24 task in chemistry. How to solve problems in chemistry, ready-made solutions. The structure of the examination work consists of two blocks

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 formed by 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) \u003d 2ν (Fe 2 O 3) \u003d 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) = 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), so 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 potassium hydroxide solution 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%

Methodology for solving problems in chemistry

When solving problems, you need to be guided by a few simple rules:

Carefully read the condition of the problem;
Write down what is given;
Convert, if necessary, units of physical quantities to SI units (some non-systemic units are allowed, such as liters);
Write down, if necessary, the reaction equation and arrange the coefficients;
Solve the problem using the concept of the amount of substance, and not the method of drawing up proportions;
Write down the answer.

In order to successfully prepare in chemistry, one should carefully consider the solutions to the problems given in the text, as well as independently solve a sufficient number of them. It is in the process of solving problems that the main theoretical provisions of the chemistry course will be fixed. It is necessary to solve problems throughout the entire time of studying chemistry and preparing for the exam.

You can use the tasks on this page, or you can download a good collection of tasks and exercises with the solution of typical and complicated tasks (M. I. Lebedeva, I. A. Ankudimova): download.

Mole, molar mass

Molar mass is the ratio of the mass of a substance to the amount of a substance, i.e.

М(х) = m(x)/ν(x), (1)

where M(x) is the molar mass of substance X, m(x) is the mass of substance X, ν(x) is the amount of substance X. The SI unit for molar mass is kg/mol, but g/mol is commonly used. The unit of mass is g, kg. The SI unit for the amount of a substance is the mole.

Any chemistry problem solved through the amount of matter. Remember the basic formula:

ν(x) = m(x)/ М(х) = V(x)/V m = N/N A , (2)

where V(x) is the volume of substance Х(l), Vm is the molar volume of gas (l/mol), N is the number of particles, N A is the Avogadro constant.

1. Determine the mass sodium iodide NaI amount of substance 0.6 mol.

Given: ν(NaI)= 0.6 mol.

Find: m(NaI) =?

Solution. The molar mass of sodium iodide is:

M(NaI) = M(Na) + M(I) = 23 + 127 = 150 g/mol

Determine the mass of NaI:

m(NaI) = ν(NaI) M(NaI) = 0.6 150 = 90 g.

2. Determine the amount of substance atomic boron contained in sodium tetraborate Na 2 B 4 O 7 weighing 40.4 g.

Given: m(Na 2 B 4 O 7) \u003d 40.4 g.

Find: ν(B)=?

Solution. The molar mass of sodium tetraborate is 202 g/mol. Determine the amount of substance Na 2 B 4 O 7:

ν (Na 2 B 4 O 7) \u003d m (Na 2 B 4 O 7) / M (Na 2 B 4 O 7) \u003d 40.4 / 202 \u003d 0.2 mol.

Recall that 1 mol of sodium tetraborate molecule contains 2 mol of sodium atoms, 4 mol of boron atoms and 7 mol of oxygen atoms (see the formula of sodium tetraborate). Then the amount of atomic boron substance is: ν (B) \u003d 4 ν (Na 2 B 4 O 7) \u003d 4 0.2 \u003d 0.8 mol.

Calculations by chemical formulas. Mass share.

The mass fraction of a substance is the ratio of the mass of a given substance in the system to the mass of the entire system, i.e. ω(X) =m(X)/m, where ω(X) is the mass fraction of substance X, m(X) is the mass of substance X, m is the mass of the entire system. Mass fraction is a dimensionless quantity. It is expressed as a fraction of a unit or as a percentage. For example, the mass fraction of atomic oxygen is 0.42, or 42%, i.e. ω(O)=0.42. The mass fraction of atomic chlorine in sodium chloride is 0.607, or 60.7%, i.e. ω(Cl)=0.607.

