Municipal budgetary educational institution

"Average comprehensive school № 37

with in-depth study of individual subjects "

Vyborg, Leningrad region

"Solving computational problems of an increased level of complexity"

(materials for preparing for the exam)

chemistry teacher

Podkladova Lyubov Mikhailovna

2015

The statistics of the Unified State Examination show that about half of the students cope with half of the tasks. Analyzing the results of checking the results of the USE in chemistry for students of our school, I came to the conclusion that it is necessary to strengthen the work on solving calculation problems, so I chose methodical theme"Problem Solving increased complexity».

Tasks are a special type of tasks that require students to apply knowledge in compiling reaction equations, sometimes several, compiling a logical chain in carrying out calculations. As a result of the decision, new facts, information, values ​​of quantities should be obtained from a certain set of initial data. If the algorithm for completing a task is known in advance, it turns from a task into an exercise, the purpose of which is to turn skills into skills, bringing them to automatism. Therefore, in the first classes in preparing students for the exam, I remind you of the values ​​\u200b\u200band units of their measurement.

Value

Designation

Units

in different systems

g, mg, kg, t, ... * (1g \u003d 10 -3 kg)

l, ml, cm 3, m 3, ...

*(1ml \u003d 1cm 3, 1 m 3 \u003d 1000l)

Density

g/ml, kg/l, g/l,…

Relative atomic mass

Relative molecular mass

Molar mass

g/mol, …

Molar volume

Vm or Vm

l / mol, ... (at n.o. - 22.4 l / mol)

Amount of substance

mole, kmol, mlmol

Relative density of one gas over another

Mass fraction of a substance in a mixture or solution

Volume fraction of a substance in a mixture or solution

Molar concentration

mol/l

Product output from theoretically possible

Avogadro constant

N A

6.02 10 23 mol -1

Temperature

t0 or

Celsius

on the Kelvin scale

Pressure

Pa, kPa, atm., mm. rt. Art.

Universal gas constant

8.31 J/mol∙K

Normal conditions

t 0 \u003d 0 0 C or T \u003d 273K

P \u003d 101.3 kPa \u003d 1 atm \u003d 760 mm. rt. Art.

Then I propose an algorithm for solving problems, which I have been using for several years in my work.

"An algorithm for solving computational problems".

V(r-ra)V(r-ra)

↓ρ ∙ Vm/ ρ

m(r-ra)m(r-ra)

↓m∙ ω m/ ω

m(in-va)m(in-va)

↓ m/ MM∙ n

n 1 (in-va)-- by ur. districts.→ n 2 (in-va)→

V(gas) / V M ↓ n∙ V M

V 1 (gas)V 2 (gas)

Formulas used to solve problems.

n = m / Mn(gas) = V(gas) / V M n = N / N A

ρ = m / V

D = M 1(gas) / M 2(gas)

D(H 2 ) = M(gas) / 2 D(air) = M(gas) / 29

(M (H 2) \u003d 2 g / mol; M (air.) \u003d 29 g / mol)

ω = m(in-va) / m(mixtures or solutions)  = V(in-va) / V(mixtures or solutions)

 = m(pract.) / m(theor.)  = n(pract.) / n(theor.)  = V(pract.) / V(theor.)

C = n / V

M (gas mixtures) = V 1 (gas) ∙ M 1(gas) + V 2 (gas) ∙ M 2(gas) / V(gas mixtures)

The Mendeleev-Clapeyron equation:

P∙ V = n∙ R∙ T

For passing the exam, where the types of tasks are quite standard (No. 24, 25, 26), the student must first of all show knowledge of standard calculation algorithms, and only in task No. 39 can he meet a task with an undefined algorithm for him.

The classification of chemical problems of increased complexity is complicated by the fact that most of them are combined problems. I divided the calculation tasks into two groups.

1. Tasks without using reaction equations. Some state of matter or a complex system is described. Knowing some characteristics of this state, it is necessary to find others. An example would be tasks:

1.1 Calculations according to the formula of the substance, the characteristics of the portion of the substance

1.2 Calculations according to the characteristics of the composition of the mixture, solution.

Tasks are found in the Unified State Examination - No. 24. For students, the solution of such problems does not cause difficulties.

2. Tasks using one or more reaction equations. To solve them, in addition to the characteristics of substances, it is necessary to use the characteristics of processes. In the tasks of this group, the following types of tasks of increased complexity can be distinguished:

2.1 Formation of solutions.

1) What mass of sodium oxide must be dissolved in 33.8 ml of water to obtain a 4% sodium hydroxide solution.

