I will solve the exam in chemistry task 33

Task number 1

Hydrogen with a volume of 3.36 L was passed with heating through the copper (II) oxide powder, while the hydrogen reacted completely. As a result of the reaction, 10.4 g of a solid residue was obtained. This residue was dissolved in concentrated sulfuric acid weighing 100 g. Determine the mass fraction of salt in the resulting solution (neglect the hydrolysis processes).

Answer: 25.4%

Explanation:

ν (H 2) = V (H 2) / V m = 3.36 L / 22.4 L / mol = 0.15 mol,

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

m (CuO) = m (solid) - m (Cu) = 10.4 g - 9.6 g = 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) = ν I (CuSO 4) + ν II (CuSO 4) = 0.01 mol + 0.15 mol = 0.16 mol.

m total (CuSO 4) = ν total. (CuSO 4) M (CuSO 4) = 0.16 mol 160 g / mol = 25.6 g

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

m (solution) = m (solid) + m (solution H 2 SO 4) - m (SO 2) = 10.4 g + 100 g - 9.6 g = 100.8 g

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

Task number 2

Hydrogen with a volume of 3.36 L (NU) 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 turning brown in air. Determine the mass fraction of nitric acid in the resulting solution (neglect the hydrolysis processes).

In the answer, write down the reaction equations that are indicated in the condition of the problem, and provide 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 nitric acid solution 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 and copper (II) oxide involved in reaction (I):

ν (H 2) = V (H 2) / V m = 3.36 L / 22.4 L / mol = 0.15 mol, ν (CuO) = 16 g / 80 g / mol = 0.2 mol

According to the reaction equation (I) ν (H 2) = ν (CuO), and according to the condition of the problem, the amount of hydrogen substance is deficient (0.15 mol H2 and 0.1 mol CuO), therefore, copper (II) oxide did not fully react ...

The calculation is carried out according to the lack of substance, therefore, ν (Cu) = ν (H 2) = 0.15 mol and ν rest. (CuO) = 0.2 mol - 0.15 mol = 0.05 mol.

To calculate the mass of the solution in the future, 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 equal to: m (solid. Rest.) = M (Cu) + m rest. (CuO) = 9.6 g + 4 g = 13.6 g

Calculate the initial mass and amount of the nitric acid substance:

m out. (HNO 3) = m (solution HNO 3) ω (HNO 3) = 535.5 g 0.2 = 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) = 2ν (CuO), therefore, ν total. (HNO 3) = ν II (HNO 3) + ν III (HNO 3) = 8/3 · 0.15 mol + 2 · 0.05 mol = 0.5 l.

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

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

In order to calculate the mass of the resulting solution, it is necessary to take into account the mass of nitrogen 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

Let's calculate the mass of the resulting solution:

m (solution) = m (solid) + m (solution HNO 3) - m (NO) = 13.6 g + 535.5 g - 3 g = 546.1 g

ω (HNO 3) = m rest. (HNO 3) / m (solution) 100% = 75.6 g / 546.1 g 100% = 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 end of the reaction, 117 g of a 10% sodium sulfide solution was poured into the solution. Determine the mass fraction of sodium sulfide in the final solution (neglect hydrolysis processes).

Answer: 5.12%

Explanation:

Fe + CuSO 4 → FeSO 4 + Cu (I)

ν (CuSO 4 5H 2 O) = m (CuSO 4 5H 2 O) / M (CuSO 4 5H 2 O) = 12.5 g / 250 g / mol = 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 sulphate substance is deficient (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)

The calculation is carried out according to the lack of substance, therefore, ν (CuSO 4 5H 2 O) = ν (Cu) = ν (FeSO 4) = 0.05 mol and ν rest. (Fe) = 0.1 mol - 0.05 mol = 0.05 mol.

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

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) = ν (CuSO 4) = 0.05 mol, therefore, m (CuSO 4) = ν (CuSO 4) M (CuSO 4) = 0.05 mol 160 g / mol = 8 g

m out. (solution CuSO 4) = m (CuSO 4) / ω (CuSO 4) 100% = 8 g / 20% 100% = 40 g

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

m out. (Na 2 S) = m ref. (solution Na 2 S) ω (Na 2 S) = 117 g 0.1 = 11.7 g

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

According to the reaction equation (II) ν (Na 2 S) = ν (FeSO 4), and according to the reaction condition, sodium sulfide in excess (0.15 mol Na 2 S and 0.05 mol FeSO 4). The calculation is carried out on the basis of a disadvantage, i.e. by the amount of iron (II) sulfate substance).

