Tasks for determining the quantitative composition of the mixture. Chemical properties of metals. material for preparing for the exam (GIA) in chemistry (Grade 11) on the topic. Tasks for mixtures and alloys of metals Necessary theoretical information

Task number 1

When a mixture of copper and copper (II) oxide is dissolved in concentrated nitric acid 18.4 g of brown gas were released and 470 g of a solution with a mass fraction of salt of 20% was obtained. Determine the mass fraction of copper oxide in the initial mixture.

Answer: 65.22%

Explanation:

When copper and copper (II) oxide are dissolved, the following reactions occur:

Cu + 4HNO 3 (conc.) = Cu(NO 3) 2 + 2NO 2 + 2H 2 O (I)

CuO + 2HNO 3 (conc.) = Cu (NO 3) 2 + H 2 O (II)

Brown gas NO 2 is released only when copper interacts with concentrated nitric acid. Let's find its quantity by the formula:

where m is the mass of the substance [g], M is molar mass substances [g/mol].

M(NO 2) = 46 g/mol

ν (NO 2) \u003d m (NO 2) / M (NO 2) \u003d 18.4 g / 46 g / mol \u003d 0.4 mol.

According to the reaction condition (I):

ν I (Cu(NO 3) 2) = ν(Cu) = 1/2ν(NO 2),

Consequently,

ν I (Cu (NO 3) 2) \u003d ν (Cu) \u003d 0.4 mol / 2 \u003d 0.2 mol.

The mass of copper nitrate formed due to the interaction of copper with concentrated nitric acid:

m I (Cu(NO 3) 2) = M(Cu(NO 3) 2) . ν(Cu(NO 3) 2) = 188 g/mol. 0.2 mol = 37.6 g

The mass of copper reacted is:

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

The total mass of copper nitrate contained in the solution is:

m total (Cu(NO 3) 2) = w(Cu(NO 3) 2) . m(r−ra)/100% = 20%. 470g / 100% = 94g

The mass of copper nitrate formed by the interaction of copper oxide with concentrated nitric acid (II):

m II (Cu (NO 3) 2) \u003d m (mixture) - m I (Cu (NO 3) 2) \u003d 94 g - 37.6 g \u003d 56.4 g.

The amount of copper nitrate formed by the interaction of copper oxide with concentrated nitric acid (II):

ν II (Cu (NO 3) 2) \u003d m II (Cu (NO 3) 2) / M II (Cu (NO 3) 2) \u003d 56.4 g / 188 g / mol \u003d 0.3 mol

According to the reaction condition:

(II) ν II (Cu (NO 3) 2) \u003d ν (CuO) \u003d 0.3 mol.

m(CuO) = M(CuO) . ν(CuO) = 80 g/mol. 0.3 mol = 24 g

m(mixtures) = m(CuO) + m(Cu) = 24 g + 12.8 g = 36.8 g

The mass fraction of copper oxide in the mixture is:

w(CuO)% = m(CuO)/m(mixtures) . 100% = 24g / 36.8g 100% = 65.22%

Task number 2

Determine the mass fraction of sodium carbonate in a solution obtained by boiling 150 g of an 8.4% sodium bicarbonate solution. What volume of a 15.6% solution of barium chloride (density 1.11 g / ml) will react with the resulting sodium carbonate? Water evaporation can be neglected.

Answer: 5.42%, 90 ml

Explanation:

The decomposition of sodium bicarbonate in solution is described by the reaction:

2NaHCO 3 → Na 2 CO 3 + CO 2 + H 2 O

The mass fraction of a solute is calculated by the formula:

w(w-in)% \u003d m(w-w) / m(r−ra) . one hundred%,

where m (in-va) is the mass of the dissolved substance, m (p−ra) is the mass of the solution).

Calculate the mass of dissolved sodium bicarbonate (NaHCO 3):

M (NaHCO 3) \u003d m (p−ra) . w(NaHCO 3) / 100% \u003d 150 g. 8.4% / 100% \u003d 12.6 g

ν (NaHCO 3) \u003d m (NaHCO 3) / M (NaHCO 3) \u003d 12.6 g / 84 g / mol \u003d 0.15 mol

According to the reaction equation:

ν (Na 2 CO 3) \u003d ν (CO 2) \u003d 1 / 2ν (NaHCO 3) \u003d 0.15 mol / 2 \u003d 0.075 mol

m(Na 2 CO 3) = M(Na 2 CO 3) . ν(Na 2 CO 3) = 106 g/mol. 0.075 mol = 7.95 g

m(CO 2) = M(CO 2) . ν(CO 2) \u003d 44 g / mol. 0.075 mol = 3.3 g

Since the evaporation of water can be neglected, we find the mass of sodium bicarbonate formed after decomposition by subtracting the mass of carbon dioxide from the mass of the initial solution:

m (r-ra) \u003d m (ref. r-ra) - m (CO 2) \u003d 150 g - 3.3 g \u003d 146.7 g

w (Na 2 CO 3)% \u003d m (Na 2 CO 3) / m (p−ra) . 100% = 7.95g/146.7g 100% = 5.42%

The reaction of interaction of solutions of barium chloride and sodium carbonate is described by the equation:

BaCl 2 + Na 2 CO 3 → BaCO 3 ↓ + 2NaCl

According to the reaction equation:

ν (BaCl 2) \u003d ν (Na 2 CO 3) \u003d 0.075 mol, therefore

m (BaCl 2) \u003d M (BaCl 2) . ν(BaCl 2) = 208 g/mol. 0.075 mol = 15.6 g

The mass of a barium chloride solution is:

m(p−ra BaCl 2) = m(BaCl 2)/w(BaCl 2) . 100% = 15.6 g / 15.6% . 100% = 100 g

The volume of the solution is calculated by the formula:

V(r-ra) = m(r-ra)/ρ(r-ra), where ρ(r-ra) is the density of the solution.

V(r-ra BaCl 2) \u003d m(r-ra BaCl 2) / ρ(r-ra)

V (solution BaCl 2) \u003d 100 g / 1.11 g / ml \u003d 90 ml

Task number 3

In what mass ratios should 10% sodium hydroxide and sulfuric acid solutions be mixed to obtain a neutral sodium sulfate solution? What is the mass fraction of salt in this solution?

Answer: m (p-ra H 2 SO 4) / m (p-ra NaOH) \u003d 1.225; w(Na 2 SO 4) = 8%Explanation:

The interaction of solutions of sodium hydroxide and sulfuric acid with the formation of an average salt (neutral solution) proceeds according to the scheme:

2NaOH + H 2 SO 4 → Na 2 SO 4 + H 2 O

The masses of sodium hydroxide and sulfuric acid are equal:

m(NaOH) = 0.1m(p−ra NaOH); m (H 2 SO 4) \u003d 0.1m (H 2 SO 4)

Accordingly, the amounts of sodium hydroxide and sulfuric acid are equal:

ν(NaOH) = m(NaOH)/M(NaOH) = 0.1m(p−ra NaOH)/40 g/mol

ν (H 2 SO 4) \u003d m (H 2 SO 4) / M (H 2 SO 4) \u003d 0.1m (p-ra H 2 SO 4) / 98 g / mol