Find:

m (Na 2 O)

Given:

V (H 2 O) = 33.8 ml

ω(NaOH) = 4%

ρ (H 2 O) \u003d 1 g / ml

M (NaOH) \u003d 40 g / mol

m (H 2 O) = 33.8 g

Na 2 O + H 2 O \u003d 2 NaOH

1 mol 2mol

Let the mass of Na 2 O = x.

n (Na 2 O) \u003d x / 62

n(NaOH) = x/31

m(NaOH) = 40x /31

m (solution) = 33.8 + x

0.04 = 40x /31 ∙ (33.8+x)

x \u003d 1.08, m (Na 2 O) \u003d 1.08 g

Answer: m (Na 2 O) \u003d 1.08 g

2) To 200 ml of sodium hydroxide solution (ρ \u003d 1.2 g / ml) with a mass fraction of alkali of 20% was added metallic sodium weighing 69 g.

What is the mass fraction of the substance in the resulting solution?

Find:

ω 2 (NaOH)

Given:

V (NaO H) solution = 200 ml

ρ (solution) = 1.2 g/ml

ω 1 (NaOH) \u003d 20%

m (Na) \u003d 69 g

M (Na) \u003d 23 g / mol

Metallic sodium interacts with water in an alkali solution.

2Na + 2H 2 O \u003d 2 NaOH + H 2

1 mol 2mol

m 1 (p-ra) = 200 ∙ 1.2 = 240 (g)

m 1 (NaOH) in-va \u003d 240 ∙ 0.2 = 48 (g)

n (Na) \u003d 69/23 \u003d 3 (mol)

n 2 (NaOH) \u003d 3 (mol)

m 2 (NaOH) \u003d 3 ∙ 40 = 120 (g)

m total (NaOH) \u003d 120 + 48 \u003d 168 (g)

n (H 2) \u003d 1.5 mol

m (H 2) \u003d 3 g

m (p-ra after p-tion) \u003d 240 + 69 - 3 \u003d 306 (g)

ω 2 (NaOH) \u003d 168 / 306 \u003d 0.55 (55%)

Answer: ω 2 (NaOH) \u003d 55%

3) What is the mass of selenium oxide (VI) should be added to 100 g of a 15% solution of selenic acid to double its mass fraction?

Find:

m (SeO 3)

Given:

m 1 (H 2 SeO 4) solution = 100 g

ω 1 (H 2 SeO 4) = 15%

ω 2 (H 2 SeO 4) = 30%

M (SeO 3) \u003d 127 g / mol

M (H 2 SeO 4) \u003d 145 g / mol

m 1 (H 2 SeO 4 ) = 15 g

SeO 3 + H 2 O \u003d H 2 SeO 4

1 mol 1mol

Let m (SeO 3) = x

n(SeO 3 ) = x/127 = 0.0079x

n 2 (H 2 SeO 4 ) = 0.0079x

m 2 (H 2 SeO 4 ) = 145 ∙ 0.079x = 1.1455x

m total . (H 2 SeO 4 ) = 1.1455x + 15

m 2 (r-ra) \u003d 100 + x

ω (NaOH) \u003d m (NaOH) / m (solution)

0.3 = (1.1455x + 1) / 100 + x

x = 17.8, m (SeO 3 ) = 17.8 g

Answer: m (SeO 3) = 17.8 g

2.2 Calculation by reaction equations when one of the substances is in excess /

1) To a solution containing 9.84 g of calcium nitrate was added a solution containing 9.84 g of sodium orthophosphate. The precipitate formed was filtered off and the filtrate was evaporated. Determine the masses of the reaction products and the composition of the dry residue in mass fractions after evaporation of the filtrate, assuming that anhydrous salts are formed.