We calculate the mass of unreacted sodium sulfide:

ν rest. (Na 2 S) = ν ref. (Na 2 S) - ν reag. (Na 2 S) = 0.15 mol - 0.05 mol = 0.1 mol

m rest (Na 2 S) = ν (Na 2 S) M (Na 2 S) = 0.1 mol 78 g / mol = 7.8 g

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

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

m (solution) = m ref. (solution CuSO 4) + m out. (Fe) - m rest. (Fe) - m (Cu) + m ref. (solution Na 2 S) - m (FeS) = 40 g + 5.6 g - 3.2 g - 2.8 g + 117 g - 4.4 g = 152.2 g

ω (Na 2 S) = m (Na 2 S) / m (solution) 100% = 7.8 g / 152.2 g 100% = 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 (neglect the 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 (dil.) → FeSO 4 + H 2 (II)

Let's calculate the amount of copper sulfate and iron substances that enter into the reaction (I):

ν (CuSO 4 5H 2 O) = m (CuSO 4 5H 2 O) / M (CuSO 4 5H 2 O) = 37.5 g / 250 g / mol = 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 deficient (0.15 mol CuSO 4 5H 2 O and 0.2 mol Fe), so the iron did not react fully.

The calculation is carried out according to the lack of substance, therefore, ν (CuSO 4 5H 2 O) = ν (Cu) = ν (FeSO 4) = 0.15 mol and ν rest. (Fe) = 0.2 mol - 0.15 mol = 0.05 mol.

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

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

m out. (solution CuSO 4) = m (CuSO 4) / ω (CuSO 4) 100% = 24 g / 20% 100% = 120 g

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

Let's calculate the mass and amount of the sulfuric acid substance:

m out. (H 2 SO 4) = m ref. (solution H 2 SO 4) ω (H 2 SO 4) = 100 g 0.2 = 20 g

ν ref. (H 2 SO 4) = m ref. (H 2 SO 4) / M (H 2 SO 4) = 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 reaction equation (II) ν (Fe) = ν (FeSO 4), then the total amount of the substance of iron sulfate (II) is the sum of the amounts formed by reactions (I) and (II), and are equal to:

ν (FeSO 4) = 0.05 mol + 0.15 mol = 0.2 mol;

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

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

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

m (solution) = m ref. (solution CuSO 4) + m out. (Fe) - m (Cu) + m ref. (solution H 2 SO 4) - m (H 2) = 120 g + 11.2 g - 9.6 g + 100 g - 0.1 g = 221.5 g

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

ω (FeSO 4) = m (FeSO 4) / m (solution) 100% = 30.4 g / 221.5 g 100% = 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 were added. After completion of the reaction, 146 g of a 25% hydrochloric acid solution were added to the resulting mixture. Calculate the mass fraction of hydrogen chloride in the resulting solution (neglect 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's calculate the amount of copper sulfate and magnesium substances that enter into the 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), therefore magnesium did not react fully.

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

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

m out. (solution CuSO 4) = m (CuSO 4) / ω (CuSO 4) 100% = 32 g / 20% 100% = 160 g

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

We calculate the mass and amount of the substance of hydrochloric acid:

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

Since ν rest. (Mg) = 0.4 mol, ν ex. (HCl) = 1 mol and ν ex. (HCl)> 2ν rest. (Mg), magnesium is in short supply and is completely soluble in hydrochloric acid.

Let's calculate the amount of the substance of hydrochloric acid that has not reacted with magnesium:

ν rest. (HCl) = ν ref. (HCl) - ν reag. (HCl) = 1 mol - 2 0.4 mol = 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) = ν (H 2) = 0.4 mol and m (H 2) = ν (H 2) M (H 2) = 0.4 mol 2 g / mol = 0.8 g

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

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

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

ω (HCl) = m rest. (HCl) / m (solution) 100% = 7.3 g / 306.8 g 100% = 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 a 30% sodium hydroxide solution was added to the resulting mixture. Determine the mass fraction of sodium hydroxide in the resulting solution (neglect the 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 sulfate substance is in deficit (0.1 mol CuSO 4 5H 2 O and 0.3 mol Zn), therefore zinc did not react fully.

The calculation is carried out according to the lack of substance, therefore, ν (CuSO 4 5H 2 O) = ν (ZnSO 4) = ν (Cu) = ν reag. (Zn) = 0.1 mol and ν rest. (Zn) = 0.3 mol - 0.1 mol = 0.2 mol.

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

m out. (solution CuSO 4) = m (CuSO 4) / ω (CuSO 4) 100% = 16 g / 10% 100% = 160 g

m out. (NaOH) = m ref. (NaOH solution) ω (NaOH) = 240 g 0.3 = 72 g

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

ν total (NaOH) = ν II (NaOH) + ν III (NaOH) = 2 0.2 mol + 4 0.1 mol = 0.8 mol

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

m rest (NaOH) = m ref. (NaOH) - m reag. (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) = ν (H 2) = 0.2 mol and m (H 2) = ν (H 2) M (H 2) = 0.2 mol 2 g / mol = 0.4 g

m (solution) = m ref. (solution CuSO 4) + m out. (Zn) - m (Cu) + m ref. (NaOH solution) - m (H 2) = 160 g + 19.5 g - 6.4 g + 240 g - 0.4 g = 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 (II) sulfate in water, the powder obtained by sintering 2.16 g of aluminum and 6.4 g of iron (III) oxide was introduced. Determine the mass fraction of copper (II) sulfate in the resulting solution (neglect hydrolysis processes).