3. Determine the mass fraction water of crystallization in barium chloride dihydrate BaCl 2 2H 2 O.

Solution: The molar mass of BaCl 2 2H 2 O is:

M (BaCl 2 2H 2 O) \u003d 137+ 2 35.5 + 2 18 \u003d 244 g / mol

From the formula BaCl 2 2H 2 O it follows that 1 mol of barium chloride dihydrate contains 2 mol of H 2 O. From this we can determine the mass of water contained in BaCl 2 2H 2 O:

m(H 2 O) \u003d 2 18 \u003d 36 g.

We find the mass fraction of water of crystallization in barium chloride dihydrate BaCl 2 2H 2 O.

ω (H 2 O) \u003d m (H 2 O) / m (BaCl 2 2H 2 O) \u003d 36/244 \u003d 0.1475 \u003d 14.75%.

4. From a rock sample weighing 25 g containing the mineral argentite Ag 2 S, silver weighing 5.4 g was isolated. Determine the mass fraction argentite in the sample.

Given: m(Ag)=5.4 g; m = 25 g.

Find: ω(Ag 2 S) =?

Solution: we determine the amount of silver substance in argentite: ν (Ag) \u003d m (Ag) / M (Ag) \u003d 5.4 / 108 \u003d 0.05 mol.

From the formula Ag 2 S it follows that the amount of argentite substance is half the amount of silver substance. Determine the amount of argentite substance:

ν (Ag 2 S) \u003d 0.5 ν (Ag) \u003d 0.5 0.05 \u003d 0.025 mol

We calculate the mass of argentite:

m (Ag 2 S) \u003d ν (Ag 2 S) M (Ag 2 S) \u003d 0.025 248 \u003d 6.2 g.

Now we determine the mass fraction of argentite in a rock sample, weighing 25 g.

ω (Ag 2 S) \u003d m (Ag 2 S) / m \u003d 6.2 / 25 \u003d 0.248 \u003d 24.8%.

Derivation of compound formulas

5. Determine the simplest compound formula potassium with manganese and oxygen, if the mass fractions of elements in this substance are 24.7, 34.8 and 40.5%, respectively.

Given: ω(K)=24.7%; ω(Mn)=34.8%; ω(O)=40.5%.

Find: compound formula.

Solution: for calculations, we select the mass of the compound, equal to 100 g, i.e. m=100 g. Masses of potassium, manganese and oxygen will be:

m (K) = m ω (K); m (K) \u003d 100 0.247 \u003d 24.7 g;

m (Mn) = m ω(Mn); m (Mn) = 100 0.348 = 34.8 g;

m (O) = m ω(O); m (O) \u003d 100 0.405 \u003d 40.5 g.

We determine the amount of substances of atomic potassium, manganese and oxygen:

ν (K) \u003d m (K) / M (K) \u003d 24.7 / 39 \u003d 0.63 mol

ν (Mn) \u003d m (Mn) / M (Mn) \u003d 34.8 / 55 \u003d 0.63 mol

ν (O) \u003d m (O) / M (O) \u003d 40.5 / 16 \u003d 2.5 mol

We find the ratio of the amounts of substances:

ν(K) : ν(Mn) : ν(O) = 0.63: 0.63: 2.5.

Dividing the right side of the equation by a smaller number (0.63) we get:

ν(K) : ν(Mn) : ν(O) = 1: 1: 4.

Therefore, the simplest formula of the KMnO 4 compound.

6. During the combustion of 1.3 g of the substance, 4.4 g of carbon monoxide (IV) and 0.9 g of water were formed. Find the molecular formula substance if its hydrogen density is 39.

Given: m(in-va) \u003d 1.3 g; m(CO 2)=4.4 g; m(H 2 O)=0.9 g; D H2 \u003d 39.

Find: the formula of the substance.

Solution: Assume that the substance you are looking for contains carbon, hydrogen and oxygen, because during its combustion, CO 2 and H 2 O were formed. Then it is necessary to find the amounts of substances CO 2 and H 2 O in order to determine the amounts of substances of atomic carbon, hydrogen and oxygen.