According to the reaction equation ν (H 2 SO 4) = 1/2ν (NaOH), therefore

0.1m (solution H 2 SO 4) / 98 = ½. 0.1m(NaOH solution)/40

m (solution H 2 SO 4) / m (solution NaOH) \u003d 98/2/40 \u003d 1.225

According to the reaction equation

ν(H 2 SO 4) = ν(Na 2 SO 4), so

ν (Na 2 SO 4) \u003d 0.1m (solution H 2 SO 4) / 98 g / mol

m (Na 2 SO 4) \u003d 0.1m (solution H 2 SO 4) / 98 g / mol. 142 g / mol \u003d 0.145 m (solution H 2 SO 4)

m (p-ra H 2 SO 4) = 1.225. m (p-ra NaOH)

The mass of the resulting sodium sulfate solution is:

m (r-ra Na 2 SO 4) \u003d m (r-ra H 2 SO 4) + m (r-ra NaOH) \u003d 2.225 m (r-ra NaOH) \u003d 2.225 (r-ra H 2 SO 4) / 1.225

w (Na 2 SO 4)% \u003d 0.145 m (p-ra H 2 SO 4). 1.225 / 2.225 m (solution H 2 SO 4) . 100% = 8.0%

Task number 4

How many liters of chlorine (N.O.) will be released if 26.1 g of manganese (IV) oxide are added to 200 ml of 35% hydrochloric acid (density 1.17 g / ml) when heated? How many grams of sodium hydroxide in a cold solution will react with this amount of chlorine?

Answer: V (Cl 2) \u003d 6.72 l; m(NaOH) = 24 g

Explanation:

When hydrochloric acid is added to manganese (IV) oxide, the reaction proceeds:

MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2 O

Calculate the masses of hydrogen chloride and the mass of hydrochloric acid in solution:

m(p-ra 35% HCl) = V(p-ra HCl) . ρ(solution HCl) = 200 ml. 1.17 g/ml = 234 g

m(HCl) = M(HCl) . w(HCl)%/100% = 234 g 35%/100% = 81.9 g

Hence the amount of HCl:

ν(HCl) = m(HCl)/M(HCl) = 81.9 g/36.5 g/mol = 2.244 mol

Amount of manganese (IV) oxide:

ν (MnO 2) \u003d m (MnO 2) / M (MnO 2) \u003d 26.1 g / 87 g / mol \u003d 0.3 mol

According to the reaction equation ν (MnO 2) = ¼ ν (HCl), and according to the condition ν (MnO 2)< ¼ν(HCl), следовательно, MnO 2 – вещество в недостатке, полностью прореагирует с соляной кислотой.

According to the reaction equation ν (MnO 2) \u003d ν (Cl 2) \u003d 0.3 mol, therefore, the volume of released at n.o. chlorine is equal to:

V (Cl 2) \u003d V m. ν (Cl 2) \u003d 22.4 l / mol. 0.3 mol = 6.72 l

In a cold solution, alkali interacts with chlorine to form hypochlorite (NaClO) and sodium chloride (NaCl) (disproportionation reaction):

Cl 2 + 2NaOH → NaClO + NaCl + H 2 O

According to the reaction equation ν (Cl 2) \u003d ½ ν (NaOH), therefore ν (NaOH) \u003d 0.3 mol. 2 = 0.6 mol

The mass of sodium hydroxide reacted with chlorine in a cold solution is:

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

Task number 5

A mixture of copper and copper (II) oxide can react with 219 g of a 10% hydrochloric acid solution or 61.25 g of an 80% sulfuric acid solution. Determine the mass fraction of copper in the mixture.

Answer: 21%

Explanation:

Since copper is in the series of metal activities to the right of hydrogen, hydrochloric acid does not interact with it. HCl only reacts with copper(II) oxide to form salt and water:

Both copper and copper (II) oxide interact with an 80% solution of sulfuric acid:

Calculate the mass and amount of HCl substance that reacts with CuO:

hence the amount of CuO reacting with HCl:

Calculate the mass and amount of substance H 2 SO 4:

Since in the reaction of copper oxide with sulfuric acid

therefore, the reaction with Cu takes 0.2 mol of H 2 SO 4 and the amount of copper substance will be equal to:

The masses of substances are equal:

Therefore, the mass fraction of copper in the initial mixture

Task number 6

When interacting 5.6 g of potassium hydroxide with 5.0 g of ammonium chloride, ammonia was obtained. It was dissolved in 50 g of water. Determine the mass fraction of ammonia in the resulting solution. Determine the volume of a 10% nitric acid solution with a density of 1.06 g/ml that will be required to neutralize the ammonia.

Answer: ω (NH 3)% \u003d 3.1%; V (solution HNO 3) = 55.5 ml

Explanation:

As a result of the exchange reaction of potassium hydroxide with ammonium chloride, ammonia is released and potassium chloride and water are formed:

KOH + NH 4 Cl → KCl + NH 3 + H 2 O

Calculate the amount of reacting KOH and NH 4 Cl:

According to the reaction equation, KOH and NH 4 Cl react in equal amounts, and according to the condition of the problem, ν(KOH) > ν(NH 4 Cl). Therefore, ammonium chloride reacts completely, and 0.09346 moles of ammonia are formed.

The mass fraction of ammonia in the resulting solution is calculated by the formula:

Ammonia reacts with nitric acid according to the reaction:

HNO 3 + NH 3 → NH 4 NO 3

Therefore, an equal amount of nitric acid reacts with ammonia, i.e.

therefore, the mass of reacted nitric acid is:

The mass of the nitric acid solution is:

The volume of a 10% solution of nitric acid is:

Task number 7

Phosphorus (V) chloride weighing 4.17 g completely reacted with water. What volume of a potassium hydroxide solution with a mass fraction of 10% (density 1.07 g/ml) is required to completely neutralize the resulting solution?

Answer: V(solution KOH) = 84 ml

Explanation:

Phosphorus (V) chloride is completely hydrolyzed with the formation of phosphoric acid and hydrogen chloride:

The resulting orthophosphoric and hydrochloric acids are neutralized with a solution of potassium hydroxide, and medium salts are formed:

Let's calculate the amount of phosphorus (V) chloride substance that reacts with water:

Therefore, (according to the reaction equation) are formed

And (5 times greater than ν(PCl 5)).

To neutralize phosphoric acid, 0.06 mol KOH is needed, and to neutralize hydrochloric acid, 0.1 mol KOH.

Therefore, the total amount of alkali required to neutralize the acid solution is equal.

The mass of alkali is:

The mass of the alkali solution is:

The volume of a 10% alkali solution is:

Task number 8

When pouring 160 g of a 10% solution of barium nitrate and 50 g of a 11% solution of potassium chromate precipitated. Calculate the mass fraction of barium nitrate in the resulting solution.

Answer: ω (Ba (NO 3) 2) = 4.24%

Explanation:

In the exchange reaction, when barium nitrate interacts with potassium chromate, potassium nitrate is formed, and a precipitate forms - barium chromate:

Ba(NO 3) 2 + K 2 CrO 4 → BaCrO 4 ↓ + 2KNO 3

Let us calculate the masses and quantities of the reacting barium nitrate and potassium chromate:

According to the salt reaction equation, Ba (NO 3) 2 and K 2 CrO 4 react 1 to 1, according to the condition of the problem ν (Ba (NO 3) 2) > ν (K 2 CrO 4), therefore, potassium chromate is in short supply and completely reacts .

Calculate the amount and mass of barium nitrate remaining after the reaction:

Since barium chromate precipitates, the mass of the solution obtained by pouring solutions of barium nitrate and potassium chromate is:

Calculate the mass of barium chromate precipitated:

Then the mass of the solution is:

Find the mass fraction of barium nitrate in the resulting solution:

Task number 9

If a mixture of potassium and calcium chlorides is added to a solution of sodium carbonate, then 10 g of a precipitate is formed. If the same mixture is added to a solution of silver nitrate, then 57.4 g of a precipitate is formed. Determine the mass fraction of potassium chloride in the initial mixture.

