
Correctly distribute your strength and do not forget about the rest.

Dare, try and you will succeed!

In our last article, we talked about the basic tasks in the exam in chemistry in 2018. Now, we have to analyze in more detail the tasks of an increased (in the USE codifier in chemistry in 2018 - a high level of complexity) level of complexity, previously referred to as part C.

Only five (5) tasks belong to tasks of an increased level of complexity - No. 30,31,32,33,34 and 35. Let's consider the topics of the tasks, how to prepare for them and how to solve difficult tasks in the Unified State Examination in Chemistry 2018.

An example of task 30 in the exam in chemistry 2018

It is aimed at testing the student's knowledge of redox reactions (ORD). The task always gives the equation of a chemical reaction with omissions of substances from either side of the reaction (left side - reagents, right side - products). A maximum of three (3) points can be awarded for this assignment. The first point is given for the correct filling of the gaps in the reaction and the correct equalization of the reaction (arrangement of coefficients). The second point can be obtained by correctly writing the OVR balance, and the last point is given for the correct determination of who is the oxidizing agent in the reaction and who is the reducing agent. Let's analyze the solution of task No. 30 from the demo version of the exam in chemistry in 2018:

Using the electron balance method, write the equation for the reaction

Na 2 SO 3 + ... + KOH à K 2 MnO 4 + ... + H 2 O

Determine the oxidizing agent and reducing agent.

The first thing to do is to place the charges on the atoms indicated in the equation, it turns out:

Na + 2 S +4 O 3 -2 + ... + K + O -2 H + à K + 2 Mn +6 O 4 -2 + ... + H + 2 O -2

Often after this action, we immediately see the first pair of elements that changed the oxidation state (CO), that is, from different sides of the reaction, the same atom has a different oxidation state. In this particular task, we do not observe this. Therefore, it is necessary to take advantage of additional knowledge, namely, on the left side of the reaction, we see potassium hydroxide ( KOH), the presence of which tells us that the reaction proceeds in an alkaline environment. On the right side, we see potassium manganate, and we know that in an alkaline reaction, potassium manganate is obtained from potassium permanganate, therefore, the gap on the left side of the reaction is potassium permanganate ( KMnO 4 ). It turns out that on the left we had manganese in CO +7, and on the right in CO +6, so we can write the first part of the OVR balance:

Mn +7 +1 e — à Mn +6

Now, we can guess what else should happen in the reaction. If manganese receives electrons, then someone had to give them to him (we observe the law of conservation of mass). Consider all the elements on the left side of the reaction: hydrogen, sodium and potassium are already in CO +1, which is the maximum for them, oxygen will not give up its electrons to manganese, which means that sulfur remains in CO +4. We conclude that sulfur gives up electrons and goes into the state of sulfur with CO +6. Now we can write the second part of the balance sheet:

S +4 -2 e — à S +6

Looking at the equation, we see that on the right side, there is no sulfur and sodium anywhere, which means they must be in the gap, and sodium sulfate is a logical compound to fill it ( NaSO 4 ).

Now the OVR balance is written (we get the first score) and the equation takes the form:

Na 2 SO 3 + KMnO 4 + KOHà K 2 MnO 4 + NaSO 4 + H 2 O

Mn +7 +1 e — à Mn +6	1	2
S +4 -2e —à S+6	2	1

It is important to immediately write in this place who is the oxidizing agent and who is the reducing agent, since students often focus on equalizing the equation and simply forget to do this part of the task, thereby losing a point. By definition, an oxidizing agent is the particle that gains electrons (in our case, manganese), and a reducing agent is the particle that donates electrons (in our case, sulfur), so we get:

Oxidizer: Mn +7 (KMnO 4 )

Reducing agent: S +4 (Na 2 SO 3 )

Here it must be remembered that we indicate the state of the particles in which they were when they began to exhibit the properties of an oxidizing or reducing agent, and not the states in which they came as a result of the redox.

Now, to get the last score, you need to correctly equalize the equation (arrange the coefficients). Using the balance, we see that in order for it to go from sulfur +4 to a state of +6, two manganese +7 must become manganese +6, and we put 2 in front of manganese:

Na 2 SO 3 + 2KMnO 4 + KOHà 2K 2 MnO 4 + NaSO 4 + H 2 O

Now we see that we have 4 potassium on the right, and only three on the left, so we need to put 2 in front of potassium hydroxide:

Na 2 SO 3 + 2KMnO 4 + 2KOHà 2K 2 MnO 4 + NaSO 4 + H 2 O

As a result, the correct answer to task number 30 is as follows:

Na 2 SO 3 + 2KMnO 4 + 2KOHà 2K 2 MnO 4 + NaSO 4 + H 2 O

Mn +7 +1e -à Mn+6	1	2
S +4 -2e —à S+6	2	1

Oxidizer: Mn +7 (KMnO 4)

Reducing agent: S +4 (Na 2 SO 3 )

The solution of task 31 in the exam in chemistry

This is a chain of inorganic transformations. To successfully complete this task, it is necessary to have a good understanding of the reactions characteristic of inorganic compounds. The task consists of four (4) reactions, for each of which, you can get one (1) point, for a total of four (4) points, you can get four (4) points for the task. It is important to remember the rules for completing the task: all equations must be equalized, even if the student wrote the equation correctly, but did not equalize, he will not receive a point; it is not necessary to solve all the reactions, you can do one and get one (1) point, two reactions and get two (2) points, etc., it is not necessary to complete the equations in strict order, for example, the student can do reaction 1 and 3, then this is what you need to do, and at the same time get two (2) points, the main thing is to indicate that these are reactions 1 and 3. Let's analyze the solution of task No. 31 from the demo version of the exam in chemistry in 2018:

Iron was dissolved in hot concentrated sulfuric acid. The resulting salt was treated with an excess of sodium hydroxide solution. The brown precipitate formed was filtered off and dried. The resulting substance was heated with iron.
Write the equations for the four described reactions.

For the convenience of the solution, on a draft, you can draw up the following scheme:

To complete the task, of course, you need to know all the proposed reactions. However, there are always hidden clues in the condition (concentrated sulfuric acid, excess sodium hydroxide, brown precipitate, calcined, heated with iron). For example, a student does not remember what happens to iron when it interacts with conc. sulfuric acid, but he remembers that the brown precipitate of iron, after treatment with alkali, is most likely iron hydroxide 3 ( Y = Fe(Oh) 3 ). Now we have the opportunity, by substituting Y in the written scheme, to try to make equations 2 and 3. The subsequent steps are purely chemical, so we will not paint them in such detail. The student must remember that heating iron hydroxide 3 leads to the formation of iron oxide 3 ( Z = Fe 2 O 3 ) and water, and heating iron oxide 3 with pure iron will bring them to the middle state - iron oxide 2 ( FeO). Substance X, which is a salt obtained after reaction with sulfuric acid, while giving iron hydroxide 3 after treatment with alkali, will be iron sulfate 3 ( X = Fe 2 (SO 4 ) 3 ). It is important not to forget to equalize the equations. As a result, the correct answer to task number 31 is as follows:

1) 2Fe + 6H 2 SO 4 (k) a Fe 2 (SO 4) 3+ 3SO 2 + 6H 2 O

2) Fe 2 (SO 4) 3+ 6NaOH (ex) à 2 Fe(OH) 3 + 3Na2SO4

3) 2Fe(OH)3à Fe 2 O 3 + 3H2O

4) Fe 2 O 3 + Fea 3FeO

Task 32 Unified State Examination in Chemistry

Very similar to task #31, only it gives a chain of organic transformations. Design requirements and solution logic are similar to task #31, the only difference is that in task #32 five (5) equations are given, which means that you can score five (5) points in total. Due to the similarity with task number 31, we will not consider it in detail.

The solution of task 33 in chemistry 2018

The calculation task, for its implementation it is necessary to know the basic calculation formulas, be able to use a calculator and draw logical parallels. Task #33 is worth four (4) points. Consider part of the solution to task No. 33 from the USE demo version in chemistry 2018:

Determine the mass fractions (in%) of iron (II) sulfate and aluminum sulfide in the mixture, if during the treatment of 25 g of this mixture with water, a gas was released that completely reacted with 960 g of a 5% solution of copper sulfate. In the answer, write down the reaction equations that specified in the condition of the problem, and give all the necessary calculations (indicate the units of the required physical quantities).

We get the first (1) point for writing the reactions that occur in the problem. Obtaining this particular score depends on the knowledge of chemistry, the remaining three (3) points can only be obtained through calculations, therefore, if a student has problems with mathematics, he must receive at least one (1) point for completing assignment No. 33:

Al 2 S 3 + 6H 2 Oà 2Al(OH)3 + 3H2S

CuSO 4 + H 2 Sà CuS + H 2 SO 4

Since further actions are purely mathematical, we will not analyze them here. You can watch the selection analysis on our YouTube channel (link to the video analysis of task No. 33).

Formulas that will be required to solve this task:

Task 34 in chemistry 2018

Estimated task, which differs from task No. 33 as follows:

- - If in task No. 33 we know which substances interact between, then in task No. 34 we must find what reacted;
  - In task No. 34, organic compounds are given, while in task No. 33, inorganic processes are most often given.

In fact, task No. 34 is the opposite of task No. 33, which means that the logic of the task is the opposite. For task No. 34, you can get four (4) points, while, as in task No. 33, only one of them (in 90% of cases) is obtained for knowledge of chemistry, the remaining 3 (less often 2) points are obtained for mathematical calculations . To successfully complete task No. 34, you must:

Know the general formulas of all the main classes of organic compounds;

Know the basic reactions of organic compounds;

Be able to write an equation in general form.

Once again, I would like to note that the theoretical bases necessary for the successful passing of the exam in chemistry in 2018 have not changed, which means that all the knowledge that your child received at school will help him pass the exam in chemistry in 2018. In our center for preparing for the Unified State Examination and the OGE Hodograph, your child will receive all necessary for the preparation of theoretical materials, and in the classroom will consolidate the knowledge gained for successful implementation all exam assignments. The best teachers who have passed a very large competition and difficult entrance tests will work with him. Classes are held in small groups, which allows the teacher to devote time to each child and form his individual strategy for completing the examination work.

We have no problems with the lack of tests of a new format, our teachers write them themselves, based on all the recommendations of the codifier, specifier and demo version of the Unified State Examination in Chemistry 2018.

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Task number 1

Hydrogen with a volume of 3.36 liters was passed when heated through copper (II) oxide powder, while the hydrogen reacted completely. The reaction gave 10.4 g of a solid residue. This residue was dissolved in 100 g of concentrated sulfuric acid. Determine the mass fraction of salt in the resulting solution (disregard hydrolysis processes).

Answer: 25.4%

Explanation:

ν (H 2) \u003d V (H 2) / V m \u003d 3.36 l / 22.4 l / mol \u003d 0.15 mol,

ν (H 2) \u003d ν (Cu) \u003d 0.15 mol, therefore, m (Cu) \u003d 0.15 mol 64 g / mol \u003d 9.6 g

m(CuO) \u003d m (solid rest.) - m (Cu) \u003d 10.4 g - 9.6 g \u003d 0.8 g

ν(CuO) = m(CuO)/M(CuO) = 0.8 g/80 g/mol = 0.01 mol

According to equation (I) ν(Cu) = ν I (CuSO 4), according to equation (II) ν (CuO) = ν II (CuSO 4), therefore, ν total. (CuSO 4) \u003d ν I (CuSO 4) + ν II (CuSO 4) \u003d 0.01 mol + 0.15 mol \u003d 0.16 mol.

m total (CuSO 4) = vtot. (CuSO 4) M (CuSO 4) \u003d 0.16 mol 160 g / mol \u003d 25.6 g

ν (Cu) \u003d ν (SO 2), therefore, ν (SO 2) \u003d 0.15 mol and m (SO 2) \u003d ν (SO 2) M (SO 2) \u003d 0.15 mol 64 g / mol = 9.6 g

m (solution) \u003d m (solid rest.) + m (solution H 2 SO 4) - m (SO 2) \u003d 10.4 g + 100 g - 9.6 g \u003d 100.8 g

ω (CuSO 4) \u003d m (CuSO 4) / m (solution) 100% \u003d 25.6 g / 100.8 g 100% \u003d 25.4%

Task number 2

Hydrogen with a volume of 3.36 l (n.o.) was passed by heating over a powder of copper (II) oxide weighing 16 g. The residue formed as a result of this reaction was dissolved in 535.5 g of 20% nitric acid, as a result of which a colorless gas that boils in air. Determine the mass fraction of nitric acid in the resulting solution (disregard hydrolysis processes).

In your answer, write down the reaction equations that are indicated in the condition of the problem, and give all the necessary calculations (indicate the units of measurement of the initial physical quantities).

Answer: 13.84%

Explanation:

When hydrogen is passed over copper (II) oxide, copper is reduced:

CuO + H 2 → Cu + H 2 O (heating) (I)

The solid residue, consisting of metallic copper and copper (II) oxide, reacts with a solution of nitric acid according to the equations:

3Cu + 8HNO 3 (20% solution) → 3Cu(NO 3) 2 + 2NO + 4H 2 O (II)

CuO + 2HNO 3 (20% solution) → Cu(NO 3) 2 + H 2 O (III)

Let us calculate the amount of hydrogen substance and copper oxide (II) participating in reaction (I):

ν (H 2) \u003d V (H 2) / V m \u003d 3.36 l / 22.4 l / mol \u003d 0.15 mol, ν (CuO) \u003d 16 g / 80 g / mol \u003d 0.2 mol

According to the reaction equation (I) ν (H 2) \u003d ν (CuO), and according to the condition of the problem, the amount of hydrogen substance is in short supply (0.15 mol H 2 and 0.1 mol CuO), therefore copper (II) oxide did not completely react .