ν (CO 2) \u003d m (CO 2) / M (CO 2) \u003d 4.4 / 44 \u003d 0.1 mol;

ν (H 2 O) \u003d m (H 2 O) / M (H 2 O) \u003d 0.9 / 18 \u003d 0.05 mol.

We determine the amount of substances of atomic carbon and hydrogen:

ν(C)= ν(CO 2); v(C)=0.1 mol;

ν(H)= 2 ν(H 2 O); ν (H) \u003d 2 0.05 \u003d 0.1 mol.

Therefore, the masses of carbon and hydrogen will be equal:

m(C) = ν(C) M(C) = 0.1 12 = 1.2 g;

m (H) \u003d ν (H) M (H) \u003d 0.1 1 \u003d 0.1 g.

We determine the qualitative composition of the substance:

m (in-va) \u003d m (C) + m (H) \u003d 1.2 + 0.1 \u003d 1.3 g.

Consequently, the substance consists only of carbon and hydrogen (see the condition of the problem). Let us now determine its molecular weight, based on the given in the condition tasks density of a substance with respect to hydrogen.

M (in-va) \u003d 2 D H2 \u003d 2 39 \u003d 78 g / mol.

ν(C) : ν(H) = 0.1: 0.1

Dividing the right side of the equation by the number 0.1, we get:

ν(C) : ν(H) = 1: 1

Let's take the number of carbon (or hydrogen) atoms as "x", then, multiplying "x" by the atomic masses of carbon and hydrogen and equating this amount to the molecular weight of the substance, we solve the equation:

12x + x \u003d 78. Hence x \u003d 6. Therefore, the formula of the substance C 6 H 6 is benzene.

Molar volume of gases. Laws of ideal gases. Volume fraction.

The molar volume of a gas is equal to the ratio of the volume of gas to the amount of substance of this gas, i.e.

Vm = V(X)/ ν(x),

where V m is the molar volume of gas - a constant value for any gas under given conditions; V(X) is the volume of gas X; ν(x) - the amount of gas substance X. The molar volume of gases under normal conditions (normal pressure p n \u003d 101 325 Pa ≈ 101.3 kPa and temperature Tn \u003d 273.15 K ≈ 273 K) is V m \u003d 22.4 l /mol.

In calculations involving gases, it is often necessary to switch from these conditions to normal conditions or vice versa. In this case, it is convenient to use the formula following from the combined gas law of Boyle-Mariotte and Gay-Lussac:

──── = ─── (3)

Where p is pressure; V is the volume; T is the temperature in the Kelvin scale; the index "n" indicates normal conditions.

The composition of gas mixtures is often expressed using a volume fraction - the ratio of the volume of a given component to the total volume of the system, i.e.

where φ(X) is the volume fraction of the X component; V(X) is the volume of the X component; V is the volume of the system. The volume fraction is a dimensionless quantity, it is expressed in fractions of a unit or as a percentage.

7. What volume takes at a temperature of 20 ° C and a pressure of 250 kPa ammonia weighing 51 g?

Given: m(NH 3)=51 g; p=250 kPa; t=20°C.

Find: V(NH 3) \u003d?

Solution: determine the amount of ammonia substance:

ν (NH 3) \u003d m (NH 3) / M (NH 3) \u003d 51/17 \u003d 3 mol.

The volume of ammonia under normal conditions is:

V (NH 3) \u003d V m ν (NH 3) \u003d 22.4 3 \u003d 67.2 l.

Using formula (3), we bring the volume of ammonia to these conditions [temperature T \u003d (273 + 20) K \u003d 293 K]:

p n TV n (NH 3) 101.3 293 67.2

V (NH 3) \u003d ──────── \u003d ────────── \u003d 29.2 l.