Answer: 4.03%

Explanation:

When sintering aluminum 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 and iron oxide (III) substances 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) = 6.4 g / 160 g / mol = 0.04 mol

According to the reaction equation (I) ν (Al) = 2ν (Fe 2 O 3) = 2ν (Al 2 O 3) and by the condition of the problem, the amount of aluminum substance is twice the amount of iron (III) oxide 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) = 2 0.04 mol = 0.08 mol

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

To calculate the mass of the final solution in the future, it is necessary to know the mass of the initial solution of copper sulfate:

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

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

m out. (solution CuSO 4) = m (CuSO 4) / ω (CuSO 4) 100% = 16 g / 20% 100% = 80 g

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

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), therefore, the iron reacted completely.

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

ν rest. (CuSO 4) = ν ref. (CuSO 4) - ν reag. (CuSO 4) = 0.1 mol - 0.08 mol = 0.02 mol

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

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

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) goes into solution in the future):

m (solution) = m ref. (solution CuSO 4) + m (Fe) - m (Cu) = 80 g + 4.48 g - 5.12 g = 79.36 g

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

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

Task number 8

182.5 g of a 20% hydrochloric acid solution was added to 18.2 g of calcium phosphide. 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 (neglect the 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 the substance of hydrochloric acid and calcium phosphide entering into the reaction (I):

m out. (HCl) = m (solution HCl) ω (HCl) = 182.5 g 0.2 = 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) = 18.2 g / 182 g / mol = 0.1 mol

According to the reaction equation (I) ν (HCl) = 6ν (Ca 3 P 2) = 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) = 1 mol - 6 0.1 mol = 0.4 mol

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

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

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

Let's 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) = 200.2 g / 286 g / mol = 0.7 mol

Both calcium chloride and hydrochloric acid react in 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) = ν (CaCl 2) + 1 / 2ν rest. (HCl) = 3ν ref. (Ca 3 P 2) + 1 / 2ν rest. (HCl) = 3 0.1 mol + 1/2 0.4 mol = 0.3 mol + 0.2 mol = 0.5 mol

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

ν rest. (Na 2 CO 3) = ν ref. (Na 2 CO 3) - ν reag. (Na 2 CO 3) = 0.7 mol - 0.5 mol = 0.2 mol

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

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

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

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

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

We calculate the mass of the resulting solution by the formula:

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

Mass fraction of sodium carbonate is equal to:

ω (Na 2 CO 3) = m rest. (Na 2 CO 3) / m (solution) 100% = 21.2 g / 355.3 g 100% = 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 (neglect the 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 reacting with each other:

m out. (H 2 SO 4) = m (solution H 2 SO 4) ω (H 2 SO 4) = 490 g 0.2 = 98 g, hence

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

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

Let us calculate the amount of sulfuric acid unreacted by reaction (I):

ν rest. I (H 2 SO 4) = ν ref. (H 2 SO 4) - 2ν ref. (Na 3 N) = 1 mol - 2 0.1 mol = 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) = 57.2 g / 286 g / mol = 0.2 mol

Since by the condition of the problem ν rest. I (H 2 SO 4) = 3ν ref. (Na 2 CO 3 10H 2 O), i.e. dilute sulfuric acid 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 out. (Na 2 CO 3) = 0.8 mol - 0.2 mol = 0.6 mol

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

ν (CO 2) = ν (Na 2 CO 3) = 0.2 mol

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

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

Mass fraction of sulfuric acid is equal to:

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

Task number 10

Lithium nitride weighing 3.5 g was dissolved in 365 g of 10% hydrochloric acid. To the solution was added 20 g of calcium carbonate. Determine the mass fraction of hydrochloric acid in the resulting solution (neglect the 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 substances reacting with each other:

m out. (HCl) = m (solution HCl) ω (HCl) = 365 g 0.1 = 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) = 3.5 g / 35 g / mol = 0.1 mol

Let us calculate the amount of hydrochloric acid unreacted by reaction (I):

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

Let's calculate the amount of calcium carbonate substance:

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

Since by the condition of the problem ν rest. I (HCl) = 3ν 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 out. (CaCO 3) = 0.6 mol - 2 · 0.2 mol = 0.2 mol

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

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

ν (CO 2) = ν (CaCO 3) = 0.2 mol

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

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

m (solution) = m ref. (solution HCl) + m out. (Li 3 N) + m (CaCO 3) - m (CO 2) = 365 g + 3.5 g + 20 g - 8.8 g = 379.7 g

Mass fraction of hydrochloric acid is equal to:

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

Task number 11

The solid residue obtained by reacting 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 (neglect the 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 (II) oxide:

ν ref. (H 2) = V (H 2) / V m = 2.24 L / 22.4 L / mol = 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) = ν (solid. Rest.) - ν ref. (H 2) = 0.15 mol - 0.1 mol = 0.05 mol

Let us calculate the masses of the 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)

Let's calculate the amount of nitric acid substance:

m out. (HNO 3) = m (solution HNO 3) ω (HNO 3) = 126 g 0.85 = 107.1 g, hence

ν ref. (HNO 3) = m ref. (HNO 3) / M (HNO 3) = 107.1 g / 63 g / mol = 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) = ν II (HNO 3) + ν III (HNO 3) = 4 · 0.1 mol + 2 · 0.05 mol = 0.5 mol.