Find:

ω (NaNO3)

ω (Na 3 PO 4)

Given:

m (Ca (NO 3) 2) \u003d 9.84 g

m (Na 3 PO 4) \u003d 9.84 g

M (Na 3 PO 4) = 164 g / mol

M (Ca (NO 3) 2) \u003d 164 g / mol

M (NaNO 3) \u003d 85 g / mol

M (Ca 3 (PO 4) 2) = 310 g / mol

2Na 3 PO 4 + 3 Сa (NO 3) 2 \u003d 6NaNO 3 + Ca 3 (PO 4) 2 ↓

2 mole 3 mole 6 mole 1 mole

n (Сa(NO 3 ) 2 ) total = n (Na 3 PO 4 ) total. = 9.84/164 =

Ca (NO 3) 2 0.06 / 3< 0,06/2 Na 3 PO 4

Na 3 PO 4 is taken in excess,

we carry out calculations for n (Сa (NO 3) 2).

n (Ca 3 (PO 4) 2) = 0.02 mol

m (Ca 3 (PO 4) 2) \u003d 310 ∙ 0.02 \u003d 6.2 (g)

n (NaNO 3) \u003d 0.12 mol

m (NaNO 3) \u003d 85 ∙ 0.12 \u003d 10.2 (g)

The composition of the filtrate includes a solution of NaNO 3 and

solution of excess Na 3 PO 4.

n proreact. (Na 3 PO 4) \u003d 0.04 mol

n rest. (Na 3 PO 4) \u003d 0.06 - 0.04 \u003d 0.02 (mol)

m rest. (Na 3 PO 4) \u003d 164 ∙ 0.02 \u003d 3.28 (g)

The dry residue contains a mixture of NaNO 3 and Na 3 PO 4 salts.

m (dry rest.) \u003d 3.28 + 10.2 \u003d 13.48 (g)

ω (NaNO 3) \u003d 10.2 / 13.48 \u003d 0.76 (76%)

ω (Na 3 PO 4) = 24%

Answer: ω (NaNO 3) = 76%, ω (Na 3 PO 4) = 24%

2) How many liters of chlorine will be released if 200 ml of 35% hydrochloric acid

(ρ \u003d 1.17 g / ml) add 26.1 g of manganese oxide (IV) ? How many grams of sodium hydroxide in a cold solution will react with this amount of chlorine?

Find:

V(Cl2)

m (NaO H)

Given:

m (MnO 2) = 26.1 g

ρ (HCl solution) = 1.17 g/ml

ω(HCl) = 35%

V (HCl) solution) = 200 ml.

M (MnO 2) \u003d 87 g / mol

M (HCl) \u003d 36.5 g / mol

M (NaOH) \u003d 40 g / mol

V (Cl 2) = 6.72 (l)

m (NaOH) = 24 (g)

MnO 2 + 4 HCl \u003d MnCl 2 + Cl 2 + 2 H 2 O

1 mol 4 mol 1 mol

2 NaO H + Cl 2 = Na Cl + Na ClO + H 2 O

2 mol 1 mol

n (MnO 2) \u003d 26.1 / 87 \u003d 0.3 (mol)

m solution (НCl) = 200 ∙ 1.17 = 234 (g)

m total (НCl) = 234 ∙ 0.35 = 81.9 (g)

n (НCl) \u003d 81.9 / 36.5 \u003d 2.24 (mol)

0,3 < 2.24 /4

HCl - in excess, calculations for n (MnO 2)

n (MnO 2) \u003d n (Cl 2) \u003d 0.3 mol

V (Cl 2) \u003d 0.3 ∙ 22.4 = 6.72 (l)

n(NaOH) = 0.6 mol

m(NaOH) = 0.6 ∙ 40 = 24 (d)

2.3 Composition of the solution obtained during the reaction.

1) In 25 ml of 25% sodium hydroxide solution (ρ \u003d 1.28 g / ml) phosphorus oxide is dissolved (V) obtained by the oxidation of 6.2 g of phosphorus. What is the composition of the salt and what is its mass fraction in solution?

Find:

ω (salts)

Given:

V (NaOH) solution = 25 ml

ω(NaOH) = 25%

m (P) = 6.2 g

ρ (NaOH) solution = 1.28 g / ml

M (NaOH) \u003d 40 g / mol

M (P) \u003d 31 g / mol

M (P 2 O 5) \u003d 142 g / mol

M (NaH 2 PO 4) \u003d 120 g / mol

4P + 5O 2 \u003d 2 P 2 O 5

4mol 2mol

6 NaO H + P 2 O 5 \u003d 2 Na 3 RO 4 + 3 H 2 O

4 NaO H + P 2 O 5 \u003d 2 Na 2 H PO 4 + H 2 O

n (P) \u003d 6.2 / 31 \u003d 0.2 (mol)

n (P 2 O 5) = 0.1 mol

m (P 2 O 5) \u003d 0.1 ∙ 142 = 14.2 (g)

m (NaO H) solution = 25 ∙ 1.28 = 32 (g)

m (NaO H) in-va \u003d 0.25 ∙ 32 = 8 (g)

n (NaO H) in-va \u003d 8/40 \u003d 0.2 (mol)

According to the quantitative ratio of NaO H and P 2 O 5

it can be concluded that the acid salt NaH 2 PO 4 is formed.