Answer: ω(KCl) = 57.3%

Explanation:

Calcium chloride enters into an exchange reaction with sodium carbonate, as a result of which calcium carbonate precipitates:

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

Potassium chloride does not react with sodium carbonate.

Therefore, the mass of precipitated calcium carbonate is 10 g. Let's calculate its amount of substance:

Both chlorides enter into an exchange reaction with silver nitrate:

CaCl 2 + 2AgNO 3 → 2AgCl↓ + Ca(NO 3) 2 (II)

KCl + AgNO 3 → AgCl↓ + KNO 3 (III)

Calculate the total amount of silver chloride substance:

According to reaction (II), and therefore

By reaction (III), therefore,

The mass of potassium chloride in the initial mixture is:

Calculate the mass fraction of potassium chloride in the mixture:

Task number 10

A mixture of sodium and sodium oxide was dissolved in water. In this case, 4.48 l (n.o.) of gas were released and 240 g of a solution with a mass fraction of sodium hydroxide of 10% was formed. Determine the mass fraction of sodium in the initial mixture.

Answer: ω(KCl) = 59.74%

Explanation:

When sodium oxide reacts with water, an alkali is formed, and when sodium reacts with water, an alkali is formed, and hydrogen is released:

2Na + 2H 2 O → 2NaOH + H 2 (I)

Na 2 O + H 2 O → 2NaOH (II)

Let us calculate the amount of hydrogen substance released during the interaction of sodium with water:

According to the reaction equation (I) ν (Na) \u003d ν (NaOH) \u003d 2ν (H 2) \u003d 0.4 mol, therefore, the mass of sodium reacting with water and the mass of alkali formed as a result of reaction (I) are equal:

m(Na) = 23 g/mol 0.4 mol = 9.2 g and m I (NaOH) = 40 g/mol 0.4 mol = 16 g

Calculate the total mass of alkali in solution:

The mass and amount of the alkali substance formed by reaction (II) are equal to:

m II (NaOH) = 24 g - 16 g = 8 g

According to the reaction equation (II) ν(Na 2 O) = 1/2ν II (NaOH), therefore,

ν(Na 2 O) \u003d 0.2 mol / 2 \u003d 0.1 mol

m(Na 2 O) \u003d M (Na 2 O) ν (Na 2 O) \u003d 62 g / mol 0.1 mol \u003d 6.2 g

Let us calculate the mass of the initial mixture consisting of sodium and sodium oxide:

m(mixtures) = m(Na) + m(Na 2 O) = 9.2 g + 6.2 g = 15.4 g

The mass fraction of sodium in the mixture is:

Task number 11

A mixture of sodium carbonate and sodium hydrogen carbonate can react with 73 g of a 20% hydrochloric acid solution and 80 g of a 10% sodium hydroxide solution. Determine the mass fraction of sodium carbonate in the initial mixture.

Answer: ω (Na 2 CO 3) = 38.7%

Explanation:

The interaction of sodium carbonate and bicarbonate with hydrochloric acid proceeds according to the reactions:

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

NaHCO 3 + HCl → NaCl + CO 2 + H 2 O (II)

Sodium bicarbonate reacts with sodium hydroxide to form an average salt:

NaHCO 3 + NaOH → Na 2 CO 3 + H 2 O (III)

Let us calculate the mass and amount of the NaOH substance reacting according to reaction (III):

m(NaOH)=80 g*0.1 = 8 g

Therefore, 0.2 mol of NaHCO 3 reacts according to reaction (III), since ν III (NaHCO 3) = ν (NaOH).

Let us calculate the total mass and amount of HCl substance reacting according to reactions (I) and (II):

v(HCl) = m(HCl)/M(HCl) = 14.6 g/36.5 g/mol = 0.4 mol

According to the reaction equation (II) ν II (NaHCO 3) = ν II (HCl), therefore, 0.2 mol of HCl reacts with sodium bicarbonate. Then with sodium carbonate according to reaction (I) interacts

ν I (HCl) \u003d 0.4 mol - 0.2 mol \u003d 0.2 mol.

According to the reaction equation (I) ν (Na 2 CO 3) \u003d 1 / 2ν (HCl) \u003d 0.2 mol / 2 \u003d 0.1 mol.

Let us calculate the masses of bicarbonate and sodium carbonate in the initial mixture:

m (NaHCO 3) \u003d M (NaHCO 3) ν (NaHCO 3) \u003d 84 g / mol 0.2 mol \u003d 16.8 g

m (Na 2 CO 3) \u003d M (Na 2 CO 3) ν (Na 2 CO 3) \u003d 106 g / mol 0.1 mol \u003d 10.6 g

The mass of the initial mixture of salts is equal to:

m (mixtures) \u003d m (NaHCO 3) + m (Na 2 CO 3) \u003d 16.8 g + 10.6 g \u003d 27.4 g

The mass fraction of sodium carbonate is equal to:

Task number 12

A mixture of aluminum sulfide and aluminum was treated with water, and 6.72 liters (N.O.) of gas were released. If the same mixture is dissolved in an excess of sodium hydroxide solution, then 3.36 liters (N.O.) of gas will be released. Determine the mass fraction of aluminum in the initial mixture.

Answer: ω(Al) = 15.25%

Explanation:

When processing a mixture of aluminum sulfide and aluminum with water, only aluminum sulfide reacts:

Al 2 S 3 + 6H 2 O → 2Al(OH) 3 ↓ + 3H 2 S (I)

As a result of this interaction, hydrogen sulfide is formed, the amount of the substance of which is:

n (H 2 S) \u003d V (H 2 S) / V m \u003d 6.72 / 22.4 \u003d 0.3 mol,

The amount of hydrogen sulfide released depends only on the mass of aluminum sulfide (it does not depend on the amount of metallic aluminum). Therefore, based on equation I, we can conclude that:

n(Al 2 S 3) \u003d n(H 2 S) ⋅ 1/3, where 1 is the coefficient in front of Al 2 S 3, and 3 is the coefficient in front of H 2 S. Then:

n(Al 2 S 3) \u003d 0.3 ⋅ 1/3 \u003d 0.1 mol.

Consequently:

m(Al 2 S 3) = n(Al 2 S 3) ⋅ M(Al 2 S 3) = 0.1 ⋅ 150 g = 15 g;

When the same mixture interacts in an excess of sodium hydroxide solution, gas is released only when sodium hydroxide interacts with aluminum:

Al 2 S 3 + 8NaOH → 2Na + 3Na 2 S (II)

Let us calculate the amount of hydrogen released during the interaction of aluminum with a NaOH solution:

n III (H 2) \u003d V (H 2) / V m \u003d 3.36 / 22.4 \u003d 0.15 mol

According to the reaction equation (III) n(Al) \u003d n III (H 2) ⋅ 2/3, therefore, n (Al) \u003d 0.1 mol

The mass of aluminum is:

m(Al) = M(Al) n(Al) = 27 g/mol 0.1 mol = 2.7 g

Therefore, the mass of the initial mixture:

m (mixtures) \u003d m (Al 2 S 3) + m (Al) \u003d 15 g + 2.7 g \u003d 17.7 g.