We carry out the calculation according to the lack of substance, therefore, ν (Cu) \u003d ν (H 2) \u003d 0.15 mol and ν rest. (CuO) \u003d 0.2 mol - 0.15 mol \u003d 0.05 mol.

To further calculate the mass of the solution, it is necessary to know the masses of the formed copper and unreacted copper (II) oxide:

m rest. (CuO) = ν(CuO) M(CuO) = 0.05 mol 80 g/mol = 4 g

The total mass of the solid residue is: m(solid rest.) = m(Cu) + m rest. (CuO) = 9.6 g + 4 g = 13.6 g

Calculate the initial mass and amount of nitric acid substance:

m ref. (HNO 3) \u003d m (p-ra HNO 3) ω (HNO 3) \u003d 535.5 g 0.2 \u003d 107.1 g

According to the reaction equation (II) ν II (HNO 3) = 8/3ν (Cu), according to the reaction equation (III) ν III (HNO 3) = 2v (CuO), therefore, ν total. (HNO 3) \u003d ν II (HNO 3) + ν III (HNO 3) \u003d 8/3 0.15 mol + 2 0.05 mol \u003d 0.5 l.

The total mass of reacted as a result of reactions (II) and (III) is equal to:

m rest. (HNO 3) = m ref. (HNO 3) – m total. (HNO 3) \u003d 107.1 g - 31.5 g \u003d 75.6 g

In order to calculate the mass of the resulting solution, it is necessary to take into account the mass of nitric oxide (II) released in reaction (II):

ν(NO) = 2/3ν(Cu), therefore, ν(NO) = 2/3 0.15 mol = 0.1 mol and m(NO) = ν(NO) M(NO) = 0, 1 mol 30 g/mol = 3 g

Calculate the mass of the resulting solution:

m (solution) \u003d m (solid rest.) + m (solution HNO 3) - m (NO) \u003d 13.6 g + 535.5 g - 3 g \u003d 546.1 g

ω(HNO 3) = m rest. (HNO 3) / m (solution) 100% \u003d 75.6 g / 546.1 g 100% \u003d 13.84%

Task number 3

To a 20% salt solution obtained by dissolving 12.5 g of copper sulfate (CuSO 4 · 5H 2 O) in water, 5.6 g of iron was added. After the completion of the reaction, 117 g of a 10% sodium sulfide solution was added to the solution. Determine the mass fraction of sodium sulfide in the final solution (disregard hydrolysis processes).

Answer: 5.12%

Explanation:

Fe + CuSO 4 → FeSO 4 + Cu (I)

ν (CuSO 4 5H 2 O) \u003d m (CuSO 4 5H 2 O) / M (CuSO 4 5H 2 O) \u003d 12.5 g / 250 g / mol \u003d 0.05 mol

ν ref. (Fe) = m ref. (Fe)/M(Fe) = 5.6 g/56 g/mol = 0.1 mol

According to the reaction equation (I), ν (Fe) = ν (CuSO 4), and according to the condition of the problem, the amount of copper sulfate substance is in short supply (0.05 mol CuSO 4 5H 2 O and 0.1 mol Fe), so the iron did not react fully.

Only iron (II) sulfate interacts with sodium sulfide:

FeSO 4 + Na 2 S → FeS↓ + Na 2 SO 4 (II)

We calculate by the lack of substance, therefore, ν (CuSO 4 5H 2 O) \u003d ν (Cu) \u003d ν (FeSO 4) \u003d 0.05 mol and ν rest. (Fe) \u003d 0.1 mol - 0.05 mol \u003d 0.05 mol.

To further calculate the mass of the final solution, it is necessary to know the masses of the formed copper, unreacted iron (reaction (I)) and the initial solution of copper sulphate:

m(Cu) = ν(Cu) M(Cu) = 0.05 mol 64 g/mol = 3.2 g

m rest. (Fe) = ν rest. (Fe) M(Fe) = 0.05 mol 56 g/mol = 2.8 g

ν (CuSO 4 5H 2 O) \u003d ν (CuSO 4) \u003d 0.05 mol, therefore, m (CuSO 4) \u003d ν (CuSO 4) M (CuSO 4) \u003d 0.05 mol 160 g / mol = 8 g

m ref. (p-ra CuSO 4) \u003d m (CuSO 4) / ω (CuSO 4) 100% \u003d 8 g / 20% 100% \u003d 40 g

Only iron (II) sulfate interacts with sodium sulfide (copper (II) sulfate reacted completely according to reaction (I)).

m ref. (Na 2 S) \u003d m ref. (p-ra Na 2 S) ω (Na 2 S) \u003d 117 g 0.1 \u003d 11.7 g

ν ref. (Na 2 S) \u003d m ref. (Na 2 S) / M (Na 2 S) \u003d 11.7 g / 78 g / mol \u003d 0.15 mol

According to the reaction equation (II), ν (Na 2 S) = ν (FeSO 4), and according to the reaction condition, sodium sulfide is in excess (0.15 mol Na 2 S and 0.05 mol FeSO 4). We calculate by deficiency, i.e. by the amount of iron sulfate (II) substance).

Calculate the mass of unreacted sodium sulfide:

ν rest. (Na 2 S) \u003d ν ref. (Na 2 S) - ν react. (Na 2 S) \u003d 0.15 mol - 0.05 mol \u003d 0.1 mol

m rest. (Na 2 S) \u003d ν (Na 2 S) M (Na 2 S) \u003d 0.1 mol 78 g / mol \u003d 7.8 g

To calculate the mass of the final solution, it is necessary to calculate the mass of iron (II) sulfide precipitated by reaction (II):

ν (FeSO 4) \u003d ν (FeS) and m (FeS) \u003d ν (FeS) M (FeS) \u003d 0.05 mol 88 g / mol \u003d 4.4 g

m (r-ra) \u003d m ref. (p-ra CuSO 4) + m ref. (Fe) - m rest. (Fe) – m(Cu) + m ref. (p-ra Na 2 S) - m (FeS) \u003d 40 g + 5.6 g - 3.2 g - 2.8 g + 117 g - 4.4 g \u003d 152.2 g

ω (Na 2 S) \u003d m (Na 2 S) / m (solution) 100% \u003d 7.8 g / 152.2 g 100% \u003d 5.12%

Task number 4

To a 20% salt solution obtained by dissolving 37.5 g of copper sulfate (CuSO 4 · 5H 2 O) in water, 11.2 g of iron was added. After completion of the reaction, 100 g of a 20% sulfuric acid solution was added to the resulting mixture. Determine the mass fraction of salt in the resulting solution (disregard hydrolysis processes).

Answer: 13.72%

Explanation:

When copper (II) sulfate interacts with iron, a substitution reaction occurs:

Fe + CuSO 4 → FeSO 4 + Cu (I)

20% sulfuric acid reacts with iron according to the equation:

Fe + H 2 SO 4 (diff.) → FeSO 4 + H 2 (II)

Let us calculate the amount of copper sulphate and iron entering into reaction (I):

ν (CuSO 4 5H 2 O) \u003d m (CuSO 4 5H 2 O) / M (CuSO 4 5H 2 O) \u003d 37.5 g / 250 g / mol \u003d 0.15 mol

ν ref. (Fe) = m ref. (Fe)/M(Fe) = 11.2 g/56 g/mol = 0.2 mol

According to the reaction equation (I), ν (Fe) = ν (CuSO 4), and according to the condition of the problem, the amount of copper sulfate substance is in short supply (0.15 mol CuSO 4 5H 2 O and 0.2 mol Fe), so the iron did not react fully.

We calculate by the lack of substance, therefore, ν (CuSO 4 5H 2 O) \u003d ν (Cu) \u003d ν (FeSO 4) \u003d 0.15 mol and ν rest. (Fe) \u003d 0.2 mol - 0.15 mol \u003d 0.05 mol.

m(Cu) = ν(Cu) M(Cu) = 0.15 mol 64 g/mol = 9.6 g

ν (CuSO 4 5H 2 O) \u003d ν (CuSO 4) \u003d 0.15 mol, therefore, m (CuSO 4) \u003d ν (CuSO 4) M (CuSO 4) \u003d 0.15 mol 160 g / mol = 24 g

m ref. (p-ra CuSO 4) \u003d m (CuSO 4) / ω (CuSO 4) 100% \u003d 24 g / 20% 100% \u003d 120 g

Dilute sulfuric acid does not react with copper, but interacts with iron according to reaction (II).

Calculate the mass and amount of sulfuric acid substance:

m ref. (H 2 SO 4) = m ref. (p-ra H 2 SO 4) ω (H 2 SO 4) \u003d 100 g 0.2 \u003d 20 g

ν ref. (H 2 SO 4) = m ref. (H 2 SO 4) / M (H 2 SO 4) \u003d 20 g / 98 g / mol ≈ 0.204 mol

Since ν rest. (Fe) = 0.05 mol, and ν ref. (H 2 SO 4) ≈ 0.204 mol, therefore, iron is in short supply and is completely dissolved by sulfuric acid.

According to the equation of reaction (II) ν (Fe) \u003d ν (FeSO 4), then the total amount of iron (II) sulfate substance is the sum of the quantities forming according to reactions (I) and (II), and are equal to:

ν (FeSO 4) \u003d 0.05 mol + 0.15 mol \u003d 0.2 mol;

m (FeSO 4) \u003d ν (FeSO 4) M (FeSO 4) \u003d 0.2 mol 152 g / mol \u003d 30.4 g

ν rest. (Fe) \u003d ν (H 2) \u003d 0.05 mol and m (H 2) \u003d ν (H 2) M (H 2) \u003d 0.05 mol 2 g / mol \u003d 0.1 g

The mass of the resulting solution is calculated by the formula (the mass of iron that did not react in reaction (I) is not taken into account, since in reaction (II) it goes into solution):

m (r-ra) \u003d m ref. (p-ra CuSO 4) + m ref. (Fe) - m(Cu) + m ref. (p-ra H 2 SO 4) - m (H 2) \u003d 120 g + 11.2 g - 9.6 g + 100 g - 0.1 g \u003d 221.5 g

The mass fraction of iron sulfate (II) in the resulting solution is equal to:

ω (FeSO 4) \u003d m (FeSO 4) / m (solution) 100% \u003d 30.4 g / 221.5 g 100% \u003d 13.72%

Task number 5

To a 20% salt solution obtained by dissolving 50 g of copper sulfate (CuSO 4 · 5H 2 O) in water, 14.4 g of magnesium was added. After completion of the reaction, 146 g of a 25% hydrochloric acid solution was added to the resulting mixture. Calculate the mass fraction of hydrogen chloride in the resulting solution (disregard hydrolysis processes).

Answer: 2.38%

Explanation:

When copper (II) sulfate interacts with magnesium, a substitution reaction occurs:

Mg + CuSO 4 → MgSO 4 + Cu(I)

25% hydrochloric acid reacts with magnesium according to the equation:

Mg + 2HCl → MgCl 2 + H 2 (II)

Let us calculate the amount of copper sulphate and magnesium substance that enter into reaction (I):

According to the reaction equation (I) ν (Mg) = ν (CuSO 4), and according to the condition of the problem, the amount of copper sulfate substance is in short supply (0.2 mol CuSO 4 5H 2 O and 0.6 mol Mg), so magnesium did not react fully.

The calculation is carried out according to the lack of substance, therefore, ν (CuSO 4 5H 2 O) \u003d ν (Cu) \u003d ν react. (Mg) = 0.2 mol and ν rest. (Mg) \u003d 0.6 mol - 0.2 mol \u003d 0.4 mol.

To further calculate the mass of the final solution, it is necessary to know the mass of the formed copper (reaction (I)) and the initial solution of copper sulphate:

m ref. (p-ra CuSO 4) \u003d m (CuSO 4) / ω (CuSO 4) 100% \u003d 32 g / 20% 100% \u003d 160 g

Hydrochloric acid does not react with copper, but interacts with magnesium according to reaction (II).

Calculate the mass and amount of hydrochloric acid substance:

m ref. (HCl) = m ref. (solution HCl) ω(HCl) = 146 g 0.25 = 36.5 g

Since ν rest. (Mg) = 0.4 mol, ν ref. (HCl) = 1 mol and ν ref. (HCl) > 2v rest. (Mg), then magnesium is deficient and completely dissolves in hydrochloric acid.