2 NaO H + P 2 O 5 + H 2 O \u003d 2 NaH 2 PO 4

2mol 1mol 2mol

0.2mol 0.1mol 0.2mol

n (NaH 2 PO 4) = 0.2 mol

m (NaH 2 PO 4) \u003d 0.2 ∙ 120 = 24 (d)

m (p-ra after p-tion) \u003d 32 + 14.2 \u003d 46.2 (g)

ω (NaH 2 PO 4) \u003d 24 / 46.2 \u003d 0 52 (52%)

Answer: ω (NaH 2 PO 4) = 52%

2) When electrolyzing 2 liters of an aqueous solution of sodium sulfate with a mass fraction of salt 4%

(ρ = 1.025 g/ml) 448 l of gas (n.o.) were released on the insoluble anode. Determine the mass fraction of sodium sulfate in the solution after electrolysis.

Find:

m (Na 2 O)

Given:

V (r-ra Na 2 SO 4) \u003d 2l \u003d 2000 ml

ω (Na 2 SO 4 ) = 4%

ρ (r-ra Na 2 SO 4) \u003d 1 g / ml

M (H 2 O) \u003d 18 g / mol

V (O 2) \u003d 448 l

V M \u003d 22.4 l / mol

During the electrolysis of sodium sulfate, water decomposes, oxygen gas is released at the anode.

2 H 2 O \u003d 2 H 2 + O 2

2 mol 1mol

n (O 2) \u003d 448 / 22.4 \u003d 20 (mol)

n (H 2 O) \u003d 40 mol

m (H 2 O ) decomp. = 40 ∙ 18 = 720 (g)

m (r-ra to el-za) = 2000 ∙ 1.025 = 2050 (g)

m (Na 2 SO 4) in-va \u003d 2050 ∙ 0.04 = 82 (g)

m (solution after el-za) \u003d 2050 - 720 \u003d 1330 (g)

ω (Na 2 SO 4 ) \u003d 82 / 1330 \u003d 0.062 (6.2%)

Answer: ω (Na 2 SO 4 ) = 0.062 (6.2%)

2.4 A mixture of a known composition enters into the reaction; it is necessary to find portions of spent reagents and / or products obtained.

1) Determine the volume of the sulfur oxide gas mixture (IV) and nitrogen, which contains 20% sulfur dioxide by mass, which must be passed through 1000 g of a 4% sodium hydroxide solution so that the mass fractions of salts formed in the solution become the same.

Find:

V (gases)

Given:

m(NaOH) = 1000 g

ω(NaOH) = 4%

m (medium salt) =

m (acid salt)

M (NaOH) \u003d 40 g / mol

Answer: V (gases) = 156.8

NaO H + SO 2 = NaHSO 3 (1)

1 mole 1 mole

2NaO H + SO 2 = Na 2 SO 3 + H 2 O (2)

2 mol 1mol

m (NaOH) in-va \u003d 1000 ∙ 0.04 = 40 (g)

n(NaOH) = 40/40 = 1 (mol)

Let n 1 (NaOH) \u003d x, then n 2 (NaOH) \u003d 1 - x

n 1 (SO 2) \u003d n (NaHSO 3) \u003d x

M (NaHSO 3) \u003d 104 x n 2 (SO 2) \u003d (1 - x) / 2 \u003d 0.5 ∙ (1-x)

m (Na 2 SO 3) \u003d 0.5 ∙ (1-x) ∙ 126 \u003d 63 (1 - x)

104 x \u003d 63 (1 - x)

x = 0.38 mol

n 1 (SO 2) \u003d 0.38 mol

n 2 (SO 2 ) = 0.31 mol

n total (SO 2 ) = 0.69 mol

m total (SO 2) \u003d 0.69 ∙ 64 \u003d 44.16 (g) - this is 20% of the mass of the gas mixture. The mass of nitrogen gas is 80%.

m (N 2) \u003d 176.6 g, n 1 (N 2) \u003d 176.6 / 28 \u003d 6.31 mol

n total (gases) \u003d 0.69 + 6.31 \u003d 7 mol

V (gases) = 7 ∙ 22.4 = 156.8 (l)

2) When dissolving 2.22 g of a mixture of iron and aluminum filings in an 18.25% hydrochloric acid solution (ρ = 1.09 g/ml) 1344 ml of hydrogen (n.o.) were released. Find the percentage of each metal in the mixture and determine the volume of hydrochloric acid required to dissolve 2.22 g of the mixture.