The mass fraction of aluminum in the initial mixture is equal to:

ω(Al) = 100% ⋅ 2.7/17.7 = 15.25%

Task number 13

Oxygen weighing 22.4 g was consumed for the complete combustion of a mixture of carbon and silicon dioxide. What volume of a 20% potassium hydroxide solution (ρ = 1.173 g / ml) can react with the initial mixture if it is known that the mass fraction of carbon in it is 70 %?

Answer: 28.6 ml

Explanation:

Silicon dioxide does not react with oxygen. When carbon is burned, carbon dioxide is produced:

C + O 2 → CO 2

Let us calculate the amount of carbon substance involved in combustion:

According to the reaction equation ν (O 2) = ν (C), therefore, ν (C) = 0.7 mol

Calculate the mass of burned carbon:

m(C) = M(C) ν(C) = 12 g/mol 0.7 mol = 8.4 g

Calculate the mass of the initial mixture of carbon and silicon dioxide:

Calculate the mass and amount of silicon dioxide substance:

m (SiO 2) \u003d m (mixtures) - m (C) \u003d 12 g - 8.4 g \u003d 3.6 g

Only silicon dioxide interacts with alkali:

2KOH + SiO 2 → K 2 SiO 3 + H 2 O

As a result of this reaction, alkali is consumed in the amount of:

ν(KOH) = 2 0.06 mol = 0.12 mol

m(KOH) = M(KOH) ν(KOH) = 56 g/mol 0.12 mol = 6.72 g

Calculate the mass and volume of the KOH alkali solution:

Task number 14

A mixture of bicarbonate and potassium carbonate with a mass fraction of carbonate in it of 73.4% can react with 40 g of a 14% potassium hydroxide solution. The initial mixture was treated with an excess of sulfuric acid solution. What volume (N.S.) of gas is released in this case?

Answer: V (CO 2) \u003d 6.72 l

Explanation:

Only an acidic salt, potassium bicarbonate, can react with alkali:

KHCO 3 + KOH → K 2 CO 3 + H 2 O

Calculate the mass and amount of potassium hydroxide substance:

According to the reaction equation ν (KOH) = ν (KHCO 3), therefore, ν (KHCO 3) = 0.1 mol

The mass of potassium bicarbonate in the mixture is:

m (KHCO 3) \u003d M (KHCO 3) ν (KHCO 3) \u003d 100 g / mol 0.1 mol \u003d 10 g

Mass fraction of bicarbonate in a mixture of salts

ω (KHCO 3) \u003d 100% - ω (K 2 CO 3) \u003d 100% - 73.4% \u003d 26.6%

Calculate the mass of the mixture of salts:

Calculate the mass and amount of sodium carbonate substance:

When salts of potassium carbonate and bicarbonate interact with an excess of sulfuric acid, an acid salt is formed - potassium hydrogen sulfate:

KHCO 3 + H 2 SO 4 → KHSO 4 + CO 2 + H 2 O

K 2 CO 3 + 2H 2 SO 4 → 2KHSO 4 + CO 2 + H 2 O

The total amount and volume of carbon dioxide released are equal to:

ν (CO 2) \u003d 0.2 mol + 0.1 mol \u003d 0.3 mol

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

Task number 15

A mixture of magnesium and zinc filings was treated with an excess of dilute sulfuric acid, and 22.4 liters (n.o.) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 13.44 liters (N.O.) of hydrogen will be released. Calculate the mass fraction of magnesium in the initial mixture.

Answer: 19.75%

Explanation:

When processing a mixture of magnesium and zinc sawdust with dilute sulfuric acid, hydrogen is released:

Zn + H 2 SO 4 (diff.) → ZnSO 4 + H 2 (I)

Mg + H 2 SO 4 (razb.) → MgSO 4 + H 2 (II)

Only Zn (amphoteric metal) reacts with an excess of NaOH solution:

Zn + 2NaOH + 2H 2 O → Na 2 + H 2 (III)

Let us calculate the amount of hydrogen released by reaction (III):

According to the reaction equation (III) ν III (H 2) = ν (Zn), therefore, ν (Zn) = 0.6 mol

The mass of zinc is:

m(Zn) = M(Zn) ν(Zn) = 65 g/mol 0.6 mol = 39 g

According to the reaction equations (I) and (II) ν I (H 2) \u003d ν (Zn) and ν II (H 2) \u003d ν (Mg), the total amount of hydrogen released is:

Therefore, the amount of hydrogen released by reaction (II):

ν II (H 2) \u003d ν (Mg) \u003d 1 mol - 0.6 mol \u003d 0.4 mol

The mass of magnesium is:

m(Mg) = M(Mg) ν(Mg) = 24 g/mol 0.4 mol = 9.6 g

Calculate the mass of magnesium and zinc:

m(mixtures) = m(Mg) + m(Zn) = 9.6 g + 39 g = 48.6 g

The mass fraction of magnesium in the initial mixture is equal to:

Task number 16

In an excess of oxygen, 8 g of sulfur were burned. The resulting gas was passed through 200 g of 8% sodium hydroxide solution. Determine the mass fractions of salts in the resulting solution.

Answer: ω (NaHSO 3) ≈ 4.81%; ω(Na 2 SO 3) ≈ 8.75%

Explanation:

When sulfur is burned in excess oxygen, sulfur dioxide is formed:

S + O 2 → SO 2 (I)

Calculate the amount of substance of burned sulfur:

According to the reaction equation ν (S) = ν (SO 2), therefore, ν (SO 2) = 0.25 mol

The mass of the released sulfur dioxide is:

m(SO 2) \u003d M (SO 2) ν (SO 2) \u003d 64 g / mol 0.25 mol \u003d 16 g

Calculate the mass and amount of sodium hydroxide substance:

The reaction between alkali and NaOH proceeds in steps: first an acid salt is formed, then it turns into an average one:

NaOH + SO 2 → NaHSO 3 (II)

NaHSO 3 + NaOH → Na 2 SO 3 + H 2 O (III)

Since ν(NaOH) > ν(SO 2) and ν(NaOH)< 2ν(SO 2), следовательно, образуются средняя и кислая соли.

As a result of reaction (II), ν (NaHSO 3) \u003d 0.25 mol is formed, and ν (NaOH) \u003d 0.4 mol - 0.25 mol \u003d 0.15 mol remains.

When sodium hydrosulfate and alkali interact (reaction (III)) sodium sulfite is formed with the amount of substance ν (Na 2 SO 3) \u003d 0.15 mol, and ν (NaHSO 3) \u003d 0.25 mol - 0.15 mol \u003d 0 remains, 1 mol.

Calculate the masses of hydrosulfite and sodium sulfite:

m (NaHSO 3) \u003d M (NaHSO 3) ν (NaHSO 3) \u003d 104 g / mol 0.1 mol \u003d 10.4 g

m (Na 2 SO 3) \u003d M (Na 2 SO 3) ν (Na 2 SO 3) \u003d 126 g / mol 0.15 mol \u003d 18.9 g

The mass of the solution obtained by passing sulfur dioxide through an alkali solution is:

m(p−ra) \u003d m(p−ra NaOH) + m(SO 2) \u003d 200 g + 16 g \u003d 216 g

The mass fractions of salts in the resulting solution are equal to:

Task number 17

A mixture of aluminum and iron filings was treated with an excess of dilute hydrochloric acid, and 8.96 L (n.o.) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 6.72 liters (N.O.) of hydrogen will be released. Calculate the mass fraction of iron in the initial mixture.