Calculate the amount of substance unreacted with magnesium hydrochloric acid:

ν rest. (HCl) = ν ref. (HCl) – ν react. (HCl) \u003d 1 mol - 2 0.4 mol \u003d 0.2 mol

m rest. (HCl) = ν rest. (HCl) M(HCl) = 0.2 mol 36.5 g/mol = 7.3 g

To calculate the mass of the final solution, it is necessary to calculate the mass of hydrogen released as a result of reaction (II):

ν rest. (Mg) \u003d ν (H 2) \u003d 0.4 mol and m (H 2) \u003d ν (H 2) M (H 2) \u003d 0.4 mol 2 g / mol \u003d 0.8 g

The mass of the resulting solution is calculated by the formula (the mass of unreacted in reaction (I) and magnesium is not taken into account, since in reaction (II) it goes into solution):

m (p-ra) \u003d m ref (p-ra CuSO 4) + m ref. (Mg) - m(Cu) + m ref. (solution HCl) - m (H 2) \u003d 160 g + 14.4 g - 12.8 g + 146 g - 0.8 g \u003d 306.8 g

The mass fraction of hydrochloric acid in the resulting solution is:

ω(HCl) = m rest. (HCl) / m (solution) 100% \u003d 7.3 g / 306.8 g 100% \u003d 2.38%

Task number 6

To a 10% salt solution obtained by dissolving 25 g of copper sulfate (CuSO 4 · 5H 2 O) in water, 19.5 g of zinc was added. After completion of the reaction, 240 g of 30% sodium hydroxide solution was added to the resulting mixture. Determine the mass fraction of sodium hydroxide in the resulting solution (disregard hydrolysis processes).

Answer: 9.69%

Explanation:

Zn + CuSO 4 → ZnSO 4 + Cu (I)

According to the reaction equation (I), ν (Zn) = ν (CuSO 4), and according to the condition of the problem, the amount of copper sulphate substance is in short supply (0.1 mol CuSO 4 5H 2 O and 0.3 mol Zn), so zinc did not react fully.

We calculate by the lack of substance, therefore, ν (CuSO 4 5H 2 O) \u003d ν (ZnSO 4) \u003d ν (Cu) \u003d ν react. (Zn) = 0.1 mol and ν rest. (Zn) \u003d 0.3 mol - 0.1 mol \u003d 0.2 mol.

To further calculate the mass of the final solution, it is necessary to know the mass of the formed copper (reaction (I)) and the initial solution of copper sulphate:

m ref. (p-ra CuSO 4) \u003d m (CuSO 4) / ω (CuSO 4) 100% \u003d 16 g / 10% 100% \u003d 160 g

m ref. (NaOH) = m ref. (p-ra NaOH) ω(NaOH) = 240 g 0.3 = 72 g

ν ref. (NaOH) = m ref. (NaOH)/M(NaOH) = 72 g/40 g/mol = 1.8 mol

vtot. (NaOH) \u003d ν II (NaOH) + ν III (NaOH) \u003d 2 0.2 mol + 4 0.1 mol \u003d 0.8 mol

m react. (NaOH) = ν react. (NaOH) M(NaOH) = 0.8 mol 40 g/mol = 32 g

m rest. (NaOH) = m ref. (NaOH) - m react. (NaOH) = 72 g - 32 g = 40 g

To calculate the mass of the final solution, it is necessary to calculate the mass of hydrogen released as a result of reaction (II):

ν rest. (Zn) \u003d ν (H 2) \u003d 0.2 mol and m (H 2) \u003d ν (H 2) M (H 2) \u003d 0.2 mol 2 g / mol \u003d 0.4 g

m (r-ra) \u003d m ref. (p-ra CuSO 4) + m ref. (Zn) - m(Cu) + m ref. (p-ra NaOH) - m (H 2) \u003d 160 g + 19.5 g - 6.4 g + 240 g - 0.4 g \u003d 412.7 g

ω(NaOH) = m rest. (NaOH)/m(solution) 100% = 40 g/412.7 g 100% = 9.69%

Task number 7

In a 20% salt solution obtained by dissolving 25 g of pentahydrate copper sulfate (II) in water, the powder obtained by sintering 2.16 g of aluminum and 6.4 g of iron oxide (III) was added. Determine the mass fraction of copper (II) sulfate in the resulting solution (disregard hydrolysis processes).

Answer: 4.03%

Explanation:

When aluminum is sintered with iron (III) oxide, the more active metal displaces the less active from its oxide:

2Al + Fe 2 O 3 → Al 2 O 3 + 2Fe (I)

Let us calculate the amount of aluminum substance and iron oxide (III) entering into reaction (I):

ν ref. (Al) = m ref. (Al)/M(Al) = 2.16 g / 27 g/mol = 0.08 mol

ν ref. (Fe 2 O 3) = m ref. (Fe 2 O 3) / M (Fe 2 O 3) \u003d 6.4 g / 160 g / mol \u003d 0.04 mol

According to the reaction equation (I) ν(Al) = 2ν(Fe 2 O 3) = 2ν(Al 2 O 3) and according to the condition of the problem, the amount of aluminum substance is twice the amount of iron oxide (III) substance, therefore, unreacted substances in reaction (I) does not remain.

The amount of substance and the mass of the formed iron are equal:

ν(Fe) = 2ν ref. (Fe 2 O 3) \u003d 2 0.04 mol \u003d 0.08 mol

m(Fe) = ν(Fe) M(Fe) = 0.08 mol 56 g/mol = 4.48 g

To further calculate the mass of the final solution, it is necessary to know the mass of the initial solution of copper sulphate:

ν (CuSO 4 5H 2 O) \u003d m (CuSO 4 5H 2 O) / M (CuSO 4 5H 2 O) \u003d 25 g / 250 g / mol \u003d 0.1 mol

ν (CuSO 4 5H 2 O) \u003d ν (CuSO 4) \u003d 0.1 mol, therefore, m (CuSO 4) \u003d ν (CuSO 4) M (CuSO 4) \u003d 0.1 mol 160 g / mol = 16 g

m ref. (p-ra CuSO 4) \u003d m (CuSO 4) / ω (CuSO 4) 100% \u003d 16 g / 20% 100% \u003d 80 g

The iron formed by reaction (I) reacts with a solution of copper sulphate:

Fe + CuSO 4 → FeSO 4 + Cu (II)

According to the reaction equation (II), ν(Fe) = ν(CuSO 4), and according to the condition of the problem, the amount of iron substance (0.1 mol CuSO 4 5H 2 O and 0.08 mol Fe), so the iron reacted completely.

Calculate the amount of substance and the mass of unreacted copper (II) sulfate:

ν rest. (CuSO 4) \u003d ν ref. (CuSO 4) - ν react. (CuSO 4) \u003d 0.1 mol - 0.08 mol \u003d 0.02 mol

m rest. (CuSO 4) \u003d ν rest. (CuSO 4) M (CuSO 4) \u003d 0.02 mol 160 g / mol \u003d 3.2 g

To calculate the mass of the final solution, it is necessary to calculate the mass of copper formed:

ν(Fe) = ν(Cu) = 0.08 mol and m(Cu) = ν(Cu) M(Cu) = 0.08 mol 64 g/mol = 5.12 g

The mass of the resulting solution is calculated by the formula (the iron formed by reaction (I) subsequently passes into the solution):

m (r-ra) \u003d m ref. (p-ra CuSO 4) + m (Fe) - m (Cu) \u003d 80 g + 4.48 g - 5.12 g \u003d 79.36 g

Mass fraction of copper (II) sulfate in the resulting solution:

ω(CuSO 4) = m rest. (CuSO 4) / m (solution) 100% \u003d 3.2 g / 79.36 g 100% \u003d 4.03%

Task number 8

18.2 g of calcium phosphide was added to 182.5 g of a 20% hydrochloric acid solution. Next, 200.2 g of Na 2 CO 3 10H 2 O was added to the resulting solution. Determine the mass fraction of sodium carbonate in the resulting solution (disregard hydrolysis processes).

Answer: 5.97%

Explanation:

Hydrochloric acid and calcium phosphide react to form calcium chloride and release phosphine:

Ca 3 P 2 + 6HCl → 3CaCl 2 + 2PH 3 (I)

Let us calculate the amount of hydrochloric acid and calcium phosphide substance entering into reaction (I):

m ref. (HCl) \u003d m (p-ra HCl) ω (HCl) \u003d 182.5 g 0.2 \u003d 36.5 g, hence

ν ref. (HCl) = m ref. (HCl)/M(HCl) = 36.5 g/36.5 g/mol = 1 mol

ν ref. (Ca 3 P 2) = m ref. (Ca 3 P 2) / M (Ca 3 P 2) \u003d 18.2 g / 182 g / mol \u003d 0.1 mol

According to the reaction equation (I), ν (HCl) \u003d 6ν (Ca 3 P 2) \u003d 2ν (CaCl 2), and according to the condition of the problem, the amount of hydrochloric acid substance is 10 times greater than the amount of calcium phosphide substance, therefore, hydrochloric acid remains unreacted.

ν rest. (HCl) = ν ref. (HCl) - 6ν (Ca 3 P 2) \u003d 1 mol - 6 0.1 mol \u003d 0.4 mol

The amount of substance and the mass of the resulting phosphine are equal to:

ν(PH 3) = 2ν ref. (Ca 3 P 2) \u003d 2 0.1 mol \u003d 0.2 mol

m(PH 3) \u003d ν (PH 3) M (PH 3) \u003d 0.2 mol 34 g / mol \u003d 6.8 g

Calculate the amount of sodium carbonate hydrate:

ν ref. (Na 2 CO 3 10H 2 O) = m ref. (Na 2 CO 3 10H 2 O) / M (Na 2 CO 3 10H 2 O) \u003d 200.2 g / 286 g / mol \u003d 0.7 mol

Both calcium chloride and hydrochloric acid interact with sodium carbonate:

Na 2 CO 3 + CaCl 2 → CaCO 3 ↓ + 2NaCl (II)

Na 2 CO 3 + 2HCl → 2NaCl + CO 2 + H 2 O (III)

Let's calculate the total amount of sodium carbonate substance interacting with hydrochloric acid and calcium chloride:

ν react. (Na 2 CO 3) \u003d ν (CaCl 2) + 1 / 2ν rest. (HCl) = 3v ref. (Ca 3 P 2) + 1/2ν rest. (HCl) \u003d 3 0.1 mol + 1/2 0.4 mol \u003d 0.3 mol + 0.2 mol \u003d 0.5 mol

The total amount of the substance and the mass of unreacted sodium carbonate are equal to:

ν rest. (Na 2 CO 3) \u003d ν ref. (Na 2 CO 3) - ν react. (Na 2 CO 3) \u003d 0.7 mol - 0.5 mol \u003d 0.2 mol

m rest. (Na 2 CO 3) \u003d ν rest. (Na 2 CO 3) M (Na 2 CO 3) \u003d 0.2 mol 106 g / mol \u003d 21.2 g

To further calculate the mass of the final solution, it is necessary to know the masses of calcium carbonate precipitated by reaction (II) and carbon dioxide emitted by reaction (III):

ν(CaCl 2) = ν(CaCO 3) = 3ν ref. (Ca 3 P 2) = 0.3 mol

m (CaCO 3) \u003d ν (CaCO 3) M (CaCO 3) \u003d 0.3 mol 100 g / mol \u003d 30 g

ν(CO 2) = 1/2ν rest. (HCl) = ½ 0.4 mol = 0.2 mol

The mass of the resulting solution is calculated by the formula:

m (r-ra) \u003d m ref. (solution HCl) + m ref. (Ca 3 P 2) - m (PH 3) + m ref. (Na 2 CO 3 10H 2 O) - m (CaCO 3) - m (CO 2) \u003d 182.5 g + 18.2 g - 6.8 g + 200.2 g - 30 g - 8.8 g = 355.3 g

The mass fraction of sodium carbonate is equal to:

ω(Na 2 CO 3) = m rest. (Na 2 CO 3) / m (solution) 100% \u003d 21.2 g / 355.3 g 100% \u003d 5.97%

Task number 9

Sodium nitride weighing 8.3 g reacted with 490 g of 20% sulfuric acid. After completion of the reaction, 57.2 g of crystalline soda (Na 2 CO 3 · 10H 2 O) was added to the resulting solution. Determine the mass fraction of sulfuric acid in the resulting solution (disregard hydrolysis processes).