Find:

ω(Fe)

ω(Al)

V (HCl) solution

Given:

m (mixtures) = 2.22 g

ρ (HCl solution) = 1.09 g/ml

ω(HCl) = 18.25%

M (Fe) \u003d 56 g / mol

M (Al) \u003d 27 g / mol

M (HCl) \u003d 36.5 g / mol

Answer: ω (Fe) = 75.7%,

ω(Al) = 24.3%,

V (HCl) solution) = 22 ml.

Fe + 2HCl \u003d 2 FeCl 2 + H 2

1 mol 2 mol 1 mol

2Al + 6HCl \u003d 2 AlCl 3 + 3H 2

2 mol 6 mol 3mol

n (H 2) \u003d 1.344 / 22.4 \u003d 0.06 (mol)

Let m (Al) \u003d x, then m (Fe) \u003d 2.22 - x;

n 1 (H 2) \u003d n (Fe) \u003d (2.22 - x) / 56

n (Al) \u003d x / 27

n 2 (H 2) \u003d 3x / 27 ∙ 2 = x / 18

x / 18 + (2.22 - x) / 56 \u003d 0.06

x \u003d 0.54, m (Al) \u003d 0.54 g

ω (Al) = 0.54 / 2.22 = 0.243 (24.3%)

ω(Fe) = 75.7%

n (Al) = 0.54 / 27 = 0.02 (mol)

m (Fe) \u003d 2.22 - 0.54 \u003d 1.68 (g)

n (Fe) \u003d 1.68 / 56 \u003d 0.03 (mol)

n 1 (НCl) = 0.06 mol

n(NaOH) = 0.05 mol

m solution (NaOH) = 0.05 ∙ 40/0.4 = 5 (g)

V (HCl) solution = 24 / 1.09 = 22 (ml)

3) The gas obtained by dissolving 9.6 g of copper in concentrated sulfuric acid was passed through 200 ml of potassium hydroxide solution (ρ =1 g/ml, ω (TO Oh) = 2.8%. What is the composition of the salt? Determine its mass.

Find:

m (salts)

Given:

m(Cu) = 9.6 g

V (KO H) solution = 200 ml

ω (KOH) \u003d 2.8%

ρ (H 2 O) \u003d 1 g / ml

M (Cu) \u003d 64 g / mol

M (KOH) \u003d 56 g / mol

M (KHSO 3) \u003d 120 g / mol

Answer: m (KHSO 3) = 12 g

Cu + 2H 2 SO 4 \u003d CuSO 4 + SO 2 + 2H 2 O

1 mole 1 mole

KO H + SO 2 \u003d KHSO 3

1 mole 1 mole

2 KO H + SO 2 \u003d K 2 SO 3 + H 2 O

2 mol 1mol

n (SO 2) \u003d n (Cu) \u003d 6.4 / 64 \u003d 0.1 (mol)

m (KO H) solution = 200 g

m (KO H) in-va \u003d 200 g ∙ 0.028 = 5.6 g

n (KO H) \u003d 5.6 / 56 \u003d 0.1 (mol)

According to the quantitative ratio of SO 2 and KOH, it can be concluded that the acid salt KHSO 3 is formed.

KO H + SO 2 \u003d KHSO 3

1 mol 1 mol

n (KHSO 3) = 0.1 mol

m (KHSO 3) = 0.1 ∙ 120 = 12 g

4) After 100 ml of a 12.33% solution of ferric chloride (II) (ρ =1.03g/ml) passed chlorine until the concentration of ferric chloride (III) in the solution did not become equal to the concentration of ferric chloride (II). Determine the volume of absorbed chlorine (N.O.)