Answer: ω(Fe) ≈ 50.91%

Explanation:

The reactions of aluminum and iron filings in an excess of dilute hydrochloric acid proceed with the release of hydrogen and the formation of salts:

Fe + 2HCl → FeCl 2 + H 2 (I)

2Al + 6HCl → 2AlCl 3 + 3H 2 (II)

Only aluminum filings react with an alkali solution, resulting in the formation of a complex salt and hydrogen:

2Al + 2NaOH + 6H 2 O → 2Na + 3H 2 (III)

Let us calculate the amount of hydrogen substance released by reaction (III):

Let us calculate the amount of hydrogen substance released by reactions (I) + (II):

According to the reaction equation (III) ν III (Al) \u003d 2 / 3 ν III (H 2), therefore, ν III (Al) \u003d 2/3 0.3 mol \u003d 0.2 mol.

The mass of aluminum filings is:

m(Al) = M(Al) ν(Al) = 27 g/mol 0.2 mol = 5.4 g

According to the reaction equation (II) ν II (Al) = 2/3ν II (H 2), therefore, ν II (H 2) = 3/2 0.2 mol = 0.3 mol

The amount of hydrogen substance released by reaction (I) is:

ν I (H 2) \u003d ν I + II (H 2) - ν II (H 2) \u003d 0.4 mol - 0.3 mol \u003d 0.1 mol.

According to the reaction equation (I) ν I (Fe) \u003d ν I (H 2), therefore, ν I (Fe) \u003d 0.1 mol

The mass of iron filings is:

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

The mass of iron and aluminum filings is equal to:

m (sawdust) \u003d m (Al) + m (Fe) \u003d 5.4 g + 5.6 g \u003d 11 g

The mass fraction of iron in the initial mixture is equal to.

Tasks for determining the quantitative composition of the mixture. Chemical properties metals.

1. A mixture of aluminum and iron filings was treated with an excess of dilute hydrochloric acid, and 8.96 L (n.o.) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 6.72 liters (N.O.) of hydrogen will be released. Calculate the mass fraction of iron in the initial mixture.

2. A mixture of magnesium and zinc filings was treated with an excess of dilute sulfuric acid, and 22.4 liters (n.o.) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 13.44 liters (N.O.) of hydrogen will be released. Calculate the mass fraction of magnesium in the initial mixture.

3. When a mixture of copper and copper(II) oxide was dissolved in concentrated nitric acid, 18.4 g of brown gas was released and 470 g of a solution with a mass fraction of salt of 20% was obtained. Determine the mass fraction of copper oxide in the initial mixture.

4. A mixture of aluminum sulfide and aluminum was treated with water, and 6.72 liters (N.O.) of gas were released. If the same mixture is dissolved in an excess of sodium hydroxide solution, then 3.36 liters (N.O.) of gas will be released. Determine the mass fraction of aluminum in the initial mixture.

5. If a mixture of potassium and calcium chlorides is added to a solution of sodium carbonate, then 10 g of a precipitate is formed. If the same mixture is added to a solution of silver nitrate, then 57.4 g of a precipitate is formed. Determine the mass fraction of potassium chloride in the initial mixture.

6. A mixture of copper and aluminum weighing 10 g was treated with 96% nitric acid, and 4.48 liters of gas (n.o.) were released. Determine the quantitative composition of the initial mixture and the mass fraction of aluminum in it.

7. A mixture of magnesium and magnesium oxide weighing 6.4 g was treated with a sufficient amount of dilute sulfuric acid, and 2.24 liters of gas (n.o.) were released. Determine the quantitative composition of the initial mixture and the mass fraction of magnesium oxide in it.

8. A mixture of copper and zinc weighing 40 g was treated with a concentrated solution of alkali. In this case, gas was released with a volume of 8.96 liters (n.o.). Calculate the mass fraction of copper in the initial mixture.

On the topic: methodological developments, presentations and notes

Lesson development contains a detailed summary of the lesson, slides for the lesson, workbook on the topic under study, instructive maps for the experiment and other didactic material....

Mixture problems are a very common type of problem in chemistry. They require a clear idea of ​​which of the substances enter into the reaction proposed in the problem and which do not.
We talk about a mixture when we have not one, but several substances (components) “poured” into one container. These substances should not interact with each other.

Typical misconceptions and errors that arise when solving problems on a mixture.

1. An attempt to write both substances in one reaction.
   It turns out something like this:
   "A mixture of calcium and barium oxides was dissolved in hydrochloric acid ..."
   The reaction equation is written as follows:
   CaO + BaO + 4HCl \u003d CaCl 2 + BaCl 2 + 2H 2 O.
   This is a mistake, because there can be any amount of each oxide in this mixture.
   And in the above equation, it is assumed that their equal amount.
2. The assumption that their molar ratio corresponds to the coefficients in the reaction equations.
   For example:
   Zn + 2HCl \u003d ZnCl 2 + H 2
   2Al + 6HCl = 2AlCl 3 + 3H 2
   The amount of zinc is taken as x, and the amount of aluminum is taken as 2x (according to the coefficient in the reaction equation). This is also incorrect. These numbers can be anything and they are not related to each other in any way.
3. Attempts to find the "amount of substance of a mixture" by dividing its mass by the sum of the molar masses of the components.
   This action doesn't make any sense at all. Each molar mass can only refer to a single substance.

Often in such problems the reaction of metals with acids is used. To solve such problems, it is necessary to know exactly which metals interact with which acids and which do not.

Necessary theoretical information.

Methods for expressing the composition of mixtures.

• Mass fraction of the component in the mixture is the ratio of the mass of the component to the mass of the entire mixture. Usually the mass fraction is expressed in %, but not necessarily.

  ω ["omega"] = m component / m mixture

• Mole fraction of a component in a mixture- the ratio of the number of moles (amount of substance) of the component to the total number of moles of all substances in the mixture. For example, if the mixture includes substances A, B and C, then:

  χ [“chi”] component A \u003d n component A / (n (A) + n (B) + n (C))

• Molar ratio of components. Sometimes in tasks for a mixture, the molar ratio of its components is indicated. For example:

  N component A: n component B = 2: 3

• Volume fraction of the component in the mixture (only for gases)- the ratio of the volume of substance A to the total volume of the entire gas mixture.

  φ ["phi"] = V component / V mixture

Electrochemical series of voltages of metals.

Li Rb K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb H Sb Bi Cu Hg Ag Pd Pt Au

Reactions of metals with acids.

1. With mineral acids, which include all soluble acids ( except for nitrogen and concentrated sulfuric, whose interaction with metals occurs in a special way), react only metals, in the electrochemical series of voltages located to (to the left of) hydrogen.
2. At the same time, metals that have several oxidation states (iron, chromium, manganese, cobalt) exhibit the lowest possible oxidation state - usually +2.
3. The interaction of metals with nitric acid leads to the formation, instead of hydrogen, of nitrogen reduction products, and with concentrated sulfuric acid— to the isolation of sulfur recovery products. Since a mixture of reduction products is actually formed, there is often a direct reference to a specific substance in the problem.

Nitric acid recovery products.

The more active the metal and the lower the acid concentration, the further nitrogen is reduced.
NO 2	NO	N2O	N 2	NH4NO3
Inactive metals (to the right of iron) + conc. acid Nonmetals + conc. acid	Inactive metals (to the right of iron) + dil. acid	Active metals (alkaline, alkaline earth, zinc) + conc. acid	Active metals (alkali, alkaline earth, zinc) + medium dilution acid	Active metals (alkaline, alkaline earth, zinc) + very dil. acid
Passivation: do not react with cold concentrated nitric acid: Al, Cr, Fe, Be, Co.
do not react with nitric acid at any concentration: Au, Pt, Pd.

Sulfuric acid recovery products.