Answer: 10.76%

Explanation:

Sodium nitride and dilute sulfuric acid react to form two medium salts - ammonium and sodium sulfate:

2Na 3 N + 4H 2 SO 4 → 3Na 2 SO 4 + (NH 4) 2 SO 4 (I)

Let's calculate the amount of sulfuric acid and sodium nitride substance reacting with each other:

m ref. (H 2 SO 4) \u003d m (solution H 2 SO 4) ω (H 2 SO 4) \u003d 490 g 0.2 \u003d 98 g, hence

ν ref. (H 2 SO 4) = m ref. (H 2 SO 4) / M (H 2 SO 4) \u003d 98 g / 98 g / mol \u003d 1 mol

ν ref. (Na 3 N) \u003d m ref. (Na 3 N) / M (Na 3 N) \u003d 8.3 g / 83 g / mol \u003d 0.1 mol

Let us calculate the amount of sulfuric acid that has not reacted in reaction (I):

ν rest. I (H 2 SO 4) \u003d ν ref. (H 2 SO 4) - 2v ref. (Na 3 N) \u003d 1 mol - 2 0.1 mol \u003d 0.8 mol

Let's calculate the amount of crystalline soda substance:

ν ref. (Na 2 CO 3 10H 2 O) = m ref. (Na 2 CO 3 10H 2 O) / M (Na 2 CO 3 10H 2 O) \u003d 57.2 g / 286 g / mol \u003d 0.2 mol

Since, according to the condition of the problem, ν rest. I (H 2 SO 4) = 3v ref. (Na 2 CO 3 10H 2 O), i.e. dilute sulfuric acid is in excess, therefore, the following reaction occurs between these substances:

H 2 SO 4 + Na 2 CO 3 → Na 2 SO 4 + CO 2 + H 2 O (II)

ν rest.II (H 2 SO 4) = ν rest.I (H 2 SO 4) - ν ref. (Na 2 CO 3) \u003d 0.8 mol - 0.2 mol \u003d 0.6 mol

m rest. II (H 2 SO 4) \u003d ν rest. II (H 2 SO 4) M (H 2 SO 4) \u003d 0.6 mol 98 g / mol \u003d 58.8 g

ν (CO 2) \u003d ν (Na 2 CO 3) \u003d 0.2 mol

m(CO 2) \u003d ν (CO 2) M (CO 2) \u003d 0.2 mol 44 g / mol \u003d 8.8 g

m (r-ra) \u003d m ref. (solution H 2 SO 4) + m ref. (Na 3 N) + m (Na 2 CO 3 10H 2 O) - m (CO 2) \u003d 490 g + 8.3 g + 57.2 g - 8.8 g \u003d 546.7 g

The mass fraction of sulfuric acid is:

ω rest. II (H 2 SO 4) \u003d m rest. II (H 2 SO 4) / m (solution) 100% \u003d 58.8 g / 546.7 g 100% \u003d 10.76%

Task number 10

Lithium nitride weighing 3.5 g was dissolved in 365 g of 10% hydrochloric acid. 20 g of calcium carbonate was added to the solution. Determine the mass fraction of hydrochloric acid in the resulting solution (disregard hydrolysis processes).

Answer: 1.92%

Explanation:

Lithium nitride and hydrochloric acid react to form two salts - lithium and ammonium chlorides:

Li 3 N + 4HCl → 3LiCl + NH 4 Cl (I)

Let's calculate the amount of hydrochloric acid and lithium nitride that react with each other:

m ref. (HCl) \u003d m (p-ra HCl) ω (HCl) \u003d 365 g 0.1 \u003d 36.5 g, hence

ν ref. (HCl) = m ref. (HCl)/M(HCl) = 36.5 g/36.5 g/mol = 1 mol

ν ref. (Li 3 N) = m ref. (Li 3 N) / M (Li 3 N) \u003d 3.5 g / 35 g / mol \u003d 0.1 mol

Let us calculate the amount of hydrochloric acid that has not reacted in reaction (I):

ν rest. I (HCl) = ν ref. (HCl) - 4v ref. (Li 3 N) \u003d 1 mol - 4 0.1 mol \u003d 0.6 mol

Calculate the amount of calcium carbonate substance:

ν ref. (CaCO 3) \u003d m ref. (CaCO 3) / M (CaCO 3) \u003d 20 g / 100 g / mol \u003d 0.2 mol

Since, according to the condition of the problem, ν rest. I (HCl) = 3v ref. (CaCO 3), an excess of hydrochloric acid interacts with calcium carbonate with the release of carbon dioxide and the formation of calcium chloride:

CaCO 3 + 2HCl → CaCl 2 + CO 2 + H 2 O (II)

ν rest.II (HCl) = ν rest.I (HCl) - ν ref. (CaCO 3) \u003d 0.6 mol - 2 0.2 mol \u003d 0.2 mol

m rest. II (HCl) \u003d ν rest. II (HCl) M (HCl) \u003d 0.2 mol 36.5 g / mol \u003d 7.3 g

To further calculate the mass of the final solution, it is necessary to know the masses of carbon dioxide emitted by reaction (II):

ν (CO 2) \u003d ν (CaCO 3) \u003d 0.2 mol

m(CO 2) \u003d ν (CO 2) M (CO 2) \u003d 0.2 mol 44 g / mol \u003d 8.8 g

The mass of the resulting solution is calculated by the formula equal to:

m (r-ra) \u003d m ref. (solution HCl) + m ref. (Li 3 N) + m (CaCO 3) - m (CO 2) \u003d 365 g + 3.5 g + 20 g - 8.8 g \u003d 379.7 g

The mass fraction of hydrochloric acid is equal to:

ω rest. II (HCl) = m rest. II (HCl) / m (solution) 100% \u003d 7.3 g / 379.7 g 100% \u003d 1.92%

Task number 11

The solid residue obtained by the interaction of 2.24 l of hydrogen with 12 g of copper (II) oxide was dissolved in 126 g of an 85% nitric acid solution. Determine the mass fraction of nitric acid in the resulting solution (disregard hydrolysis processes).

Answer: 59.43%

Explanation:

When hydrogen is passed over copper (II) oxide, copper is reduced:

CuO + H 2 → Cu + H 2 O (heating) (I)

Let us calculate the amount of hydrogen substance involved in the reduction of copper oxide (II):

ν ref. (H 2) \u003d V (H 2) / V m \u003d 2.24 l / 22.4 l / mol \u003d 0.1 mol,

ν ref. (CuO) = 12 g/80 g/mol = 0.15 mol

According to equation (I) ν (CuO) = ν (H 2) = ν (Cu), therefore, 0.1 mol of copper is formed and ν remains. (CuO) \u003d ν (solid rest.) - ν ref. (H 2) \u003d 0.15 mol - 0.1 mol \u003d 0.05 mol

Let us calculate the masses of formed copper and unreacted copper (II) oxide:

m rest. (CuO) = ν rest. (CuO) M(CuO) = 0.05 mol 80 g/mol = 4 g

m(Cu) = ν(Cu) M(Cu) = 0.1 mol 64 g/mol = 6.4 g

The solid residue, consisting of metallic copper and unreacted copper (II) oxide, reacts with nitric acid according to the equations:

Cu + 4HNO 3 → Cu(NO 3) 2 + 2NO 2 + 2H 2 O (II)

CuO + 2HNO 3 → Cu(NO 3) 2 + H 2 O (III)

Calculate the amount of nitric acid substance:

m ref. (HNO 3) \u003d m (p-ra HNO 3) ω (HNO 3) \u003d 126 g 0.85 \u003d 107.1 g, hence

ν ref. (HNO 3) = m ref. (HNO 3) / M (HNO 3) \u003d 107.1 g / 63 g / mol \u003d 1.7 mol

According to equation (II) ν II (HNO 3) = 4ν (Cu), according to equation (III) ν III (HNO 3) = 2ν rest. (CuO), therefore, ν total. (HNO 3) \u003d ν II (HNO 3) + ν III (HNO 3) \u003d 4 0.1 mol + 2 0.05 mol \u003d 0.5 mol.

Let us calculate the total mass of nitric acid reacting according to reactions (II) and (III):

m total (HNO 3) = vtot. (HNO 3) M (HNO 3) \u003d 0.5 mol 63 g / mol \u003d 31.5 g

Calculate the mass of unreacted nitric acid:

m rest. (HNO 3) = m ref. (HNO 3) - m total. (HNO 3) \u003d 107.1 g - 31.5 g \u003d 75.6

In order to calculate the mass of the resulting solution, it is necessary to take into account the mass of nitrogen dioxide released in reaction (II):

ν (NO 2) \u003d 2m (Cu), therefore, ν (NO 2) \u003d 0.2 mol and m (NO 2) \u003d ν (NO 2) M (NO 2) \u003d 0.2 mol 46 g / mol = 9.2 g

Calculate the mass of the resulting solution:

m (solution) \u003d m (solution HNO 3) + m (Cu) + m (CuO) - m (NO 2) \u003d 126 g + 6.4 g + 4 g - 9.2 g \u003d 127, 2 g

The mass fraction of nitric acid in the resulting solution is equal to:

ω(HNO 3) = m rest. (HNO 3) / m (solution) 100% \u003d 75.6 g / 127.2 g 100% \u003d 59.43%

Task number 12

To a 10% salt solution obtained by dissolving 28.7 g of zinc sulfate (ZnSO 4 · 7H 2 O) in water was added 7.2 g of magnesium. After completion of the reaction, 120 g of 30% sodium hydroxide solution was added to the resulting mixture. Determine the mass fraction of sodium hydroxide in the resulting solution (disregard hydrolysis processes).

Answer: 7.21%

Explanation:

Mg + ZnSO 4 → MgSO 4 + Zn (I)

ν ref. (ZnSO 4 7H 2 O) \u003d ν (ZnSO 4) \u003d m ref. (ZnSO 4 7H 2 O) / M (ZnSO 4 7H 2 O) \u003d 28.7 g / 287 g / mol \u003d 0.1 mol

ν ref. (Mg) = m ref. (Mg)/M(Mg) = 7.2 g/24 g/mol = 0.3 mol

According to the reaction equation (I) ν ref. (Mg) \u003d ν (ZnSO 4), and according to the condition of the problem, the amount of zinc sulfate substance (0.1 mol ZnSO 4 7H 2 O and 0.3 mol Mg), so magnesium did not completely react.

The calculation is carried out according to the lack of substance, therefore, ν ref. (ZnSO 4 7H 2 O) \u003d ν (MgSO 4) \u003d ν (Zn) \u003d ν reactive. (Mg) = 0.1 mol and ν rest. (Mg) \u003d 0.3 mol - 0.1 mol \u003d 0.2 mol.

To further calculate the mass of the final solution, it is necessary to know the mass of unreacted magnesium (reaction (I)) and the initial solution of zinc sulfate:

m rest. (Mg) = ν rest. (Mg) M(Mg) = 0.2 mol 24 g/mol = 4.8 g

ν ref. (ZnSO 4 7H 2 O) = ν ref. (ZnSO 4) \u003d 0.1 mol, therefore, m (ZnSO 4) \u003d ν (ZnSO 4) M (ZnSO 4) \u003d 0.1 mol 161 g / mol \u003d 16.1 g

m ref. (p-ra ZnSO 4) \u003d m (ZnSO 4) / ω (ZnSO 4) 100% \u003d 16.1 g / 10% 100% \u003d 161 g

Magnesium sulfate and magnesium formed by reaction (I) react with a solution of sodium hydroxide:

Zn + 2NaOH + 2H 2 O → Na 2 + H 2 (II)

MgSO 4 + 2NaOH → Mg(OH) 2 ↓ + Na 2 SO 4 (III)

Calculate the mass and amount of sodium hydroxide substance:

m ref. (NaOH) = m ref. (p-ra NaOH) ω(NaOH) = 120 g 0.3 = 36 g

ν ref. (NaOH) = m ref. (NaOH)/M(NaOH) = 36 g/40 g/mol = 0.9 mol

According to the reaction equations (II) and (III) ν II (NaOH) = 2ν (Zn) and ν III (NaOH) = 2ν (MgSO 4), therefore, the total amount and mass of the reacting alkali are:

vtot. (NaOH) \u003d ν II (NaOH) + ν III (NaOH) \u003d 2ν (Zn) + 2ν (MgSO 4) \u003d 2 0.1 mol + 2 0.1 mol \u003d 0.4 mol

To calculate the final solution, we calculate the mass of magnesium hydroxide:

ν (MgSO 4) \u003d ν (Mg (OH) 2) \u003d 0.1 mol

m (Mg (OH) 2) \u003d ν (Mg (OH) 2) M (Mg (OH) 2) \u003d 0.1 mol 58 g / mol \u003d 5.8 g

Calculate the mass of unreacted alkali:

m rest. (NaOH) = m ref. (NaOH) - m react. (NaOH) = 36 g - 16 g = 20 g

To calculate the mass of the final solution, it is necessary to calculate the mass of hydrogen released as a result of reaction (II):

ν (Zn) \u003d ν (H 2) \u003d 0.1 mol and m (H 2) \u003d ν (H 2) M (H 2) \u003d 0.1 mol 2 g / mol \u003d 0.2 g

The mass of the resulting solution is calculated by the formula:

m (r-ra) \u003d m ref. (p-ra ZnSO 4) + m ref. (Mg) - m rest. (Mg)+ m ref. (p-ra NaOH) - m (Mg (OH) 2) - m (H 2) \u003d 161 g + 7.2 g - 4.8 g + 120 g - 5.8 g - 0.2 g \u003d 277, 4 g

The mass fraction of alkali in the resulting solution is equal to:

ω(NaOH) = m rest. (NaOH)/m(solution) 100% = 20 g/277.4 g 100% = 7.21%

Task number 13

To a 20% salt solution obtained by dissolving 57.4 g of zinc sulfate crystalline hydrate (ZnSO 4 · 7H 2 O) in water was added 14.4 g of magnesium. After completion of the reaction, 292 g of 25% hydrochloric acid was added to the resulting mixture. Determine the mass fraction of hydrogen chloride in the resulting solution (disregard hydrolysis processes).

Answer: 6.26%

Explanation:

When zinc sulfate interacts with magnesium, a substitution reaction occurs:

Mg + ZnSO 4 → MgSO 4 + Zn (I)

Let us calculate the amount of zinc and magnesium sulfate substance entering into reaction (I):

ν ref. (ZnSO 4 7H 2 O) \u003d ν (ZnSO 4) \u003d m ref. (ZnSO 4 7H 2 O) / M (ZnSO 4 7H 2 O) \u003d 57.4 g / 287 g / mol \u003d 0.2 mol

ν ref. (Mg) = m ref. (Mg)/M(Mg) = 14.4 g/24 g/mol = 0.6 mol

According to the reaction equation (I) ν ref. (Mg) \u003d ν (ZnSO 4), and according to the condition of the problem, the amount of zinc sulfate substance (0.2 mol ZnSO 4 7H 2 O and 0.6 mol Mg), so magnesium did not completely react.