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

m total (HNO 3) = ν total. (HNO 3) M (HNO 3) = 0.5 mol 63 g / mol = 31.5 g

Let's calculate the mass of unreacted nitric acid:

m rest (HNO 3) = m ref. (HNO 3) - m total. (HNO 3) = 107.1 g - 31.5 g = 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) = 2m (Cu), therefore, ν (NO 2) = 0.2 mol and m (NO 2) = ν (NO 2) M (NO 2) = 0.2 mol 46 g / mol = 9.2 g

Let's calculate the mass of the resulting solution:

m (solution) = m (solution HNO 3) + m (Cu) + m (CuO) - m (NO 2) = 126 g + 6.4 g + 4 g - 9.2 g = 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% = 75.6 g / 127.2 g 100% = 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, 7.2 g of magnesium was added. 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 (neglect the hydrolysis processes).

Answer: 7.21%

Explanation:

Mg + ZnSO 4 → MgSO 4 + Zn (I)

ν ref. (ZnSO 4 7H 2 O) = ν (ZnSO 4) = m out. (ZnSO 4 7H 2 O) / M (ZnSO 4 7H 2 O) = 28.7 g / 287 g / mol = 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) = 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), therefore magnesium did not fully react.

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

To calculate the mass of the final solution in the future, 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) = 0.1 mol, therefore m (ZnSO 4) = ν (ZnSO 4) M (ZnSO 4) = 0.1 mol 161 g / mol = 16.1 g

m out. (solution ZnSO 4) = m (ZnSO 4) / ω (ZnSO 4) 100% = 16.1 g / 10% 100% = 161 g

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

Zn + 2NaOH + 2H 2 O → Na 2 + H 2 (II)

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

Let's calculate the mass and amount of sodium hydroxide substance:

m out. (NaOH) = m ref. (NaOH solution) 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 equal:

ν total (NaOH) = ν II (NaOH) + ν III (NaOH) = 2ν (Zn) + 2ν (MgSO 4) = 2 0.1 mol + 2 0.1 mol = 0.4 mol

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

ν (MgSO 4) = ν (Mg (OH) 2) = 0.1 mol

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

Calculate the mass of unreacted alkali:

m rest (NaOH) = m ref. (NaOH) - m reag. (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) = ν (H 2) = 0.1 mol and m (H 2) = ν (H 2) M (H 2) = 0.1 mol 2 g / mol = 0.2 g

We calculate the mass of the resulting solution by the formula:

m (solution) = m ref. (solution ZnSO 4) + m out. (Mg) - m rest. (Mg) + m out. (NaOH solution) - m (Mg (OH) 2) - m (H 2) = 161 g + 7.2 g - 4.8 g + 120 g - 5.8 g - 0.2 g = 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 crystalline zinc sulfate hydrate (ZnSO 4 · 7H 2 O) in water, 14.4 g of magnesium was added. 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 (neglect the hydrolysis processes).

Answer: 6.26%

Explanation:

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

Mg + ZnSO 4 → MgSO 4 + Zn (I)

We calculate the amount of zinc and magnesium sulfate substances that enter into reaction (I):

ν ref. (ZnSO 4 7H 2 O) = ν (ZnSO 4) = m out. (ZnSO 4 7H 2 O) / M (ZnSO 4 7H 2 O) = 57.4 g / 287 g / mol = 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) = 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), therefore, magnesium did not fully react.

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

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

M (ZnSO 4) = 0.2 mol 161 g / mol = 32.2 g

m out. (solution ZnSO 4) = m (ZnSO 4) / ω (ZnSO 4) 100% = 32.2 g / 20% 100% = 161 g

Zn + 2HCl → ZnCl 2 + H 2 (II)

We calculate the mass and amount of the substance of hydrogen chloride:

m out. (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

ν total (HCl) = ν II (HCl) + ν III (HCl) = 2ν (Zn) + 2ν (Mg) = 2 0.2 mol + 2 0.4 mol = 1.2 mol

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

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

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

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

We calculate the mass of the resulting solution by the formula:

m (solution) = m ref. (solution ZnSO 4) + m out. (Mg) + m out. (solution HCl) - m total. (H 2) = 161 g + 14.4 g + 292 g - 1.2 g = 466.2 g

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

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

Task number 14

Zinc oxide weighing 16.2 g was heated and carbon monoxide with a volume of 1.12 l was passed through it. The carbon monoxide has completely reacted. 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 (neglect the 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 out. (CO) / V m = 1.12 L / 22.4 L / mol = 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), therefore, zinc oxide did not fully react.