Reactions of metals with water and alkalis.

1. Dissolve in water at room temperature only metals, which correspond to soluble bases (alkalis). This alkali metals(Li, Na, K, Rb, Cs), as well as Group IIA metals: Ca, Sr, Ba. This produces alkali and hydrogen. Boiling water can also dissolve magnesium.
2. Only amphoteric metals can dissolve in alkali: aluminum, zinc and tin. In this case, hydroxo complexes are formed and hydrogen is released.

Examples of problem solving.

Consider three examples of problems in which mixtures of metals react with hydrochloric acid:

Example 1When a mixture of copper and iron weighing 20 g was exposed to an excess of hydrochloric acid, 5.6 liters of gas (n.o.) were released. Determine the mass fractions of metals in the mixture.

In the first example, copper does not react with hydrochloric acid, that is, hydrogen is released when the acid reacts with iron. Thus, knowing the volume of hydrogen, we can immediately find the amount and mass of iron. And, accordingly, the mass fractions of substances in the mixture.

Example 1 solution.

1. Finding the amount of hydrogen:
   n \u003d V / V m \u003d 5.6 / 22.4 \u003d 0.25 mol.
2. According to the reaction equation:

   The amount of iron is also 0.25 mol. You can find its mass:
   m Fe \u003d 0.25 56 \u003d 14 g.

3. Now you can calculate the mass fractions of metals in the mixture:

   ω Fe \u003d m Fe / m of the whole mixture \u003d 14 / 20 \u003d 0.7 \u003d 70%

Answer: 70% iron, 30% copper.

Example 2Under the action of an excess of hydrochloric acid on a mixture of aluminum and iron weighing 11 g, 8.96 liters of gas (n.o.) were released. Determine the mass fractions of metals in the mixture.

In the second example, the reaction is both metal. Here, hydrogen is already released from the acid in both reactions. Therefore, direct calculation cannot be used here. In such cases, it is convenient to solve using a very simple system of equations, taking for x - the number of moles of one of the metals, and for y - the amount of substance of the second.

Example 2 solution.

1. Finding the amount of hydrogen:
   n \u003d V / V m \u003d 8.96 / 22.4 \u003d 0.4 mol.
2. Let the amount of aluminum be x mol, and iron y mol. Then we can express in terms of x and y the amount of hydrogen released:

   It is much more convenient to solve such systems by the subtraction method, multiplying the first equation by 18:
   27x + 18y = 7.2
   and subtracting the first equation from the second:

   (56 - 18)y \u003d 11 - 7.2
   y \u003d 3.8 / 38 \u003d 0.1 mol (Fe)
   x = 0.2 mol (Al)

3. Next, we find the masses of metals and their mass fractions in the mixture:

   M Fe = n M = 0.1 56 = 5.6 g
   m Al = 0.2 27 = 5.4 g
   ω Fe = m Fe / m mixture = 5.6 / 11 = 0.50909 (50.91%),

   respectively,
   ω Al \u003d 100% - 50.91% \u003d 49.09%

Answer: 50.91% iron, 49.09% aluminum.

Example 316 g of a mixture of zinc, aluminum and copper was treated with an excess of hydrochloric acid solution. In this case, 5.6 l of gas (n.o.) was released and 5 g of the substance did not dissolve. Determine the mass fractions of metals in the mixture.

In the third example, two metals react, but the third metal (copper) does not react. Therefore, the remainder of 5 g is the mass of copper. The amounts of the other two metals - zinc and aluminum (note that their total mass is 16 - 5 = 11 g) can be found using a system of equations, as in example No. 2.

Answer to Example 3: 56.25% zinc, 12.5% ​​aluminum, 31.25% copper.

The next three examples of tasks (No. 4, 5, 6) contain the reactions of metals with nitric and sulfuric acids. The main thing in such tasks is to correctly determine which metal will dissolve in it, and which will not.

Example 4A mixture of iron, aluminum and copper was treated with an excess of cold concentrated sulfuric acid. At the same time, part of the mixture dissolved, and 5.6 liters of gas (n.o.) were released. The remaining mixture was treated with an excess of sodium hydroxide solution. 3.36 liters of gas evolved and 3 g of undissolved residue remained. Determine the mass and composition of the initial mixture of metals.

In this example, remember that cold concentrated sulfuric acid does not react with iron and aluminum (passivation), but reacts with copper. In this case, sulfur oxide (IV) is released.
With alkali reacts only aluminum- amphoteric metal (in addition to aluminum, zinc and tin also dissolve in alkalis, and beryllium can still be dissolved in hot concentrated alkali).

Example 4 solution.

1. Only copper reacts with concentrated sulfuric acid, the number of moles of gas:
   n SO 2 \u003d V / Vm \u003d 5.6 / 22.4 \u003d 0.25 mol
2. Number of moles of hydrogen:
   n H 2 \u003d 3.36 / 22.4 \u003d 0.15 mol,
   the molar ratio of aluminum and hydrogen is 2:3 and, therefore,
   nAl \u003d 0.15 / 1.5 \u003d 0.1 mol.
   Aluminum weight:
   m Al \u003d n M \u003d 0.1 27 \u003d 2.7 g
3. The remainder is iron, weighing 3 g. You can find the mass of the mixture:
   m mixture \u003d 16 + 2.7 + 3 \u003d 21.7 g.
4. Mass fractions of metals:

   ω Cu \u003d m Cu / m mixture \u003d 16 / 21.7 \u003d 0.7373 (73.73%)
   ω Al = 2.7 / 21.7 = 0.1244 (12.44%)
   ω Fe = 13.83%

Answer: 73.73% copper, 12.44% aluminum, 13.83% iron.

Example 521.1 g of a mixture of zinc and aluminum was dissolved in 565 ml of a nitric acid solution containing 20 wt. % HNO 3 and having a density of 1.115 g/ml. The volume of released gas, which is a simple substance and the only product of the reduction of nitric acid, amounted to 2.912 l (n.o.). Determine the composition of the resulting solution in mass percent. (RCTU)

In the text of this problem, the product of nitrogen reduction is clearly indicated - "simple substance". Since nitric acid does not give hydrogen with metals, it is nitrogen. Both metals dissolved in acid.
The problem asks not the composition of the initial mixture of metals, but the composition of the solution obtained after the reactions. This makes the task more difficult.

Example 5 solution.

1. Determine the amount of gas substance:
   n N 2 \u003d V / Vm \u003d 2.912 / 22.4 \u003d 0.13 mol.
2. We determine the mass of the nitric acid solution, the mass and amount of the dissolved HNO3 substance:

   M solution \u003d ρ V \u003d 1.115 565 \u003d 630.3 g
   m HNO 3 \u003d ω m solution \u003d 0.2 630.3 \u003d 126.06 g
   n HNO 3 \u003d m / M \u003d 126.06 / 63 \u003d 2 mol

   Note that since the metals are completely dissolved, it means - just enough acid(these metals do not react with water). Accordingly, it will be necessary to check Is there too much acid?, and how much of it remains after the reaction in the resulting solution.

3. We compose the reaction equations ( do not forget about the electronic balance) and, for convenience of calculations, we take as 5x the amount of zinc, and for 10y the amount of aluminum. Then, in accordance with the coefficients in the equations, nitrogen in the first reaction will be x mol, and in the second - 3y mol:

   5x		x
   5Zn	+ 12HNO 3 = 5Zn(NO 3) 2 +	N 2	+ 6H2O

   Zn 0 − 2e = Zn 2+	|	5
   2N+5+10e=N2		1

   It is convenient to solve this system by multiplying the first equation by 90 and subtracting the first equation from the second.