The calculation is carried out according to the lack of substance, therefore, ν ref. (ZnSO 4 7H 2 O) \u003d ν (MgSO 4) \u003d ν (Zn) \u003d ν reactive. (Mg) = 0.2 mol and ν rest. (Mg) \u003d 0.6 mol - 0.2 mol \u003d 0.4 mol.

ν ref. (ZnSO 4 7H 2 O) = ν ref. (ZnSO 4) \u003d 0.2 mol, therefore, m (ZnSO 4) \u003d ν (ZnSO 4)

M (ZnSO 4) \u003d 0.2 mol 161 g / mol \u003d 32.2 g

m ref. (p-ra ZnSO 4) \u003d m (ZnSO 4) / ω (ZnSO 4) 100% \u003d 32.2 g / 20% 100% \u003d 161 g

Zn + 2HCl → ZnCl 2 + H 2 (II)

Calculate the mass and amount of hydrogen chloride substance:

m ref. (HCl) = m ref. (solution HCl) ω(HCl) = 292 g 0.25 = 73 g

ν ref. (HCl) = m ref. (HCl)/M(HCl) = 73 g/36.5 g/mol = 2 mol

vtot. (HCl) \u003d ν II (HCl) + ν III (HCl) \u003d 2ν (Zn) + 2ν (Mg) \u003d 2 0.2 mol + 2 0.4 mol \u003d 1.2 mol

m react. (HCl) = ν react. (HCl) M(HCl) = 1.2 mol 36.5 g/mol = 43.8 g

m rest. (HCl) = m ref. (HCl) - m react. (HCl) = 73 g - 43.8 g = 29.2 g

ν (Zn) \u003d ν II (H 2) \u003d 0.2 mol and m II (H 2) \u003d ν II (H 2) M (H 2) \u003d 0.2 mol 2 g / mol \u003d 0.4 G

m total (H 2) \u003d m II (H 2) + m III (H 2) \u003d 0.4 g + 0.8 g \u003d 1.2 g

The mass of the resulting solution is calculated by the formula:

m (r-ra) \u003d m ref. (p-ra ZnSO 4) + m ref. (Mg) + m ref. (p-ra HCl) - m total. (H 2) \u003d 161 g + 14.4 g + 292 g - 1.2 g \u003d 466.2 g

The mass fraction of hydrogen chloride in the resulting solution is equal to:

ω(HCl) = m rest. (HCl) / m (solution) 100% \u003d 29.2 g / 466.2 g 100% \u003d 6.26%

Task number 14

Zinc oxide weighing 16.2 g was heated and carbon monoxide with a volume of 1.12 liters was passed through it. The carbon monoxide reacted completely. The resulting solid residue was dissolved in 60 g of 40% sodium hydroxide solution. Determine the mass fraction of sodium hydroxide in the resulting solution (disregard hydrolysis processes).

Answer: 10.62%

Explanation:

Zn + 2NaOH + 2H 2 O → Na 2 + H 2 (II)

ZnO + 2NaOH + H 2 O → Na 2 (III)

ν ref. (ZnO) = m ref. (ZnO)/M(ZnO) = 16.2 g / 81 g/mol = 0.2 mol

ν ref. (CO) = V ref. (CO) / V m \u003d 1.12 l / 22.4 l / mol \u003d 0.05 mol

According to the reaction equation (I) ν . (ZnO) = ν(CO), and according to the condition of the problem, the amount of carbon monoxide substance is 4 times less than the amount of zinc oxide substance (0.05 mol CO and 0.2 mol ZnO), so zinc oxide did not react completely.

The calculation is carried out according to the lack of substance, therefore, ν ref. (ZnO) = 0.2 mol and ν rest. (ZnO) \u003d 0.2 mol - 0.05 mol \u003d 0.15 mol.

m rest. (ZnO) = ν rest. (ZnO) M(ZnO) = 0.15 mol 81 g/mol = 12.15 g

m(Zn) = ν(Zn) M(Zn) = 0.05 mol 65 g/mol = 3.25 g

Calculate the mass and amount of sodium hydroxide substance:

m ref. (NaOH) = m ref. (p-ra NaOH) ω(NaOH) = 60 g 0.4 = 24 g

ν ref. (NaOH) = m ref. (NaOH)/M(NaOH) = 24 g/40 g/mol = 0.6 mol

According to the reaction equations (II) and (III) ν II (NaOH) = 2ν (Zn) and ν III (NaOH) = 2ν rest. (ZnO), therefore, the total amount and mass of the reacting alkali are:

vtot. (NaOH) = ν II (NaOH) + ν III (NaOH) = 2ν(Zn) + 2ν rest. (ZnO) = 2 0.05 mol + 2 0.15 mol = 0.4 mol

m react. (NaOH) = ν react. (NaOH) M(NaOH) = 0.4 mol 40 g/mol = 16 g

m rest. (NaOH) = m ref. (NaOH) - m react. (NaOH) = 24 g - 16 g = 8 g

To calculate the mass of the final solution, it is necessary to calculate the mass of hydrogen released as a result of reaction (II):

ν rest. (Zn) \u003d ν (H 2) \u003d 0.05 mol and m (H 2) \u003d ν (H 2) M (H 2) \u003d 0.05 mol 2 g / mol \u003d 0.1 g

The mass of the resulting solution is calculated by the formula:

m (r-ra) \u003d m ref. (p-ra NaOH) + m(Zn) + m rest. (ZnO) - m (H 2) \u003d 60 g + 12.15 g + 3.25 g - 0.1 g \u003d 75.3 g

The mass fraction of alkali in the resulting solution is equal to:

ω(NaOH) = m rest. (NaOH)/m(solution) 100% = 8 g/75.3 g 100% = 10.62%

Task number 15

To a 10% salt solution obtained by dissolving 37.9 g of lead sugar ((CH 3 COO) 2 Pb 3H 2 O) in water was added 7.8 g of zinc. After completion of the reaction, 156 g of a 10% sodium sulfide solution was added to the resulting mixture. Determine the mass fraction of sodium sulfide in the resulting solution (disregard hydrolysis processes).

Answer: 1.71%

Explanation:

When zinc sulfate interacts with magnesium, a substitution reaction occurs:

ν ref. ((CH 3 COO) 2 Pb 3H 2 O) = ν ref. ((CH 3 COO) 2 Pb) = m ref. ((CH 3 COO) 2 Pb 3H 2 O) / M ((CH 3 COO) 2 Pb 3H 2 O) \u003d 37.9 g / 379 g / mol \u003d 0.1 mol

ν ref. (Zn) = m ref. (Zn)/M(Zn) = 7.8 g/65 g/mol = 0.12 mol

According to the reaction equation (I), ν(Zn) = ν((CH 3 COO) 2 Pb), and according to the condition of the problem, the amount of lead acetate substance is less than the amount of zinc substance (0.1 mol (CH 3 COO) 2 Pb 3H 2 O and 0.12 mol Zn), so the zinc did not react completely.

The calculation is carried out according to the lack of substance, therefore, ν ref. ((CH 3 COO) 2 Pb 3H 2 O) = ν((CH 3 COO) 2 Zn) = ν(Pb) = ν react. (Zn) = 0.1 mol and ν rest. (Zn) \u003d 0.12 mol - 0.1 mol \u003d 0.02 mol.

m(Pb) = ν(Pb) M(Pb) = 0.1 mol 207 g/mol = 20.7 g

m rest. (Zn) = ν rest. (Zn) M(Zn) = 0.02 mol 65 g/mol = 1.3 g

ν ref. ((CH 3 COO) 2 Pb 3H 2 O) = ν ref. ((CH 3 COO) 2 Pb) = 0.1 mol, therefore,

m ((CH 3 COO) 2 Pb) \u003d ν ((CH 3 COO) 2 Pb) M ((CH 3 COO) 2 Pb) \u003d 0.1 mol 325 g / mol \u003d 32.5 g

m ref. (p-ra CH 3 COO) 2 Pb) \u003d m ((CH 3 COO) 2 Pb) / ω ((CH 3 COO) 2 Pb) 100% \u003d 32.5 g / 10% 100% \u003d 325 g

Calculate the mass and amount of sodium sulfide substance:

m ref. (Na 2 S) \u003d m ref. (p-ra Na 2 S) ω (Na 2 S) \u003d 156 g 0.1 \u003d 15.6 g

ν ref. (Na 2 S) \u003d m ref. (Na 2 S) / M (Na 2 S) \u003d 15.6 g / 78 g / mol \u003d 0.2 mol

ν rest. (Na 2 S) \u003d ν ref. (Na 2 S) - ν react. (Na 2 S) \u003d 0.2 mol - 0.1 mol \u003d 0.1 mol

m rest. (Na 2 S) \u003d ν react. (Na 2 S) M (Na 2 S) \u003d 0.1 mol 78 g / mol \u003d 7.8 g

ν((CH 3 COO) 2 Zn) = ν(ZnS) = 0.1 mol and m(ZnS) = ν(ZnS) M(ZnS) = 0.1 mol 97 g/mol = 9.7 g

The mass of the resulting solution is calculated by the formula:

m (r-ra) \u003d m ref. (solution (CH 3 COO) 2 Pb) + m ref. (Zn) – m rest. (Zn) – m(Pb) + m ref. (p-ra Na 2 S) - m (ZnS) \u003d 325 g + 7.8 g - 1.3 g - 20.7 g + 156 g - 9.7 g \u003d 457.1 g

The mass fraction of sodium sulfide in the resulting solution is equal to:

ω(Na 2 S) = m rest. (Na 2 S) / m (solution) 100% \u003d 7.8 g / 457.1 g 100% \u003d 1.71%

Task number 16

Zinc oxide weighing 32.4 g was heated and carbon monoxide was passed through it with a volume of 2.24 liters. The carbon monoxide reacted completely. The resulting solid residue was dissolved in 224 g of a 40% potassium hydroxide solution. Determine the mass fraction of potassium hydroxide in the resulting solution (disregard hydrolysis processes).

Answer: 17.6%

Explanation:

When zinc oxide interacts with carbon monoxide, a redox reaction occurs:

ZnO + CO → Zn + CO 2 (heating) (I)

The formed zinc and unreacted zinc oxide react with sodium hydroxide solution:

ZnO + 2KOH + H 2 O → K 2 (III)

Let us calculate the amount of zinc oxide and carbon monoxide substance entering into reaction (I):

ν ref. (ZnO) = m ref. (ZnO)/M(ZnO) = 32.4 g / 81 g/mol = 0.4 mol

ν ref. (CO) = V ref. (CO) / V m \u003d 2.24 l / 22.4 l / mol \u003d 0.1 mol

According to the reaction equation (I) ν . (ZnO) \u003d ν (CO), and according to the condition of the problem, the amount of carbon monoxide substance is 4 times less than the amount of zinc oxide substance (0.1 mol CO and 0.4 mol ZnO), therefore zinc oxide did not react completely.

The calculation is carried out according to the lack of substance, therefore, ν ref. (ZnO) = 0.4 mol and ν rest. (ZnO) \u003d 0.4 mol - 0.1 mol \u003d 0.3 mol.

To further calculate the mass of the final solution, it is necessary to know the masses of the resulting zinc and unreacted zinc oxide:

m rest. (ZnO) = ν rest. (ZnO) M(ZnO) = 0.3 mol 81 g/mol = 24.3 g

m(Zn) = ν(Zn) M(Zn) = 0.1 mol 65 g/mol = 6.5 g

Calculate the mass and amount of sodium hydroxide substance:

m ref. (KOH) = m ref. (solution KOH) ω(KOH) = 224 g 0.4 = 89.6 g

ν ref. (KOH) = m ref. (KOH)/M(KOH) = 89.6 g/56 g/mol = 1.6 mol

According to the reaction equations (II) and (III) ν II (KOH) = 2ν (Zn) and ν III (KOH) = 2ν rest. (ZnO), therefore, the total amount and mass of the reacting alkali are:

vtot. (KOH) = ν II (KOH) + ν III (KOH) = 2ν(Zn) + 2ν rest. (ZnO) = 2 0.1 mol + 2 0.3 mol = 0.8 mol

m react. (KOH) = ν react. (KOH) M(KOH) = 0.8 mol 56 g/mol = 44.8 g

Calculate the mass of unreacted alkali:

m rest. (KOH) = m ref. (KOH) - m react. (KOH) = 89.6 g - 44.8 g = 44.8 g

To calculate the mass of the final solution, it is necessary to calculate the mass of hydrogen released as a result of reaction (II):

The mass of the resulting solution is calculated by the formula:

m (r-ra) \u003d m ref. (solution KOH) + m(Zn) + m rest. (ZnO) - m (H 2) \u003d 224 g + 6.5 g + 24.3 g - 0.2 g \u003d 254.6 g

The mass fraction of alkali in the resulting solution is equal to:

ω(KOH) = m rest. (KOH)/m(solution) 100% = 44.8 g/254.6 g 100% = 17.6%

Task number 17

To a 10% salt solution obtained by dissolving 75.8 g of lead sugar ((CH 3 COO) 2 Pb 3H 2 O) in water was added 15.6 g of zinc. After completion of the reaction, 312 g of a 10% sodium sulfide solution was added to the resulting mixture. Determine the mass fraction of sodium sulfide in the resulting solution (disregard hydrolysis processes).