Find:

V(Cl2)

Given:

V (FeCl 2) = 100 ml

ω (FeCl 2) = 12.33%

ρ (r-ra FeCl 2) \u003d 1.03 g / ml

M (FeCl 2) \u003d 127 g / mol

M (FeCl 3) \u003d 162.5 g / mol

V M \u003d 22.4 l / mol

m (FeCl 2) solution = 1.03 ∙ 100 = 103 (g)

m (FeCl 2) p-in-va \u003d 103 ∙ 0.1233 = 12.7 (g)

2FeCl 2 + Cl 2 = 2 FeCl 3

2 mol 1 mol 2 mol

Let n (FeCl 2) proreact. \u003d x, then n (FeCl 3) arr. = x;

m (FeCl 2) proreact. = 127x

m (FeCl 2) rest. = 12.7 - 127x

m (FeCl 3) arr. = 162.5x

According to the condition of the problem m (FeCl 2) rest. \u003d m (FeCl 3)

12.7 - 127x = 162.5x

x \u003d 0.044, n (FeCl 2) proreact. = 0.044 mol

n (Cl 2) \u003d 0.022 mol

V (Cl 2) \u003d 0.022 ∙ 22.4 = 0.5 (l)

Answer: V (Cl 2) \u003d 0.5 (l)

5) After calcining a mixture of magnesium and calcium carbonates, the mass of the released gas turned out to be equal to the mass of the solid residue. Determine the mass fractions of substances in the initial mixture. What volume of carbon dioxide (N.O.) can be absorbed by 40 g of this mixture, which is in the form of a suspension.

Find:

ω (MgCO 3)

ω (CaCO 3)

Given:

m (solid product) \u003d m (gas)

m ( mixtures of carbonates)=40g

M (MgO) \u003d 40 g / mol

M CaO = 56 g/mol

M (CO 2) \u003d 44 g / mol

M (MgCO 3) \u003d 84 g / mol

M (CaCO 3) \u003d 100 g / mol

1) We will carry out calculations using 1 mol of a mixture of carbonates.

MgCO 3 \u003d MgO + CO 2

1mol 1mol 1mol

CaCO 3 \u003d CaO + CO 2

1 mole 1 mole 1 mole

Let n (MgCO 3) \u003d x, then n (CaCO 3) \u003d 1 - x.

n (MgO) = x, n (CaO) = 1 - x

m(MgO) = 40x

m (СаO) = 56 ∙ (1 - x) \u003d 56 - 56x

From a mixture taken in an amount of 1 mol, carbon dioxide is formed in an amount of 1 mol.

m (CO 2) = 44.g

m (tv.prod.) = 40x + 56 - 56x = 56 - 16x

56 - 16x = 44

x = 0.75,

n (MgCO 3) = 0.75 mol

n (CaCO 3) = 0.25 mol

m (MgCO 3) \u003d 63 g

m (CaCO 3) = 25 g

m (mixtures of carbonates) = 88 g

ω (MgCO 3) \u003d 63/88 \u003d 0.716 (71.6%)

ω (CaCO 3) = 28.4%

2) A suspension of a mixture of carbonates, when carbon dioxide is passed through, turns into a mixture of hydrocarbons.

MgCO 3 + CO 2 + H 2 O \u003d Mg (HCO 3) 2 (1)

1 mole 1 mole

CaCO 3 + CO 2 + H 2 O \u003d Ca (HCO 3) 2 (2)

1 mol 1mol

m (MgCO 3) \u003d 40 ∙ 0.75 = 28.64(g)

n 1 (CO 2) \u003d n (MgCO 3) \u003d 28.64 / 84 \u003d 0.341 (mol)

m (CaCO 3) = 11.36 g

n 2 (CO 2) \u003d n (CaCO 3) \u003d 11.36 / 100 \u003d 0.1136 mol

n total (CO 2) \u003d 0.4546 mol

V (CO 2) = n total (CO2) ∙ V M = 0.4546 ∙ 22.4 = 10.18 (l)

Answer: ω (MgCO 3) = 71.6%, ω (CaCO 3) = 28.4%,

V (CO 2 ) \u003d 10.18 liters.

6) A mixture of powders of aluminum and copper weighing 2.46 g was heated in a stream of oxygen. The resulting solid was dissolved in 15 ml of a sulfuric acid solution (acid mass fraction 39.2%, density 1.33 g/ml). The mixture completely dissolved without evolution of gas. To neutralize the excess acid, 21 ml of sodium bicarbonate solution with a concentration of 1.9 mol/l was required. Calculate the mass fractions of metals in the mixture and the volume of oxygen (N.O.) that reacted.