8. Determine volume, which will take under normal conditions a gas mixture containing hydrogen, weighing 1.4 g and nitrogen, weighing 5.6 g.

Given: m(N 2)=5.6 g; m(H2)=1.4; well.

Find: V(mixture)=?

Solution: find the amount of substance hydrogen and nitrogen:

ν (N 2) \u003d m (N 2) / M (N 2) \u003d 5.6 / 28 \u003d 0.2 mol

ν (H 2) \u003d m (H 2) / M (H 2) \u003d 1.4 / 2 \u003d 0.7 mol

Since under normal conditions these gases do not interact with each other, the volume of the gas mixture will be equal to the sum of the volumes of gases, i.e.

V (mixtures) \u003d V (N 2) + V (H 2) \u003d V m ν (N 2) + V m ν (H 2) \u003d 22.4 0.2 + 22.4 0.7 \u003d 20.16 l.

Calculations by chemical equations

Calculations according to chemical equations (stoichiometric calculations) are based on the law of conservation of the mass of substances. However, in real chemical processes, due to an incomplete reaction and various losses of substances, the mass of the resulting products is often less than that which should be formed in accordance with the law of conservation of the mass of substances. The yield of the reaction product (or the mass fraction of the yield) is the ratio of the mass of the actually obtained product, expressed as a percentage, to its mass, which should be formed in accordance with the theoretical calculation, i.e.

η = /m(X) (4)

Where η is the product yield, %; m p (X) - the mass of the product X obtained in the real process; m(X) is the calculated mass of substance X.

In those tasks where the product yield is not specified, it is assumed that it is quantitative (theoretical), i.e. η=100%.

9. What mass of phosphorus should be burned for getting phosphorus oxide (V) weighing 7.1 g?

Given: m(P 2 O 5) \u003d 7.1 g.

Find: m(P) =?

Solution: we write the equation for the combustion reaction of phosphorus and arrange the stoichiometric coefficients.

4P+ 5O 2 = 2P 2 O 5

We determine the amount of substance P 2 O 5 obtained in the reaction.

ν (P 2 O 5) \u003d m (P 2 O 5) / M (P 2 O 5) \u003d 7.1 / 142 \u003d 0.05 mol.

It follows from the reaction equation that ν (P 2 O 5) \u003d 2 ν (P), therefore, the amount of phosphorus substance required in the reaction is:

ν (P 2 O 5) \u003d 2 ν (P) \u003d 2 0.05 \u003d 0.1 mol.

From here we find the mass of phosphorus:

m(Р) = ν(Р) М(Р) = 0.1 31 = 3.1 g.

10. Magnesium weighing 6 g and zinc weighing 6.5 g were dissolved in an excess of hydrochloric acid. What volume hydrogen, measured under normal conditions, stand out wherein?

Given: m(Mg)=6 g; m(Zn)=6.5 g; well.

Find: V(H 2) =?

Solution: we write down the reaction equations for the interaction of magnesium and zinc with hydrochloric acid and arrange the stoichiometric coefficients.

Zn + 2 HCl \u003d ZnCl 2 + H 2

Mg + 2 HCl \u003d MgCl 2 + H 2

We determine the amount of magnesium and zinc substances that reacted with hydrochloric acid.

ν(Mg) \u003d m (Mg) / M (Mg) \u003d 6/24 \u003d 0.25 mol

ν (Zn) \u003d m (Zn) / M (Zn) \u003d 6.5 / 65 \u003d 0.1 mol.

It follows from the reaction equations that the amount of the substance of the metal and hydrogen are equal, i.e. ν (Mg) \u003d ν (H 2); ν (Zn) \u003d ν (H 2), we determine the amount of hydrogen resulting from two reactions:

ν (Н 2) \u003d ν (Mg) + ν (Zn) \u003d 0.25 + 0.1 \u003d 0.35 mol.