Answer: 1.71%

Explanation:

When zinc sulfate interacts with magnesium, a substitution reaction occurs:

Zn + (CH 3 COO) 2 Pb → (CH 3 COO) 2 Zn + Pb↓ (I)

Let us calculate the amount of lead and zinc acetate substances that enter into reaction (I):

ν ref. ((CH 3 COO) 2 Pb 3H 2 O) = ν ref. ((CH 3 COO) 2 Pb) = m ref. ((CH 3 COO) 2 Pb 3H 2 O) / M ((CH 3 COO) 2 Pb 3H 2 O) \u003d 75.8 g / 379 g / mol \u003d 0.2 mol

ν ref. (Zn) = m ref. (Zn)/M(Zn) = 15.6 g/65 g/mol = 0.24 mol

According to the reaction equation (I) ν (Zn) = ν ((CH 3 COO) 2 Pb), and according to the condition of the problem, the amount of lead acetate substance is less than the amount of zinc substance (0.2 mol (CH 3 COO) 2 Pb 3H 2 O and 0.24 mol Zn), so the zinc did not react completely.

The calculation is carried out according to the lack of substance, therefore, ν ref. ((CH 3 COO) 2 Pb 3H 2 O) = ν((CH 3 COO) 2 Zn) = ν(Pb) = ν react. (Zn) = 0.2 mol and ν rest. (Zn) \u003d 0.24 mol - 0.2 mol \u003d 0.04 mol.

To further calculate the mass of the final solution, it is necessary to know the masses of the formed lead, unreacted zinc and the initial solution of lead sugar:

m rest. (Pb) = ν rest. (Pb) M(Pb) = 0.2 mol 207 g/mol = 41.4 g

m rest. (Zn) = ν rest. (Zn) M(Zn) = 0.04 mol 65 g/mol = 2.6 g

ν ref. ((CH 3 COO) 2 Pb 3H 2 O) = ν ref. ((CH 3 COO) 2 Pb) = 0.2 mol, therefore,

m ((CH 3 COO) 2 Pb) \u003d ν ((CH 3 COO) 2 Pb) M ((CH 3 COO) 2 Pb) \u003d 0.2 mol 325 g / mol \u003d 65 g

m ref. (p-ra CH 3 COO) 2 Pb) \u003d m ((CH 3 COO) 2 Pb) / ω ((CH 3 COO) 2 Pb) 100% \u003d 65 g / 10% 100% \u003d 650 g

The zinc acetate formed by reaction (I) reacts with a solution of sodium sulfide:

(CH 3 COO) 2 Zn + Na 2 S → ZnS↓ + 2CH 3 COONa (II)

Calculate the mass and amount of sodium sulfide substance:

m ref. (Na 2 S) \u003d m ref. (p-ra Na 2 S) ω (Na 2 S) \u003d 312 g 0.1 \u003d 31.2 g

ν ref. (Na 2 S) \u003d m ref. (Na 2 S) / M (Na 2 S) \u003d 31.2 g / 78 g / mol \u003d 0.4 mol

According to the reaction equation (II) ν ((CH 3 COO) 2 Zn) = ν (Na 2 S), therefore, the amount of unreacted sodium sulfide substance is:

ν rest. (Na 2 S) \u003d ν ref. (Na 2 S) - ν react. (Na 2 S) \u003d 0.4 mol - 0.2 mol \u003d 0.2 mol

m rest. (Na 2 S) \u003d ν react. (Na 2 S) M (Na 2 S) \u003d 0.2 mol 78 g / mol \u003d 15.6 g

To calculate the mass of the final solution, it is necessary to calculate the mass of zinc sulfide:

ν ((CH 3 COO) 2 Zn) \u003d ν (ZnS) \u003d 0.2 mol and m (ZnS) \u003d ν (ZnS) M (ZnS) \u003d 0.2 mol 97 g / mol \u003d 19.4 g

The mass of the resulting solution is calculated by the formula:

m (r-ra) \u003d m ref. (solution (CH 3 COO) 2 Pb) + m ref. (Zn) – m rest. (Zn) – m(Pb) + m ref. (p-ra Na 2 S) - m (ZnS) \u003d 650 g + 15.6 g - 2.6 g - 41.4 g + 312 g - 19.4 g \u003d 914.2 g

The mass fraction of sodium sulfide in the resulting solution is equal to:

ω(Na 2 S) = m rest. (Na 2 S) / m (solution) 100% \u003d 15.6 g / 914.2 g 100% \u003d 1.71%

Task number 18

To a 10% salt solution obtained by dissolving 50 g of copper sulfate (CuSO 4 · 5H 2 O) in water, 19.5 g of zinc was added. After completion of the reaction, 200 g of 30% sodium hydroxide solution was added to the resulting mixture. Determine the mass fraction of sodium hydroxide in the resulting solution (disregard hydrolysis processes).

Answer: 3.8%

Explanation:

When copper (II) sulfate interacts with zinc, a substitution reaction occurs:

Zn + CuSO 4 → ZnSO 4 + Cu (I)

Let's calculate the amount of substance of copper sulphate and zinc entering into reaction (I):

ν (CuSO 4 5H 2 O) \u003d m (CuSO 4 5H 2 O) / M (CuSO 4 5H 2 O) \u003d 50 g / 250 g / mol \u003d 0.2 mol

ν(Zn) = m(Zn)/M(Zn) = 19.5 g/65 g/mol = 0.3 mol

According to the reaction equation (I) ν (Zn) = ν (CuSO 4), and according to the condition of the problem, the amount of copper sulphate substance is in short supply (0.2 mol CuSO 4 5H 2 O and 0.3 mol Zn), so zinc did not react fully.

We calculate by the lack of substance, therefore, ν (CuSO 4 5H 2 O) \u003d ν (ZnSO 4) \u003d ν (Cu) \u003d ν react. (Zn) = 0.2 mol and ν rest. (Zn) \u003d 0.3 mol - 0.2 mol \u003d 0.1 mol.

To further calculate the mass of the final solution, it is necessary to know the mass of the formed copper (reaction (I)) and the initial solution of copper sulphate:

m(Cu) = ν(Cu) M(Cu) = 0.2 mol 64 g/mol = 12.8 g

ν (CuSO 4 5H 2 O) \u003d ν (CuSO 4) \u003d 0.2 mol, therefore, m (CuSO 4) \u003d ν (CuSO 4) M (CuSO 4) \u003d 0.2 mol 160 g / mol = 32 g

m ref. (p-ra CuSO 4) \u003d m (CuSO 4) / ω (CuSO 4) 100% \u003d 32 g / 10% 100% \u003d 320 g

Zinc that has not completely reacted in reaction (I) and zinc sulfate react with a solution of sodium hydroxide to form a complex salt, sodium tetrahydroxozincate:

Zn + 2NaOH + 2H 2 O → Na 2 + H 2 (II)

ZnSO 4 + 4NaOH → Na 2 + Na 2 SO 4 (III)

Calculate the mass and amount of sodium hydroxide substance:

m ref. (NaOH) = m ref. (p-ra NaOH) ω(NaOH) = 200 g 0.3 = 60 g

ν ref. (NaOH) = m ref. (NaOH)/M(NaOH) = 60 g/40 g/mol = 1.5 mol

According to the reaction equations (II) and (III) ν II (NaOH) = 2ν rest. (Zn) and ν III (NaOH) = 4ν (ZnSO 4), therefore, the total amount and mass of the reacting alkali are:

vtot. (NaOH) \u003d ν II (NaOH) + ν III (NaOH) \u003d 2 0.1 mol + 4 0.2 mol \u003d 1 mol

m react. (NaOH) = ν react. (NaOH) M(NaOH) = 1 mol 40 g/mol = 40 g

Calculate the mass of unreacted alkali:

m rest. (NaOH) = m ref. (NaOH) - m react. (NaOH) = 60 g - 40 g = 20 g

To calculate the mass of the final solution, it is necessary to calculate the mass of hydrogen released as a result of reaction (II):

ν rest. (Zn) \u003d ν (H 2) \u003d 0.1 mol and m (H 2) \u003d ν (H 2) M (H 2) \u003d 0.1 mol 2 g / mol \u003d 0.2 g

The mass of the resulting solution is calculated by the formula (the mass of zinc unreacted in reaction (I) is not taken into account, since it goes into solution in reactions (II) and (III):

m (r-ra) \u003d m ref. (p-ra CuSO 4) + m ref. (Zn) - m(Cu) + m ref. (p-ra NaOH) - m (H 2) \u003d 320 g + 19.5 g - 12.8 g + 200 g - 0.2 g \u003d 526.5 g

The mass fraction of alkali in the resulting solution is equal to:

ω(NaOH) = m rest. (NaOH)/m(solution) 100% = 20 g/526.5 g 100% = 3.8%

Task #19

As a result of the dissolution of a mixture of powders of copper and copper (II) oxide in concentrated sulfuric acid, sulfur dioxide with a volume of 8.96 liters was released and a solution weighing 400 g was formed with a mass fraction of copper (II) sulfate of 20%. Calculate the mass fraction of copper (II) oxide in the initial mixture.

Answer: 23.81%

Explanation:

When copper and copper (II) oxide interact with concentrated sulfuric acid, the following reactions occur:

Cu + 2H 2 SO 4 → CuSO 4 + SO 2 + 2H 2 O (I)

CuO + H 2 SO 4 → CuSO 4 + H 2 O (II)

Calculate the mass and amount of copper (II) sulfate substance:

m (CuSO 4) \u003d m (CuSO 4) ω (CuSO 4) \u003d 400 g 0.2 \u003d 80 g

ν (CuSO 4) \u003d m (CuSO 4) / M (CuSO 4) \u003d 80 g / 160 g / mol \u003d 0.5 mol

Calculate the amount of sulfur dioxide substance:

ν (SO 2) \u003d V (SO 2) / V m \u003d 8.96 l / 22.4 l / mol \u003d 0.4 mol

According to the reaction equation (I) ν (Cu) \u003d ν (SO 2) \u003d ν I (CuSO 4), therefore, ν (Cu) \u003d ν I (CuSO 4) \u003d 0.4 mol.

Since ν total. (CuSO 4) = ν I (CuSO 4) + ν II (CuSO 4), then ν II (CuSO 4) = ν total. (CuSO 4) - ν I (CuSO 4) \u003d 0.5 mol - 0.4 mol \u003d 0.1 mol.

According to the reaction equation (II) ν II (CuSO 4) = ν (CuO), therefore, ν (CuO) = 0.1 mol.

Calculate the masses of copper and copper oxide (II):

m(Cu) = M(Cu) ∙ ν(Cu) = 64 g/mol ∙ 0.4 mol = 25.6 g

m(CuO) = M(CuO) ∙ ν(CuO) = 80 g/mol ∙ 0.1 mol = 8 g

The total mixture, consisting of copper and copper (II) oxide, is equal to:

m(mixtures) = m(CuO) + m(Cu) = 25.6 g + 8 g = 33.6 g

Calculate the mass fraction of copper oxide (II):

ω(CuO) = m(CuO)/m(mixtures) ∙ 100% = 8 g/33.6 g ∙ 100% = 23.81%

Task number 20

As a result of heating 28.4 g of a mixture of zinc and zinc oxide powders in air, its mass increased by 4 g. Calculate the volume of a solution of potassium hydroxide with a mass fraction of 40% and a density of 1.4 g/ml, which will be required to dissolve the initial mixture.

Answer: 80 ml

Explanation:

When zinc is heated in air, zinc oxidizes and turns into an oxide:

2Zn + O 2 → 2ZnO(I)

Since the mass of the mixture increased, this increase occurred due to the mass of oxygen:

ν (O 2) \u003d m (O 2) / M (O 2) \u003d 4 g / 32 g / mol \u003d 0.125 mol, therefore, the amount of zinc is twice the amount of substance and the mass of oxygen, therefore

ν (Zn) \u003d 2ν (O 2) \u003d 2 0.125 mol \u003d 0.25 mol

m(Zn) = M(Zn) ν(Zn) = 0.25 mol 65 g/mol = 16.25 g

Calculate the mass and amount of zinc oxide substance is equal to:

m(ZnO) = m(mixtures) - m(Zn) = 28.4 g - 16.25 g = 12.15 g

ν(ZnO) = m(ZnO)/M(ZnO) = 12.15 g/81 g/mol = 0.15 mol

Both zinc and zinc oxide interact with potassium hydroxide:

Zn + 2KOH + 2H 2 O → K 2 + H 2 (II)

ZnO + 2KOH + H 2 O → K 2 (III)

According to the equations of reactions (II) and (III) ν I (KOH) = 2ν (Zn) and ν II (KOH) = 2ν (ZnO), therefore, the total amount of substance and the mass of potassium hydroxide are equal:

ν(KOH) = 2ν(Zn) + 2ν(ZnO) = 2 ∙ 0.25 mol + 2 ∙ 0.15 mol = 0.8 mol

m(KOH) = M(KOH) ∙ ν(KOH) = 56 g/mol ∙ 0.8 mol = 44.8 g

Calculate the mass of the potassium hydroxide solution:

m(solution KOH) = m(KOH)/ω(KOH) ∙ 100% = 44.8 g/40% ∙ 100% = 112 g

The volume of potassium hydroxide solution is:

V(solution KOH) \u003d m (KOH) / ρ (KOH) \u003d 112 g / 1.4 g / mol \u003d 80 ml

Task number 21

A mixture of magic oxide and magnesium carbonate weighing 20.5 g was heated to constant weight, while the mass of the mixture decreased by 5.5 g. After that, the solid residue completely reacted with a solution of sulfuric acid with a mass fraction of 28% and a density of 1.2 g / ml . Calculate the volume of sulfuric acid solution required to dissolve this residue.