   X \u003d 0.04, which means n Zn \u003d 0.04 5 \u003d 0.2 mol
   y \u003d 0.03, which means that n Al \u003d 0.03 10 \u003d 0.3 mol

   Let's check the mass of the mixture:
   0.2 65 + 0.3 27 \u003d 21.1 g.
   

4. Now let's move on to the composition of the solution. It will be convenient to rewrite the reactions again and write down over the reactions the amounts of all reacted and formed substances (except water):

   0,2	0,48	0,2	0,03
   5Zn	+ 12HNO 3 =	5Zn(NO 3) 2	+N2+	6H2O

   0,3	1,08	0,3	0,09
   10Al	+ 36HNO 3 =	10Al(NO 3) 3	+ 3N 2 +	18H2O

5. The next question is: did nitric acid remain in the solution and how much is left?
   According to the reaction equations, the amount of acid that reacted:
   n HNO 3 \u003d 0.48 + 1.08 \u003d 1.56 mol,
   those. the acid was in excess and you can calculate its remainder in solution:
   n HNO 3 rest. \u003d 2 - 1.56 \u003d 0.44 mol.
6. So in final solution contains:

   Zinc nitrate in an amount of 0.2 mol:
   m Zn (NO 3) 2 \u003d n M \u003d 0.2 189 \u003d 37.8 g
   aluminum nitrate in the amount of 0.3 mol:
   m Al (NO 3) 3 \u003d n M \u003d 0.3 213 \u003d 63.9 g
   an excess of nitric acid in an amount of 0.44 mol:
   m HNO 3 rest. = n M = 0.44 63 = 27.72 g

7. What is the mass of the final solution?
   Recall that the mass of the final solution consists of those components that we mixed (solutions and substances) minus those reaction products that left the solution (precipitates and gases):

   Then for our task:

   M new solution \u003d mass of acid solution + mass of metal alloy - mass of nitrogen
   m N 2 \u003d n M \u003d 28 (0.03 + 0.09) \u003d 3.36 g
   m new solution \u003d 630.3 + 21.1 - 3.36 \u003d 648.04 g

8. Now you can calculate the mass fractions of substances in the resulting solution:

   ωZn (NO 3) 2 \u003d m in-va / m solution \u003d 37.8 / 648.04 \u003d 0.0583
   ωAl (NO 3) 3 \u003d m in-va / m solution \u003d 63.9 / 648.04 \u003d 0.0986
   ω HNO 3 rest. \u003d m in-va / m solution \u003d 27.72 / 648.04 \u003d 0.0428

Answer: 5.83% zinc nitrate, 9.86% aluminum nitrate, 4.28% nitric acid.

Example 6When processing 17.4 g of a mixture of copper, iron and aluminum with an excess of concentrated nitric acid, 4.48 liters of gas (n.o.) were released, and when this mixture was exposed to the same mass of excess hydrochloric acid, 8.96 l of gas (n.o.). u.). Determine the composition of the initial mixture. (RCTU)

When solving this problem, we must remember, firstly, that concentrated nitric acid with an inactive metal (copper) gives NO 2, and iron and aluminum do not react with it. Hydrochloric acid, on the other hand, does not react with copper.

Answer for example 6: 36.8% copper, 32.2% iron, 31% aluminum.

Tasks for independent solution.

1. Simple problems with two mixture components.

1-1. A mixture of copper and aluminum weighing 20 g was treated with a 96% solution of nitric acid, and 8.96 liters of gas (n.a.) were released. Determine the mass fraction of aluminum in the mixture.

1-2. A mixture of copper and zinc weighing 10 g was treated with a concentrated alkali solution. In this case, 2.24 liters of gas (n.y.) were released. Calculate the mass fraction of zinc in the initial mixture.

1-3. A mixture of magnesium and magnesium oxide weighing 6.4 g was treated with a sufficient amount of dilute sulfuric acid. At the same time, 2.24 liters of gas (n.o.) were released. Find the mass fraction of magnesium in the mixture.

1-4. A mixture of zinc and zinc oxide weighing 3.08 g was dissolved in dilute sulfuric acid. Zinc sulfate weighing 6.44 g was obtained. Calculate the mass fraction of zinc in the initial mixture.

1-5. Under the action of a mixture of iron and zinc powders weighing 9.3 g on an excess of copper (II) chloride solution, 9.6 g of copper was formed. Determine the composition of the initial mixture.

1-6. What mass of a 20% hydrochloric acid solution will be required to completely dissolve 20 g of a mixture of zinc with zinc oxide, if hydrogen is released in the amount of 4.48 liters (n.o.)?

1-7. When dissolved in dilute nitric acid, 3.04 g of a mixture of iron and copper releases nitric oxide (II) with a volume of 0.896 l (n.o.). Determine the composition of the initial mixture.

1-8. When dissolving 1.11 g of a mixture of iron and aluminum filings in a 16% hydrochloric acid solution (ρ = 1.09 g/ml), 0.672 l of hydrogen (n.o.) was released. Find the mass fractions of metals in the mixture and determine the volume of hydrochloric acid consumed.

2. Tasks are more complex.

2-1. A mixture of calcium and aluminum weighing 18.8 g was calcined without access to air with an excess of graphite powder. The reaction product was treated with dilute hydrochloric acid, and 11.2 liters of gas (n.o.) were released. Determine the mass fractions of metals in the mixture.

2-2. To dissolve 1.26 g of an alloy of magnesium with aluminum, 35 ml of a 19.6% sulfuric acid solution (ρ = 1.1 g/ml) was used. The excess acid reacted with 28.6 ml of a 1.4 mol/L potassium hydrogen carbonate solution. Determine the mass fractions of metals in the alloy and the volume of gas (N.O.) released during the dissolution of the alloy.

2-3. When dissolving 27.2 g of a mixture of iron and iron oxide (II) in sulfuric acid and evaporating the solution to dryness, 111.2 g of ferrous sulfate, iron sulfate heptahydrate (II), was formed. Determine the quantitative composition of the initial mixture.

2-4. The interaction of iron weighing 28 g with chlorine formed a mixture of iron (II) and (III) chlorides weighing 77.7 g. Calculate the mass of iron (III) chloride in the resulting mixture.

2-5. What was the mass fraction of potassium in its mixture with lithium, if as a result of the treatment of this mixture with an excess of chlorine, a mixture was formed in which the mass fraction of potassium chloride was 80%?

2-6. After treatment with an excess of bromine of a mixture of potassium and magnesium with a total mass of 10.2 g, the mass of the resulting mixture of solids was 42.2 g. This mixture was treated with an excess of sodium hydroxide solution, after which the precipitate was separated and calcined to constant mass. Calculate the mass of the resulting residue.

2-7. A mixture of lithium and sodium with a total mass of 7.6 g was oxidized with an excess of oxygen, a total of 3.92 liters (n.o.) was consumed. The resulting mixture was dissolved in 80 g of a 24.5% sulfuric acid solution. Calculate the mass fractions of substances in the resulting solution.

2-8. An alloy of aluminum and silver was treated with an excess of a concentrated solution of nitric acid, the residue was dissolved in acetic acid. The volumes of gases released in both reactions, measured under the same conditions, turned out to be equal to each other. Calculate the mass fractions of metals in the alloy.