We calculate the volume of hydrogen released as a result of the reaction:

V (H 2) \u003d V m ν (H 2) \u003d 22.4 0.35 \u003d 7.84 l.

11. When passing hydrogen sulfide with a volume of 2.8 liters (normal conditions) through an excess of copper (II) sulfate solution, a precipitate was formed weighing 11.4 g. Determine the exit reaction product.

Given: V(H 2 S)=2.8 l; m(precipitate)= 11.4 g; well.

Find: η =?

Solution: we write the reaction equation for the interaction of hydrogen sulfide and copper (II) sulfate.

H 2 S + CuSO 4 \u003d CuS ↓ + H 2 SO 4

Determine the amount of hydrogen sulfide substance involved in the reaction.

ν (H 2 S) \u003d V (H 2 S) / V m \u003d 2.8 / 22.4 \u003d 0.125 mol.

It follows from the reaction equation that ν (H 2 S) \u003d ν (СuS) \u003d 0.125 mol. So you can find the theoretical mass of CuS.

m(CuS) \u003d ν (CuS) M (CuS) \u003d 0.125 96 \u003d 12 g.

Now we determine the product yield using formula (4):

η = /m(X)= 11.4 100/ 12 = 95%.

12. What weight ammonium chloride is formed by the interaction of hydrogen chloride weighing 7.3 g with ammonia weighing 5.1 g? What gas will be left in excess? Determine the mass of the excess.

Given: m(HCl)=7.3 g; m(NH 3) \u003d 5.1 g.

Find: m(NH 4 Cl) =? m(excess) =?

Solution: write the reaction equation.

HCl + NH 3 \u003d NH 4 Cl

This task is for "excess" and "deficiency". We calculate the amount of hydrogen chloride and ammonia and determine which gas is in excess.

ν(HCl) \u003d m (HCl) / M (HCl) \u003d 7.3 / 36.5 \u003d 0.2 mol;

ν (NH 3) \u003d m (NH 3) / M (NH 3) \u003d 5.1 / 17 \u003d 0.3 mol.

Ammonia is in excess, so the calculation is based on the deficiency, i.e. by hydrogen chloride. It follows from the reaction equation that ν (HCl) \u003d ν (NH 4 Cl) \u003d 0.2 mol. Determine the mass of ammonium chloride.

m (NH 4 Cl) \u003d ν (NH 4 Cl) M (NH 4 Cl) \u003d 0.2 53.5 \u003d 10.7 g.

We determined that ammonia is in excess (according to the amount of substance, the excess is 0.1 mol). Calculate the mass of excess ammonia.

m (NH 3) \u003d ν (NH 3) M (NH 3) \u003d 0.1 17 \u003d 1.7 g.

13. Technical calcium carbide weighing 20 g was treated with excess water, obtaining acetylene, passing through which through an excess of bromine water formed 1,1,2,2-tetrabromoethane weighing 86.5 g. Determine mass fraction SaS 2 in technical carbide.

Given: m = 20 g; m(C 2 H 2 Br 4) \u003d 86.5 g.

Find: ω (CaC 2) =?

Solution: we write down the equations of interaction of calcium carbide with water and acetylene with bromine water and arrange the stoichiometric coefficients.

CaC 2 +2 H 2 O \u003d Ca (OH) 2 + C 2 H 2

C 2 H 2 +2 Br 2 \u003d C 2 H 2 Br 4

Find the amount of substance tetrabromoethane.

ν (C 2 H 2 Br 4) \u003d m (C 2 H 2 Br 4) / M (C 2 H 2 Br 4) \u003d 86.5 / 346 \u003d 0.25 mol.

It follows from the reaction equations that ν (C 2 H 2 Br 4) \u003d ν (C 2 H 2) \u003d ν (CaC 2) \u003d 0.25 mol. From here we can find the mass of pure calcium carbide (without impurities).

m (CaC 2) \u003d ν (CaC 2) M (CaC 2) \u003d 0.25 64 \u003d 16 g.