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

Let's calculate the mass and amount of sodium hydroxide substance:

m out. (NaOH) = m ref. (NaOH solution) 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 equal:

ν total (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 reag. (NaOH) M (NaOH) = 0.4 mol 40 g / mol = 16 g

m rest (NaOH) = m ref. (NaOH) - m reag. (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) = ν (H 2) = 0.05 mol and m (H 2) = ν (H 2) M (H 2) = 0.05 mol 2 g / mol = 0.1 g

We calculate the mass of the resulting solution by the formula:

m (solution) = m ref. (NaOH solution) + m (Zn) + m rest. (ZnO) - m (H 2) = 60 g + 12.15 g + 3.25 g - 0.1 g = 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, 7.8 g of zinc was added. 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 (neglect the 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 out. ((CH 3 COO) 2 Pb 3H 2 O) / M ((CH 3 COO) 2 Pb 3H 2 O) = 37.9 g / 379 g / mol = 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 fully reacted.

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 reag. (Zn) = 0.1 mol and ν rest. (Zn) = 0.12 mol - 0.1 mol = 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) = ν ((CH 3 COO) 2 Pb) M ((CH 3 COO) 2 Pb) = 0.1 mol 325 g / mol = 32.5 g

m out. (solution CH 3 COO) 2 Pb) = m ((CH 3 COO) 2 Pb) / ω ((CH 3 COO) 2 Pb) 100% = 32.5 g / 10% 100% = 325 g

Let's calculate the mass and amount of sodium sulfide substance:

m out. (Na 2 S) = m ref. (solution Na 2 S) ω (Na 2 S) = 156 g 0.1 = 15.6 g

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

ν rest. (Na 2 S) = ν ref. (Na 2 S) - ν reag. (Na 2 S) = 0.2 mol - 0.1 mol = 0.1 mol

m rest (Na 2 S) = ν reag. (Na 2 S) M (Na 2 S) = 0.1 mol 78 g / mol = 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

We calculate the mass of the resulting solution by the formula:

m (solution) = m ref. (solution (CH 3 COO) 2 Pb) + m out. (Zn) - m rest. (Zn) - m (Pb) + m ref. (solution Na 2 S) - m (ZnS) = 325 g + 7.8 g - 1.3 g - 20.7 g + 156 g - 9.7 g = 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% = 7.8 g / 457.1 g 100% = 1.71%

Task number 16

Zinc oxide weighing 32.4 g was heated and carbon monoxide with a volume of 2.24 l was passed through it. The carbon monoxide has completely reacted. The resulting solid residue was dissolved in 224 g of 40% potassium hydroxide solution. Determine the mass fraction of potassium hydroxide in the resulting solution (neglect the 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 that enter into the reaction (I):

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

ν ref. (CO) = V out. (CO) / V m = 2.24 L / 22.4 L / mol = 0.1 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 (0.1 mol CO and 0.4 mol ZnO), therefore, zinc oxide did not fully react.

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

To calculate the mass of the final solution in the future, it is necessary to know the masses of the formed 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

Let's calculate the mass and amount of sodium hydroxide substance:

m out. (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 equal:

ν total (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 reag. (KOH) M (KOH) = 0.8 mol 56 g / mol = 44.8 g

Let's calculate the mass of unreacted alkali:

m rest (KOH) = m ref. (KOH) - m reag. (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):

We calculate the mass of the resulting solution by the formula:

m (solution) = m ref. (solution KOH) + m (Zn) + m rest. (ZnO) - m (H 2) = 224 g + 6.5 g + 24.3 g - 0.2 g = 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, 15.6 g of zinc was added. 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 (neglect the 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 out. ((CH 3 COO) 2 Pb 3H 2 O) / M ((CH 3 COO) 2 Pb 3H 2 O) = 75.8 g / 379 g / mol = 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 of Zn), so the zinc did not completely reacted.

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 reag. (Zn) = 0.2 mol and ν rest. (Zn) = 0.24 mol - 0.2 mol = 0.04 mol.

To calculate the mass of the final solution in the future, 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) = ν ((CH 3 COO) 2 Pb) M ((CH 3 COO) 2 Pb) = 0.2 mol 325 g / mol = 65 g

m out. (solution CH 3 COO) 2 Pb) = m ((CH 3 COO) 2 Pb) / ω ((CH 3 COO) 2 Pb) 100% = 65 g / 10% 100% = 650 g

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

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

Let's calculate the mass and amount of sodium sulfide substance:

m out. (Na 2 S) = m ref. (solution Na 2 S) ω (Na 2 S) = 312 g 0.1 = 31.2 g

ν ref. (Na 2 S) = m ref. (Na 2 S) / M (Na 2 S) = 31.2 g / 78 g / mol = 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) = ν ref. (Na 2 S) - ν reag. (Na 2 S) = 0.4 mol - 0.2 mol = 0.2 mol

m rest (Na 2 S) = ν reag. (Na 2 S) M (Na 2 S) = 0.2 mol 78 g / mol = 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) = ν (ZnS) = 0.2 mol and m (ZnS) = ν (ZnS) M (ZnS) = 0.2 mol 97 g / mol = 19.4 g

We calculate the mass of the resulting solution by the formula:

m (solution) = m ref. (solution (CH 3 COO) 2 Pb) + m out. (Zn) - m rest. (Zn) - m (Pb) + m ref. (solution Na 2 S) - m (ZnS) = 650 g + 15.6 g - 2.6 g - 41.4 g + 312 g - 19.4 g = 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% = 15.6 g / 914.2 g 100% = 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 a 30% sodium hydroxide solution was added to the resulting mixture. Determine the mass fraction of sodium hydroxide in the resulting solution (neglect the 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 copper sulfate and zinc substances that enter into the reaction (I):

ν (CuSO 4 5H 2 O) = m (CuSO 4 5H 2 O) / M (CuSO 4 5H 2 O) = 50 g / 250 g / mol = 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 sulfate substance is in short supply (0.2 mol CuSO 4 5H 2 O and 0.3 mol Zn), therefore zinc did not react fully.