Answer: 109.375 ml

Explanation:

When heated, magnesium carbonate decomposes to magnesium oxide and carbon dioxide:

MgCO 3 → MgO + CO 2 (I)

Magnesium oxide reacts with a solution of sulfuric acid according to the equation:

MgO + H 2 SO 4 → MgSO 4 + H 2 O (II)

The mass of the mixture of oxide and magnesium carbonate decreased due to the released carbon dioxide.

Calculate the amount of carbon dioxide formed:

ν (CO 2) \u003d m (CO 2) / M (CO 2) \u003d 5.5 g / 44 g / mol \u003d 0.125 mol

According to the reaction equation (I) ν (CO 2) \u003d ν I (MgO), therefore, ν I (MgO) \u003d 0.125 mol

Calculate the mass of reacted magnesium carbonate:

m (MgCO 3) \u003d ν (MgCO 3) ∙ M (MgCO 3) \u003d 84 g / mol ∙ 0.125 mol \u003d 10.5 g

Calculate the mass and amount of magnesium oxide substance in the initial mixture:

m(MgO) \u003d m (mixtures) - m (MgCO 3) \u003d 20.5 g - 10.5 g \u003d 10 g

ν(MgO) = m(MgO)/M(MgO) = 10 g/40 g/mol = 0.25 mol

The total amount of magnesium oxide is:

vtot. (MgO) \u003d ν I (MgO) + ν (MgO) \u003d 0.25 mol + 0.125 mol \u003d 0.375 mol

According to the reaction equation (II) ν total. (MgO) \u003d ν (H 2 SO 4), therefore, ν (H 2 SO 4) \u003d 0.375 mol.

Calculate the mass of sulfuric acid:

m (H 2 SO 4) \u003d ν (H 2 SO 4) ∙ M (H 2 SO 4) \u003d 0.375 mol ∙ 98 g / mol \u003d 36.75 g

Calculate the mass and volume of the sulfuric acid solution:

m (p-ra H 2 SO 4) \u003d m (H 2 SO 4) / ω (H 2 SO 4) ∙ 100% = 36.75 g / 28% ∙ 100% = 131.25 g

V (solution H 2 SO 4) \u003d m (solution H 2 SO 4) / ρ (solution H 2 SO 4) \u003d 131.25 g / 1.2 g / ml \u003d 109.375 ml

Task #22

Hydrogen with a volume of 6.72 liters (n.o.) was passed over the heated powder of copper (II) oxide, while the hydrogen reacted completely. This resulted in 20.8 g of a solid residue. This residue was dissolved in concentrated sulfuric acid weighing 200 g. Determine the mass fraction of salt in the resulting solution (disregard hydrolysis processes).

Answer: 25.4%

Explanation:

When hydrogen is passed over copper (II) oxide, copper is reduced:

CuO + H 2 → Cu + H 2 O (heating) (I)

The solid residue, consisting of metallic copper and unreacted copper (II) oxide, reacts with concentrated sulfuric acid according to the equations:

Cu + 2H 2 SO 4 (conc.) → CuSO 4 + SO 2 + 2H 2 O (II)

CuO + H 2 SO 4 → CuSO 4 + H 2 O (III)

Let us calculate the amount of hydrogen substance involved in the reduction of copper oxide (II):

ν (H 2) \u003d V (H 2) / V m \u003d 6.72 l / 22.4 l / mol \u003d 0.3 mol,

ν (H 2) \u003d ν (Cu) \u003d 0.3 mol, therefore, m (Cu) \u003d 0.3 mol 64 g / mol \u003d 19.2 g

Let us calculate the mass of unreacted CuO, knowing the mass of the solid residue:

m(CuO) \u003d m (solid rest.) - m (Cu) \u003d 20.8 g - 19.2 g \u003d 1.6 g

Calculate the amount of copper (II) oxide substance:

ν(CuO) = m(CuO)/M(CuO) = 1.6 g/80 g/mol = 0.02 mol

According to equation (I) ν(Cu) = ν I (CuSO 4), according to equation (II) ν (CuO) = ν II (CuSO 4), therefore, ν total. (CuSO 4) \u003d ν II (CuSO 4) + ν III (CuSO 4) \u003d 0.3 mol + 0.02 mol \u003d 0.32 mol.

Calculate the total mass of copper (II) sulfate:

m total (CuSO 4) = vtot. (CuSO 4) M (CuSO 4) \u003d 0.32 mol 160 g / mol \u003d 51.2 g

In order to calculate the mass of the resulting solution, it is necessary to take into account the mass of sulfur dioxide released in reaction (II):

ν (Cu) \u003d ν (SO 2), therefore, ν (SO 2) \u003d 0.3 mol and m (SO 2) \u003d ν (SO 2) M (SO 2) \u003d 0.3 mol 64 g / mol = 19.2 g

Calculate the mass of the resulting solution:

m (solution) \u003d m (solid rest.) + m (solution H 2 SO 4) - m (SO 2) \u003d 20.8 g + 200 g - 19.2 g \u003d 201.6 g

The mass fraction of copper (II) sulfate in the resulting solution is:

ω (CuSO 4) \u003d m (CuSO 4) / m (solution) 100% \u003d 51.2 g / 201.6 g 100% \u003d 25.4%

Task #23

To a 10% salt solution obtained by dissolving 114.8 g of zinc sulfate crystalline hydrate (ZnSO 4 · 7H 2 O) in water was added 12 g of magnesium. After completion of the reaction, 365 g of 20% hydrochloric acid was added to the resulting mixture. Determine the mass fraction of hydrogen chloride in the resulting solution (disregard hydrolysis processes).

Answer: 3.58%

Explanation:

When zinc sulfate interacts with magnesium, a substitution reaction occurs:

Mg + ZnSO 4 → MgSO 4 + Zn (I)

Let us calculate the amount of zinc and magnesium sulfate substance entering into reaction (I):

ν ref. (ZnSO 4 7H 2 O) \u003d ν (ZnSO 4) \u003d m ref. (ZnSO 4 7H 2 O) / M (ZnSO 4 7H 2 O) \u003d 114.8 g / 287 g / mol \u003d 0.4 mol

ν ref. (Mg) = m ref. (Mg)/M(Mg) = 12 g/24 g/mol = 0.5 mol

According to the reaction equation (I) ν ref. (Mg) \u003d ν (ZnSO 4), and according to the condition of the problem, the amount of zinc sulfate substance (0.4 mol ZnSO 4 7H 2 O and 0.5 mol Mg), so magnesium did not completely react.

The calculation is carried out according to the lack of substance, therefore, ν ref. (ZnSO 4 7H 2 O) \u003d ν (MgSO 4) \u003d ν (Zn) \u003d ν reactive. (Mg) = 0.4 mol and ν rest. (Mg) = 0.5 mol - 0.4 mol = 0.1 mol.

To further calculate the mass of the initial solution of zinc sulfate:

ν ref. (ZnSO 4 7H 2 O) = ν ref. (ZnSO 4) \u003d 0.4 mol, therefore, m (ZnSO 4) \u003d ν (ZnSO 4) M (ZnSO 4) \u003d 0.4 mol 161 g / mol \u003d 64.4 g

m ref. (p-ra ZnSO 4) \u003d m (ZnSO 4) / ω (ZnSO 4) 100% \u003d 64.4 g / 10% 100% \u003d 644 g

Magnesium and zinc can react with hydrochloric acid solution:

Zn + 2HCl → ZnCl 2 + H 2 (II)

Mg + 2HCl → MgCl 2 + H 2 (III)

Calculate the mass of hydrogen chloride in solution:

m ref. (HCl) = m ref. (solution HCl) ω(HCl) = 365 g 0.2 = 73 g

According to the reaction equations (II) and (III), ν II (HCl) = 2ν (Zn) and ν III (HCl) = 2ν (Mg), therefore, the total amount and mass of the reacting hydrogen chloride are:

ν react. (HCl) \u003d ν II (HCl) + ν III (HCl) \u003d 2ν (Zn) + 2ν (Mg) \u003d 2 0.1 mol + 2 0.4 mol \u003d 1 mol

m react. (HCl) = ν react. (HCl) M(HCl) = 1 mol 36.5 g/mol = 36.5 g

Calculate the mass of unreacted hydrochloric acid:

m rest. (HCl) = m ref. (HCl) - m react. (HCl) = 73 g - 36.5 g = 36.5 g

To calculate the mass of the final solution, it is necessary to calculate the mass of hydrogen released as a result of reactions (II) and (III):

ν (Zn) \u003d ν II (H 2) \u003d 0.1 mol and m II (H 2) \u003d ν II (H 2) M (H 2) \u003d 0.1 mol 2 g / mol \u003d 0.2 G

ν rest. (Mg) \u003d ν III (H 2) \u003d 0.4 mol and m III (H 2) \u003d ν III (H 2) M (H 2) \u003d 0.4 mol 2 g / mol \u003d 0.8 g

m total (H 2) \u003d m II (H 2) + m III (H 2) \u003d 0.2 g + 0.8 g \u003d 1 g

The mass of the resulting solution is calculated by the formula:

m (r-ra) \u003d m ref. (p-ra ZnSO 4) + m ref. (Mg) + m ref. (p-ra HCl) - m total. (H 2) \u003d 644 g + 12 g + 365 g - 1 g \u003d 1020 g

The mass fraction of hydrochloric acid in the resulting solution is:

ω(HCl) = m rest. (HCl) / m (solution) 100% \u003d 36.5 g / 1020 g 100% \u003d 3.58%

The content of the block "Organic substances" is a system of knowledge about the most important concepts and theories of organic chemistry, the characteristic chemical properties of the studied substances belonging to various classes of organic compounds, the relationship of these substances. This block includes 9 tasks. Assimilation of the content elements of this block is checked by tasks of basic (tasks 11–15 and 18), advanced (tasks 16 and 17) and high (task 33) levels of complexity. These tasks also tested the formation of skills and activities similar to those that were named in relation to the elements of the content of the "Inorganic substances" block.

Consider the tasks of the "Organic substances" block.

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Consider task 33 of a high level of complexity, which tests the assimilation of the relationship of organic compounds of various classes.

Task 33

Write the reaction equations that can be used to carry out the following transformations:

When writing reaction equations, use the structural formulas of organic substances.

Possible answer:

At a temperature of 180 °C in the presence of concentrated sulfuric acid, propanol-1 undergoes dehydration with the formation of propene:

Propene, interacting with hydrogen chloride, forms mainly 2-chloropropane in accordance with Markovnikov's rule:

Under the action of an aqueous solution of alkali, 2-chloropropane is hydrolyzed to form propanol-2:

Further, propene (X 1) must be obtained again from propanol-2, which can be carried out as a result of an intramolecular dehydration reaction at a temperature of 180 ° C under the action of concentrated sulfuric acid:

The product of propene oxidation with an aqueous solution of potassium permanganate in the cold is the dihydric alcohol propanediol-1,2, potassium permanganate is reduced to manganese(IV) oxide, which forms a brown precipitate:

In 2018, 41.1% of examinees were able to complete this task completely correctly.

The manual contains training tasks of basic and advanced levels of complexity, grouped by topic and type. The tasks are arranged in the same sequence as proposed in the examination version of the exam. At the beginning of each type of task are the content elements to be checked - topics that should be studied before proceeding with the implementation. The manual will be useful for teachers of chemistry, as it makes it possible to effectively organize the educational process in the classroom, conduct ongoing monitoring of knowledge, and prepare students for the exam.

Municipal budgetary educational institution

"Secondary school No. 4 of Shebekino, Belgorod region"

Features of solving and evaluating tasks 30-35 USE in chemistry

Prepared by: Arnautova Natalya Zakharovna,

chemistry and biology teacher

MBOU "Secondary School No. 4 of Shebekino, Belgorod Region"

2017

Methods for evaluating tasks with a detailed answer (basic approaches to determining the criteria and assessment scale for completing tasks)

The basis of the methodology for evaluating tasks with a detailed answer is a number of general provisions. The most important among them are the following:

Verification and evaluation of tasks with a detailed answer is carried out only by independent examination based on the method of element-by-element analysis of the answers of the examinees.

The use of the element-by-element analysis method makes it necessary to ensure a clear correspondence between the statement of the task condition and the content elements being checked. The list of content elements checked by any task is consistent with the requirements of the standard for the level of preparation of secondary school graduates.