We determine the mass fraction of CaC 2 in technical carbide.

ω (CaC 2) \u003d m (CaC 2) / m \u003d 16/20 \u003d 0.8 \u003d 80%.

Solutions. Mass fraction of the solution component

14. Sulfur weighing 1.8 g was dissolved in benzene with a volume of 170 ml. The density of benzene is 0.88 g / ml. Determine mass fraction sulfur in solution.

Given: V(C 6 H 6) =170 ml; m(S) = 1.8 g; ρ(C 6 C 6)=0.88 g/ml.

Find: ω(S) =?

Solution: to find the mass fraction of sulfur in the solution, it is necessary to calculate the mass of the solution. Determine the mass of benzene.

m (C 6 C 6) \u003d ρ (C 6 C 6) V (C 6 H 6) \u003d 0.88 170 \u003d 149.6 g.

Find the total mass of the solution.

m (solution) \u003d m (C 6 C 6) + m (S) \u003d 149.6 + 1.8 \u003d 151.4 g.

Calculate the mass fraction of sulfur.

ω(S) =m(S)/m=1.8 /151.4 = 0.0119 = 1.19%.

15. Iron sulfate FeSO 4 7H 2 O weighing 3.5 g was dissolved in water weighing 40 g. Determine mass fraction of iron sulfate (II) in the resulting solution.

Given: m(H 2 O)=40 g; m (FeSO 4 7H 2 O) \u003d 3.5 g.

Find: ω(FeSO 4) =?

Solution: find the mass of FeSO 4 contained in FeSO 4 7H 2 O. To do this, calculate the amount of substance FeSO 4 7H 2 O.

ν (FeSO 4 7H 2 O) \u003d m (FeSO 4 7H 2 O) / M (FeSO 4 7H 2 O) \u003d 3.5 / 278 \u003d 0.0125 mol

From the formula of ferrous sulfate it follows that ν (FeSO 4) \u003d ν (FeSO 4 7H 2 O) \u003d 0.0125 mol. Calculate the mass of FeSO 4:

m (FeSO 4) \u003d ν (FeSO 4) M (FeSO 4) \u003d 0.0125 152 \u003d 1.91 g.

Given that the mass of the solution consists of the mass of ferrous sulfate (3.5 g) and the mass of water (40 g), we calculate the mass fraction of ferrous sulfate in the solution.

ω (FeSO 4) \u003d m (FeSO 4) / m \u003d 1.91 / 43.5 \u003d 0.044 \u003d 4.4%.

Tasks for independent solution

50 g of methyl iodide in hexane were treated with sodium metal, and 1.12 liters of gas, measured under normal conditions, were released. Determine the mass fraction of methyl iodide in the solution. Answer: 28,4%.
Some alcohol was oxidized to form a monobasic carboxylic acid. When burning 13.2 g of this acid, carbon dioxide was obtained, for the complete neutralization of which it took 192 ml of a KOH solution with a mass fraction of 28%. The density of the KOH solution is 1.25 g/ml. Determine the formula for alcohol. Answer: butanol.
The gas obtained by the interaction of 9.52 g of copper with 50 ml of an 81% solution of nitric acid, with a density of 1.45 g / ml, was passed through 150 ml of a 20% NaOH solution with a density of 1.22 g / ml. Determine the mass fractions of dissolved substances. Answer: 12.5% NaOH; 6.48% NaNO 3 ; 5.26% NaNO 2 .
Determine the volume of gases released during the explosion of 10 g of nitroglycerin. Answer: 7.15 l.
A sample of organic matter weighing 4.3 g was burned in oxygen. The reaction products are carbon monoxide (IV) with a volume of 6.72 liters (normal conditions) and water with a mass of 6.3 g. The vapor density of the starting substance for hydrogen is 43. Determine the formula of the substance. Answer: C 6 H 14 .

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!