The calculation is carried out according to the lack of substance, therefore, ν (CuSO 4 5H 2 O) = ν (ZnSO 4) = ν (Cu) = ν reag. (Zn) = 0.2 mol and ν rest. (Zn) = 0.3 mol - 0.2 mol = 0.1 mol.

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

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

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

m out. (solution CuSO 4) = m (CuSO 4) / ω (CuSO 4) 100% = 32 g / 10% 100% = 320 g

With a solution of sodium hydroxide, zinc and zinc sulfate that have not completely reacted in reaction (I) react with the formation of 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)

Let's calculate the mass and amount of sodium hydroxide substance:

m out. (NaOH) = m ref. (NaOH solution) 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 equal:

ν total (NaOH) = ν II (NaOH) + ν III (NaOH) = 2 0.1 mol + 4 0.2 mol = 1 mol

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

Calculate the mass of unreacted alkali:

m rest (NaOH) = m ref. (NaOH) - m reag. (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) = ν (H 2) = 0.1 mol and m (H 2) = ν (H 2) M (H 2) = 0.1 mol 2 g / mol = 0.2 g

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

m (solution) = m ref. (solution CuSO 4) + m out. (Zn) - m (Cu) + m ref. (NaOH solution) - m (H 2) = 320 g + 19.5 g - 12.8 g + 200 g - 0.2 g = 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 number 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 with a mass fraction of copper (II) sulfate of 20% was formed. 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)

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

m (CuSO 4) = m (CuSO 4) ω (CuSO 4) = 400 g 0.2 = 80 g

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

Let's calculate the amount of sulfur dioxide gas:

ν (SO 2) = V (SO 2) / V m = 8.96 L / 22.4 L / mol = 0.4 mol

According to the reaction equation (I) ν (Cu) = ν (SO 2) = ν I (CuSO 4), therefore, ν (Cu) = ν I (CuSO 4) = 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) = 0.5 mol - 0.4 mol = 0.1 mol.

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

Let's calculate the masses of copper and copper (II) oxide:

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

Let's calculate the mass fraction of copper (II) oxide:

ω (CuO) = m (CuO) / m (mixture) ∙ 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 potassium hydroxide solution with a mass fraction of 40% and a density of 1.4 g / ml, which is required to dissolve the initial mixture.

Answer: 80 ml

Explanation:

When zinc is heated in air, zinc is oxidized and converted into oxide:

2Zn + O 2 → 2ZnO (I)

As the mass of the mixture increased, this increase was due to the mass of oxygen:

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

ν (Zn) = 2ν (O 2) = 2 · 0.125 mol = 0.25 mol

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

We calculate the mass and amount of the zinc oxide substance is:

m (ZnO) = m (mixture) - 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 the 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

We 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) = m (KOH) / ρ (KOH) = 112 g / 1.4 g / mol = 80 ml

Task number 21

A mixture of magic oxide and magnesium carbonate weighing 20.5 g was heated to constant mass, 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 sulfuric acid solution 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 evolved carbon dioxide.

Let's calculate the amount of carbon dioxide formed:

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

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

Let's calculate the mass of the reacted magnesium carbonate:

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

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

m (MgO) = m (mixture) - m (MgCO 3) = 20.5 g - 10.5 g = 10 g

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

The total amount of magnesium oxide is:

ν total (MgO) = ν I (MgO) + ν (MgO) = 0.25 mol + 0.125 mol = 0.375 mol

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

Let's calculate the mass of sulfuric acid:

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

Let's calculate the mass and volume of the sulfuric acid solution:

m (solution H 2 SO 4) = 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) = m (solution H 2 SO 4) / ρ (solution H 2 SO 4) = 131.25 g / 1.2 g / ml = 109.375 ml

Task number 22

Hydrogen with a volume of 6.72 L (NU) was passed over the heated powder of copper (II) oxide, while the hydrogen reacted completely. As a result, 20.8 g of a solid residue were obtained. This residue was dissolved in concentrated sulfuric acid weighing 200 g. Determine the mass fraction of salt in the resulting solution (neglect the 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 (II) oxide:

ν (H 2) = V (H 2) / V m = 6.72 L / 22.4 L / mol = 0.3 mol,

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

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

m (CuO) = m (solid) - m (Cu) = 20.8 g - 19.2 g = 1.6 g

Let's 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) = ν II (CuSO 4) + ν III (CuSO 4) = 0.3 mol + 0.02 mol = 0.32 mol.