3. Three metals and complex tasks.

3-1. When processing 8.2 g of a mixture of copper, iron and aluminum with an excess of concentrated nitric acid, 2.24 liters of gas were released. The same volume of gas is also released when the same mixture of the same mass is treated with an excess of dilute sulfuric acid (N.O.). Determine the composition of the initial mixture in mass percent.

3-2. 14.7 g of a mixture of iron, copper and aluminum, interacting with an excess of dilute sulfuric acid, releases 5.6 liters of hydrogen (n.o.). Determine the composition of the mixture in mass percent if chlorination of the same sample of the mixture requires 8.96 liters of chlorine (n.o.).

3-3. Iron, zinc and aluminum filings are mixed in a molar ratio of 2:4:3 (in the order listed). 4.53 g of this mixture was treated with an excess of chlorine. The resulting mixture of chlorides was dissolved in 200 ml of water. Determine the concentration of substances in the resulting solution.

3-4. An alloy of copper, iron and zinc weighing 6 g (the masses of all components are equal) was placed in an 18.25% solution of hydrochloric acid weighing 160 g. Calculate the mass fractions of the substances in the resulting solution.

3-5. 13.8 g of a mixture consisting of silicon, aluminum and iron, was treated with excess sodium hydroxide while heating, while 11.2 liters of gas (n.o.) were released. When exposed to such a mass of a mixture of excess hydrochloric acid, 8.96 liters of gas (n.o.) are released. Determine the masses of substances in the initial mixture.

3-6. When a mixture of zinc, copper and iron was treated with an excess of a concentrated alkali solution, gas was released, and the mass of the undissolved residue turned out to be 2 times less than the mass of the initial mixture. This residue was treated with an excess of hydrochloric acid, and the volume of the evolved gas turned out to be equal to the volume of the gas evolved in the first case (the volumes were measured under the same conditions). Calculate the mass fractions of metals in the initial mixture.

3-7. There is a mixture of calcium, calcium oxide and calcium carbide with a molar ratio of components 3:2:5 (in the order listed). What is the minimum volume of water that can enter into chemical interaction with such a mixture of mass 55.2 g?

3-8. A mixture of chromium, zinc and silver with a total weight of 7.1 g was treated with dilute hydrochloric acid, the mass of the undissolved residue was 3.2 g. The mass of the formed precipitate turned out to be 12.65 g. Calculate the mass fractions of metals in the initial mixture.

Answers and comments to tasks for independent solution.

1-1. 36% (aluminum does not react with concentrated nitric acid);

1-2. 65% (only amphoteric metal, zinc, dissolves in alkali);

1-3. 37,5%;

3-1. 39% Cu, 3.4% Al;

3-2. 38.1% Fe, 43.5% Cu;

3-3. 1.53% FeCl 3 , 2.56% ZnCl 2 , 1.88% AlCl 3 (iron reacts with chlorine to the +3 oxidation state);

3-4. 2.77% FeCl 2, 2.565% ZnCl 2, 14.86% HCl (do not forget that copper does not react with hydrochloric acid, so its mass is not included in the mass of the new solution);

3-5. 2.8 g Si, 5.4 g Al, 5.6 g Fe (silicon is a non-metal, it reacts with an alkali solution, forming sodium silicate and hydrogen; it does not react with hydrochloric acid);

3-6. 6.9% Cu, 43.1% Fe, 50% Zn;

3-8. 45.1% Ag, 36.6% Cr, 18.3% Zn barium)

Tasks on the mixture (USE-2017, No. 33)

Tasks with a free answer are evaluated with a maximum of 4 points

A mixture of magnesium and magnesium oxide weighing 6.4 g was treated with a sufficient amount of dilute sulfuric acid. At the same time, 2.24 liters of gas (n.o.) were released. Find the mass fraction of magnesium in the mixture.

When dissolved in dilute nitric acid, 3.04 g of a mixture of iron and copper releases nitric oxide (II) with a volume of 0.986 l (n.o.). Determine the composition of the initial mixture.

The interaction of iron weighing 28 g with chlorine formed a mixture of iron (II) and (III) chlorides weighing 77.7 g. Calculate the mass of iron (III) chloride in the resulting mixture.

When processing 8.2 g of a mixture of copper, iron and aluminum with an excess of concentrated nitric acid, 2.24 liters of gas were released. The same volume of gas is also released when the same mixture of the same mass is treated with an excess of dilute sulfuric acid (N.O.). Determine the composition of the initial mixture in mass percent.

When a mixture of copper and copper oxide 2 was dissolved in concentrated nitric acid, 18.4 g of brown gas was released and 470 solutions were obtained with a mass fraction of salt of 20%.
Determine the mass fraction of copper oxide in the initial mixture.

Assignments with multiple choice answers are evaluated with a maximum of 2 points (USE 2017)

1. (USE №5) Establish a correspondence between the formula of the substance and the class / group to which this substance belongs

FORMULA OF SUBSTANCE CLASS/GROUP

A) CaH 2 1) salt-forming oxide

B) NaH 2 PO 4 2) non-salt-forming oxide

C) H 3 N 3) medium salt

D) SeO 3 4) acid

5) sour salt

6) Binary connection

2. (USE No. 7) What properties can the following substances exhibit?

FORMULA OF SUBSTANCE PROPERTIES

A) HNO 3 1) properties of bases

B) NaOH 2) properties of salts

C) Fe (OH) 2 3) properties of acids

D) Zn (NO 3) 2 4) properties of acids and salts

5) properties of salts and bases

3. (USE No. 7) Establish a correspondence between a solid substance and the products of its interaction with water:

A) BaBr 2 1) Ba (OH) 2 + HBr

B) Al 2 S 3 2) Ba (OH) 2 + NH 3

B) KH 2 3) Ba 2+ + 2Br -

D) Ba 3 N 2 4) H 2 + KOH

5) Al(OH) 3 + H 2 S

6) does not interact

4. (USE No. 5) Establish a correspondence between oxide and its corresponding hydroxide

A) N 2 O 3 1) HPO 3

B) SeO 2 2) CuOH

C) Cu 2 O 3) H 2 SeO 3

D) P 2 O 3 4) H 2 SeO 4

5. (USE No. 9, No. 17) The following scheme of transformations of substances is given:

Sr === X ==== NH 3 === Y

Determine which of the given substances are substances X and Y.

Write in the table the numbers of the selected substances under the corresponding letters.

6. (USE №22) Establish a correspondence between the salt formula and the electrolysis products of an aqueous solution of this salt, which were released on inert electrodes: for each position indicated by a letter, select the corresponding position indicated by a number.

SALT FORMULA ELECTROLYSIS PRODUCTS

A) Na 3 PO4 1) O 2, H 2

B) KF 2) O 2, Mg

C) MgBr 2 3) H 2, Mg

D) Mg (NO 3) 2 4) Na, O 2

Write in the table the selected numbers under the corresponding

7. (USE No. 27) Calculate the mass of copper sulfate (CuSO 4 * 5H 2 O) that must be dissolved in water to obtain 240 g of a 10% copper sulfate solution.

8. (USE №27) Two solutions were mixed, the mass of the first solution is 80 g, with a mass fraction of sodium sulfate 5%, the mass of the second solution is 40 g, with a mass fraction of sodium sulfate 16%. Determine the mass fraction of sodium sulfate in the newly obtained solution.

Answer: ___________________ g. (Write down the number to tenths.)

9. (USE 35 (5)) Using the electron balance method, write the reaction equation:

CH 3 -CH \u003d CH-CH 3 + KMnO 4 + H 2 O \u003d\u003d  CH3-CH (OH) -CH (OH) -CH 3 + MnO 2 + KOH