The criterion for evaluating the performance of a task by the method of element-by-element analysis is to establish the presence in the answers of the examined elements of the answer given
in the response model. However, another model of the answer proposed by the examinee can be accepted if it does not distort the essence of the chemical component of the task condition.

The task performance rating scale is set depending on the number of content elements included in the response model, and taking into account factors such as:

The level of complexity of the checked content;

A certain sequence of actions that should be carried out when performing a task;

Unambiguous interpretation of the conditions of the task and possible options for formulating the answer;

Compliance of the task conditions with the proposed evaluation criteria for individual elements of the content;

Approximately the same level of difficulty of each of the content elements tested by the task.

When developing assessment criteria, the features of the content elements of all five tasks with a detailed answer included in the examination paper are taken into account. It also takes into account the fact that the notes of the answers of the examinees can be either very general, streamlined and not specific, or too short.
and insufficiently substantiated. Close attention is paid to highlighting the elements of the answer, estimated at one point. This takes into account the inevitability of a gradual increase in the difficulty of obtaining each subsequent score.
for a well-formulated content element.

When compiling the assessment scale for calculation tasks (33 and 34), the possibility of various ways of solving them is taken into account, and, consequently, the presence in the answer of the examinee of the main stages and results of completing the tasks indicated
in the evaluation criteria. Let us illustrate the methodology for evaluating tasks with a detailed answer using specific examples.

2017-2018 academic year

Tasks

Maximum score

Job level

Task 30

2016-2017

Tasks 30 are focused on testing the ability to determine the degree of oxidation of chemical elements, determine the oxidizing agent and reducing agent, predict the products of redox reactions, establish the formulas of substances omitted in the reaction scheme, draw up an electronic balance, and on its basis set the coefficients in the reaction equations.

The scale for assessing the performance of such tasks includes the following elements:

 an electronic balance has been drawn up - 1 point;

 the oxidizing agent and reducing agent are indicated - 1 point.

 the formulas of the missing substances are determined and the coefficients are placed
in the redox reaction equation - 1 point.

Job example:

Using the electron balance method, write the equation for the reaction

Na 2 SO 3 + ... + KOH K 2 MnO 4 + ... + H 2 O

Determine the oxidizing agent and reducing agent.

Points

Possible answer

Mn +7 + ē → Mn +6

S +4 – 2ē → S +6

Sulfur in the +4 oxidation state (or sodium sulfite due to sulfur in the +4 oxidation state) is a reducing agent.

Manganese in oxidation state +7 (or potassium permanganate due to manganese
in the oxidation state +7) - an oxidizing agent.

Na 2 SO 3 + 2KMnO 4 + 2KOH \u003d Na 2 SO 4 + 2K 2 MnO 4 + H 2 O

The answer is correct and complete:

the degree of oxidation of elements, which are, respectively, an oxidizing agent and a reducing agent in the reaction, is determined;

the processes of oxidation and reduction are recorded, and on their basis an electronic (electron-ionic) balance is compiled;

substances missing in the reaction equation are determined, all coefficients are placed

Maximum score

When evaluating the answer of the examinee, it must be taken into account that there are no uniform requirements for the design of the answer to this task. As a result, the compilation of both electronic and electron-ionic balances is accepted as the correct answer, and the indication of the oxidizing agent and reducing agent can be done in any unambiguously understandable way. However, if the answer contains elements of the answer that are mutually exclusive in meaning, then they cannot be considered correct.

Tasks of the 2018 format

1. Task 30 (2 points)

To complete the task, use the following list of substances: potassium permanganate, hydrogen chloride, sodium chloride, sodium carbonate, potassium chloride. The use of aqueous solutions of substances is acceptable.

From the proposed list of substances, select substances between which a redox reaction is possible, and write down the equation for this reaction. Make an electronic balance, indicate the oxidizing agent and reducing agent.

Explanation.

Let's write the reaction equation:

Let's make an electronic balance:

Chlorine in the −1 oxidation state is a reducing agent. Manganese in the +7 oxidation state is an oxidizing agent.TOTAL 2 points

substances are selected, the equation of the redox reaction is written, all the coefficients are placed.

the processes of oxidation and reduction are recorded, and on their basis an electronic (electron-ionic) balance is compiled; which are, respectively, an oxidizing agent and a reducing agent in the reaction;

An error was made in only one of the above response elements

Errors were made in two of the above response elements

All elements of the answer are written incorrectly

Maximum score

Tasks of the 2018 format

1. Task 31 (2 points)

To complete the task, use the following list of substances: potassium permanganate, potassium bicarbonate, sodium sulfite, barium sulfate, potassium hydroxide. The use of aqueous solutions of substances is acceptable.

Explanation.

Possible answer:

2. Task 31

To complete the task, use the following list of substances: hydrogen chloride, silver (I) nitrate, potassium permanganate, water, nitric acid. The use of aqueous solutions of substances is acceptable.

From the proposed list of substances, select substances between which an ion exchange reaction is possible. Write down the molecular, full, and abbreviated ionic equations for this reaction.

Explanation.

Possible answer:

Task 32. Tasks of the 2018 format

In the condition of task 32, testing the knowledge of the genetic relationship of various classes of inorganic substances, a description of a specific chemical experiment is proposed, the course of which the examinees will have to illustrate by means of the equations of the corresponding chemical reactions. The task grading scale remains, as in 2016, equal to 4 points: each correctly written reaction equation is estimated at 1 point.

Job example:

Write the equations for the four described reactions.

Correct Answer Content and Grading Instructions(Other formulations of the answer are allowed that do not distort its meaning)

Points

Possible answer

Four equations of the described reactions are written:

1) 2Fe + 6H 2 SO 4
Fe 2 (SO 4) 3 + 3SO 2 + 6H 2 O

2) Fe 2 (SO 4 ) 3 + 6NaOH = 2Fe(OH) 3 + 3Na 2 SO 4

3) 2Fe(OH) 3
Fe 2 O 3 + 3H 2 O

4) Fe 2 O 3 + Fe = 3FeO

All reaction equations are written incorrectly

Maximum score

It should be noted that the absence of coefficients (at least one) in front of the formulas of substances in the reaction equations is considered an error. The score for such an equation is not set.

Task 33. Tasks of the 2018 format

Tasks 33 test the assimilation of knowledge about the relationship of organic substances and provide for the verification of five elements of the content: the correctness of writing the five reaction equations corresponding to the scheme - the "chain" of transformations. When writing reaction equations, examinees must use the structural formulas of organic substances. The presence of each checked content element in the response is estimated at 1 point. The maximum number of points for completing such tasks is 5.

Job example:

Write the reaction equations that can be used to carry out the following transformations:

When writing reaction equations, use the structural formulas of organic substances.

Correct Answer Content and Grading Instructions
other formulations of the answer are allowed that do not distort its meaning)

Points

Possible answer

Five reaction equations were written corresponding to the transformation scheme:

Correctly written five reaction equations

Correctly written four reaction equations

Three reaction equations are written correctly

Correctly written two reaction equations

Correctly written one reaction equation

All elements of the answer are written incorrectly

Maximum score

Note that in the answer of the examinee, it is permissible to use structural formulas of various types (expanded, abbreviated, skeletal), unambiguously reflecting the order of bonding of atoms and the relative position of substituents and functional groups
in an organic molecule.

Task 34. Tasks of the 2018 format

Tasks 34 are calculation tasks. Their implementation requires knowledge of the chemical properties of substances and involves the implementation of a certain set of actions to ensure that the correct answer is obtained. Among these actions are the following:

– compilation of equations of chemical reactions (according to the conditions of the problem) necessary to perform stoichiometric calculations;

– performing calculations necessary to find answers to the set
questions in the condition of the problem;

- formulating a logically justified answer to all the questions posed in the task condition (for example, to establish a molecular formula).

However, it should be borne in mind that not all of these actions must necessarily be present when solving any calculation problem, and in some cases some of them can be used repeatedly.

The maximum score for completing the task is 4 points. When checking, you should first of all pay attention to the logical validity of the actions performed, since some tasks can be solved in several ways. At the same time, in order to objectively evaluate the proposed method for solving the problem, it is necessary to check the correctness of the intermediate results that were used to obtain the answer.

Job example:

Determine the mass fractions (in%) of iron (II) sulfate and aluminum sulfide
in a mixture, if during the treatment of 25 g of this mixture with water, gas was released, which completely reacted with 960 g of a 5% solution of copper sulfate.

In your answer, write down the reaction equations that are indicated in the condition of the problem,
and give all the necessary calculations (indicate the units of measurement of the required physical quantities).

Points

Possible answer

The reaction equations are composed:

The amount of hydrogen sulfide substance was calculated:

The amount of substance and mass of aluminum sulfide and iron(II) sulfate were calculated:

The mass fractions of iron(II) sulfate and aluminum sulfide in the initial mixture were determined:

ω(FeSO 4 ) \u003d 10 / 25 \u003d 0.4, or 40%

ω (Al 2 S 3) \u003d 15 / 25 \u003d 0.6, or 6 0%

The answer is correct and complete:

in the answer, the reaction equations corresponding to the task condition are correctly written;

calculations are correctly made that use the necessary physical quantities specified in the task condition;

demonstrated a logically justified relationship of physical quantities, on the basis of which calculations are carried out;

in accordance with the condition of the assignment, the required physical quantity is determined

An error was made in only one of the above response elements

All elements of the answer are written incorrectly

Maximum score

When checking the answer, the examinee must take into account the fact that in the case when the answer contains an error in the calculations in one of the three elements (second, third or fourth), which led to an incorrect answer, the mark for completing the task is reduced by only 1 point.

Task 35. Tasks of the 2018 format

Tasks 35 involve determining the molecular formula of a substance. The fulfillment of this task includes the following sequential operations: carrying out the calculations necessary to establish the molecular formula of an organic substance, writing the molecular formula of an organic substance, compiling a structural formula of a substance that unambiguously reflects the order of bonding of atoms in its molecule, writing a reaction equation that meets the condition of the task.

The grading scale for task 35 in part 2 of the examination paper will be 3 points.

Tasks 35 use a combination of verifiable content elements - calculations, on the basis of which they come to determine the molecular formula of a substance, draw up a general formula of a substance, and then determine, on its basis, the molecular and structural formula of a substance.

All these actions can be performed in a different sequence. In other words, the examinee can come to the answer in any logical way available to him. Therefore, when evaluating a task, the main attention is paid to the correctness of the chosen method for determining the molecular formula of a substance.

Job example:

When burning a sample of some organic compound weighing 14.8 g, 35.2 g of carbon dioxide and 18.0 g of water were obtained.

It is known that the relative hydrogen vapor density of this substance is 37. During the study of the chemical properties of this substance, it was found that the interaction of this substance with copper(II) oxide forms a ketone.

Based on these conditions of the assignment:

1) make the calculations necessary to establish the molecular formula of organic matter (indicate the units of measurement of the required physical quantities);

write down the molecular formula of the original organic matter;

2) make a structural formula of this substance, which unambiguously reflects the order of bonding of atoms in its molecule;

3) write the equation for the reaction of this substance with copper(II) oxide using the structural formula of the substance.

Correct Answer Content and Grading Instructions

(Other formulations of the answer are allowed that do not distort its meaning)

Points

Possible answer

The amount of substance of combustion products was found:

The general formula of a substance is C x H y O z

n (CO 2) \u003d 35.2 / 44 \u003d 0.8 mol; n (C) = 0.8 mol

n(H 2 O) \u003d 18.0 / 18 \u003d 1.0 mol; n(H) = 1.0 ∙ 2 = 2.0 mol

m (O) \u003d 14.8 - 0.8 ∙ 12 - 2 \u003d 3.2 g; n (O) \u003d 3.2 ⁄ 16 \u003d 0.2 mol

The molecular formula of the substance is determined:

x : y : z = 0.8: 2: 0.2 = 4: 10: 1

The simplest formula is C 4 H 10 O

M simple (C 4 H 10 O) = 74 g/mol

M ist (C x H y O z) \u003d 37 ∙ 2 \u003d 74 g / mol

The molecular formula of the starting substance is C 4 H 10 O

The structural formula of the substance has been compiled:

The equation for the reaction of a substance with copper oxide (II) is written:

The answer is correct and complete:

the calculations necessary to establish the molecular formula of a substance are correctly made; the molecular formula of the substance is written down;

the structural formula of the organic substance is written, which reflects the bond order and the mutual arrangement of substituents and functional groups in the molecule in accordance with the assignment condition;

the reaction equation is written, which is indicated in the task condition, using the structural formula of organic matter

An error was made in only one of the above response elements

Errors were made in two of the above response elements

Errors were made in three of the above response elements

All elements of the answer are written incorrectly

Maximum score

TOTAL part 2

2+2+ 4+5+4 +3=20 points

Bibliography

1. Methodological materials for chairmen and members of the subject commissions of the constituent entities of the Russian Federation to check the completion of tasks with a detailed answer to the examination papers of the USE in 2017. Article "Methodological recommendations for assessing the performance of USE assignments with a detailed question." Moscow, 2017.

2. FIPI project of control and measuring materials for the Unified State Examination in 2018.

3. Demo versions, specifications, USE 2018 codifiers. FIPI website.

4. Certificate of planned changes in KIM 2018. FIPI website.

5. Website "I will solve the exam": chemistry, to the expert.