We calculate the total mass of copper (II) sulfate:

m total (CuSO 4) = ν total. (CuSO 4) M (CuSO 4) = 0.32 mol 160 g / mol = 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) = ν (SO 2), therefore, ν (SO 2) = 0.3 mol and m (SO 2) = ν (SO 2) M (SO 2) = 0.3 mol 64 g / mol = 19.2 g

Let's calculate the mass of the resulting solution:

m (solution) = m (solid) + m (solution H 2 SO 4) - m (SO 2) = 20.8 g + 200 g - 19.2 g = 201.6 g

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

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

Task number 23

To a 10% salt solution obtained by dissolving 114.8 g of crystalline zinc sulfate hydrate (ZnSO 4 · 7H 2 O) in water, 12 g of magnesium were added. 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 (neglect the hydrolysis processes).

Answer: 3.58%

Explanation:

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

Mg + ZnSO 4 → MgSO 4 + Zn (I)

We calculate the amount of zinc and magnesium sulfate substances that enter into reaction (I):

ν ref. (ZnSO 4 7H 2 O) = ν (ZnSO 4) = m out. (ZnSO 4 7H 2 O) / M (ZnSO 4 7H 2 O) = 114.8 g / 287 g / mol = 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) = 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), therefore, magnesium did not fully react.

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

To calculate in the future the mass of the initial solution of zinc sulfate:

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

m out. (solution ZnSO 4) = m (ZnSO 4) / ω (ZnSO 4) 100% = 64.4 g / 10% 100% = 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)

Let's calculate the mass of hydrogen chloride in solution:

m out. (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 equal:

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

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

We calculate the mass of unreacted hydrochloric acid:

m rest (HCl) = m ref. (HCl) - m reag. (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) = ν II (H 2) = 0.1 mol and m II (H 2) = ν II (H 2) M (H 2) = 0.1 mol 2 g / mol = 0.2 G

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

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

We calculate the mass of the resulting solution by the formula:

m (solution) = m ref. (solution ZnSO 4) + m out. (Mg) + m out. (solution HCl) - m total. (H 2) = 644 g + 12 g + 365 g - 1 g = 1020 g

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

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

The content of the "Organic substances" block 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 different classes of organic compounds, the relationship of these substances. This block includes 9 tasks. The assimilation of the elements of the content of this block is checked by the tasks of the basic (tasks 11-15 and 18), advanced (tasks 16 and 17) and high (task 33) difficulty levels. These tasks also checked the formation of skills and types of activities, similar to those that were named in relation to the elements of the content of the block "Inorganic substances".

Let's consider the tasks of the "Organic matter" block.

# ADVERTISING_INSERT #

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

Task 33

Write down the reaction equations with which you can 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, in accordance with Markovnikov's rule, mainly 2-chloropropane:

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

Further, propene (X 1) must be re-obtained 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 the oxidation of propene with an aqueous solution of potassium permanganate in the cold is the dihydric alcohol propanediol-1,2, potassium permanganate is reduced in this case 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 difficulty, grouped by topics and types. The tasks are arranged in the same sequence as suggested in the exam version of the USE. At the beginning of each type of assignment, the content items to be tested are indicated - topics that should be studied before proceeding. The manual will be useful for teachers of chemistry, as it makes it possible to effectively organize the educational process in the classroom, conduct current control of knowledge, as well as prepare students for the exam.

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 the increased (in the USE codifier in chemistry of 2018 - the high level of complexity) level of complexity, previously referred to as part C.

The tasks of an increased level of complexity include only five (5) tasks - 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 complex tasks in the exam in chemistry in 2018.

Example of assignment 30 in the exam in chemistry in 2018

Aimed at testing the student's knowledge of redox reactions (ORR). The task always gives the equation of a chemical reaction with missing substances from either side of the reaction (left side - reagents, right side - products). A maximum of three (3) points can be obtained for this task. The first point is given for the correct filling of the gaps in the reaction and the correct equalization of the reaction (placing the coefficients). The second point can be obtained by correctly describing the ORP balance, and the last point is given for correctly determining who is the oxidizing agent and who is the reducing agent in the reaction. Let's analyze the solution to task number 30 from the demo version of the exam in chemistry in 2018:

Using the electronic balance method, write the reaction equation

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

Determine the oxidizing agent and reducing agent.

The first thing that needs to be done is to arrange the charges of the atoms indicated in the equation, it turns out:

Na + 2 S +4 O 3 -2 +… + K + O -2 H + a K + 2 Mn +6 O 4 -2 +… + H + 2 O -2

Often after this action, we immediately see the first pair of elements, which changed the oxidation state (CO), that is, from different sides of the reaction, at the same atom, a different oxidation state. In this particular assignment, we do not see anything like that. Therefore, it is necessary to use additional knowledge, namely, on the left side of the reaction, we see potassium hydroxide ( KOH), the presence of which tells us that the reaction takes place in an alkaline medium. On the right side, we see potassium manganate, and we know that in an alkaline reaction medium, potassium manganate is obtained from potassium permanganate, therefore, the omission 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, which means 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 should have given 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 maximum for them, oxygen will not donate its electrons to manganese, which means that sulfur remains in CO +4. We conclude that we donate electrons to sulfur and transforms into the sulfur state with CO +6. Now we can write the second part of the balance:

S +4 -2 e — à S +6

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

Now the OVR balance is written (we get the first point) 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, in this place, to immediately write who is an oxidizing agent and who is a reducing agent, since students often focus on equalizing the equation and simply forget to do this part of the task, thereby losing point. By definition, an oxidizer is a particle that receives electrons (in our case, manganese), and a reducing agent is a particle that gives up electrons (in our case, sulfur), so we get:

Oxidizing agent: Mn +7 (KMnO 4 )

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

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

Now, to get the last point, you need to correctly equalize the equation (place the coefficients). Using the balance, we see that in order for sulfur +4 to go into the +6 state, two manganese +7 must become manganese +6, and that means 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, which means we need to put 2 in front of the 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 # 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

Oxidizing agent: Mn +7 (KMnO 4)

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

Solution of task 31 in the exam in chemistry

This is a chain of inorganic transformations. To successfully complete this task, you must be well versed in the reactions characteristic of inorganic compounds. The task consists of four (4) reactions, for each of which you can get one (1) point, in total for the task you can get four (4) points. It is important to remember the rules for the design of the assignment: 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. 3, which means that you need to do this, and get two (2) points at the same time, the main thing is to indicate that these are reactions 1 and 3. Let's analyze the solution to task number 31 from the demo version of the USE in chemistry in 2018:

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

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

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, the student does not remember what happens to the iron when interacting with the end. 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, substituting Y into the written scheme, to try to make equations 2 and 3. Subsequent actions are purely chemical, so we will not describe them in such detail. The student should 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, which gives iron hydroxide 3 after alkali treatment, will be iron sulfate 3 ( X = Fe 2 (SO 4 ) 3 ). It is important to remember 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 (g) à 2 Fe (OH) 3 + 3Na 2 SO 4

3) 2Fe (OH) 3à Fe 2 O 3 + 3H 2 O

4) Fe 2 O 3 + Fe à 3FeO

Assignment 32 Unified State Exam in Chemistry

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

Solution of task 33 in chemistry of 2018

A computational task, for its implementation it is necessary to know the basic computational formulas, be able to use a calculator and draw logical parallels. For task number 33, you can get four (4) points. Consider a part of the solution to task number 33 from the demo version of the exam in chemistry in 2018:

Determine the mass fractions (in%) of iron (II) sulfate and aluminum sulfide in the mixture if, when treating 25 g of this mixture with water, a gas was released, which completely reacted with 960 g of a 5% solution of copper sulfate. In response, write down the reaction equations that are indicated in the problem statement, and provide all the necessary calculations (indicate the units of measurement of the sought physical quantities).

The first (1) score we get for writing the reactions that occur in the problem. The receipt of this particular point depends on the knowledge of chemistry, the remaining three (3) points can be obtained only through calculations, therefore, if a student has problems with mathematics, he must receive at least one (1) point for completing task # 33:

Al 2 S 3 + 6H 2 Oà 2Al (OH) 3 + 3H 2 S

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 a selection of the analysis on our YouTube channel (link to the video of the analysis of the task number 33).

Formulas that will be required to solve this task:

Chemistry Challenge 34 2018

Estimated task, which differs from task number 33 as follows:

- - If in task number 33 we know between which substances the interaction takes place, then in task number 34 we must find what reacted;
  - In task number 34, organic compounds are given, while in task number 33, inorganic processes are most often given.

In fact, task # 34 is the opposite of task # 33, which means that the logic of the task is the opposite. For task number 34, you can get four (4) points, while, as in task number 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 ... For the successful completion of task number 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 successfully passing the exam in chemistry in 2018 have not practically 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 of preparation for the Unified State Exam and the OGE Godograph, your child will receive all theoretical materials necessary for preparation, and in the classroom will consolidate the knowledge gained for successful implementation of all examination tasks. 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 performing the examination work.

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

Call today and tomorrow your child will thank you!

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

There are no changes in the KIM USE 2020 in chemistry.

Don't postpone your preparation until later.

When starting to analyze the 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. will guide you in the study of the main topics and determine what knowledge and skills will be required when completing the USE tasks in chemistry. For the successful completion of the exam in chemistry, theory is most important.
Theory needs to be backed up practice constantly solving tasks. 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 such an opportunity to decide and find out the answers. But don't rush to pry. First, decide for yourself and see how many points you scored.

Points for each chemistry task

1 point - for tasks 1-6, 11-15, 19-21, 26-28.
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 involve a short answer (in the form of a number or a word) - tasks 1-29.
Problems with detailed answers - tasks 30-35.

3.5 hours (210 minutes) are allotted for the performance of the examination work in chemistry.

There will be three cheat sheets on the exam. And you need to understand them

This is 70% of the information that will help you successfully pass the chemistry exam. The remaining 30% is the ability to use the presented 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.
Distribute your strength correctly and do not forget about rest.

Dare, try and you will succeed!