Find:

ω(Al); ω(Cu)

V(O2)

Given:

m (mixes) = 2.46 g

V (NaHCO 3 ) = 21 ml =

0.021 l

V (H 2 SO 4 ) = 15 ml

ω(H 2 SO 4 ) = 39.2%

ρ (H 2 SO 4 ) \u003d 1.33 g / ml

C (NaHCO 3) \u003d 1.9 mol / l

M (Al) \u003d 27 g / mol

М(Cu)=64 g/mol

M (H 2 SO 4) \u003d 98 g / mol

V m \u003d 22.4 l / mol

Answer: ω (Al ) = 21.95%;

ω ( Cu) = 78.05%;

V (O 2) = 0,672

4Al + 3O 2 = 2Al 2 O 3

4mol 3mol 2mol

2Cu + O 2 = 2CuO

2mol 1mol 2mol

Al 2 O 3 + 3H 2 SO 4 = Al 2 (SO 4 ) 3 + 3H 2 O(1)

1 mole 3 mole

CuO + H 2 SO 4 = CuSO 4 + H 2 O(2)

1 mole 1 mole

2 NaHCO 3 + H 2 SO 4 = Na 2 SO 4 + 2H 2 O+ SO 2 (3)

2 mol 1 mol

m (H 2 SO 4) solution = 15 ∙ 1.33 = 19.95 (g)

m (H 2 SO 4) in-va = 19.95 ∙ 0.393 = 7.8204 (g)

n ( H 2 SO 4) total = 7.8204/98 = 0.0798 (mol)

n (NaHCO 3) = 1,9 ∙ 0.021 = 0.0399 (mol)

n 3 (H 2 SO 4 ) = 0,01995 ( mole )

n 1+2 (H 2 SO 4 ) =0,0798 – 0,01995 = 0,05985 ( mole )

4) Let n (Al) = x, . m(Al) = 27x

n (Cu) = y, m (Cu) = 64y

27x + 64y = 2.46

n(Al 2 O 3 ) = 1.5x

n(CuO) = y

1.5x + y = 0.0585

x = 0.02; n(Al) = 0.02 mole

27x + 64y = 2.46

y=0.03; n(Cu)=0.03 mole

m(Al) = 0.02∙ 27 = 0,54

ω (Al) = 0.54 / 2.46 = 0.2195 (21.95%)

ω (Cu) = 78.05%

n 1 (O 2 ) = 0.015 mole

n 2 (O 2 ) = 0.015 mole

n common . (O 2 ) = 0.03 mole

V(O 2 ) = 22,4 ∙ 0 03 = 0,672 ( l )

7) When dissolving 15.4 g of an alloy of potassium with sodium in water, 6.72 liters of hydrogen (n.o.) were released. Determine the molar ratio of metals in the alloy.

Find:

n (K) : n( Na)

m (Na 2 O)

Given:

m(alloy) = 15.4 g

V (H 2) = 6.72 l

M ( Na) =23 g/mol

M (K) \u003d 39 g/mol

n (K) : n ( Na) = 1: 5

2K + 2 H 2 O= 2 K Oh+ H 2

2 mol 1 mol

2Na + 2H 2 O = 2 NaOH+ H 2

2 mol 1 mol

Let n(K) = x, n ( Na) = y, then

n 1 (H 2) = 0.5 x; n 2 (H 2) \u003d 0.5y

n (H 2) \u003d 6.72 / 22.4 \u003d 0.3 (mol)

m(K) = 39 x; m (Na) = 23 y

39x + 23y = 15.4

x = 0.1, n(K) = 0.1 mol;

0.5x + 0.5y = 0.3

y = 0.5, n( Na) = 0.5 mol

8) When processing 9 g of a mixture of aluminum with aluminum oxide with a 40% sodium hydroxide solution (ρ \u003d 1.4 g / ml) 3.36 l of gas (n.o.) were released. Determine the mass fractions of substances in the initial mixture and the volume of the alkali solution that entered into the reaction.

Find:

ω (Al)

ω (Al 2 O 3)

V r-ra ( NaOH)

Given:

M(see) = 9 g

V(H 2) = 33.8ml

ω (NaOH) = 40%

M( Al) = 27 g/mol

M( Al 2 O 3) = 102 g/mol

M( NaOH) = 40 g/mol

2Al + 2NaOH + 6H 2 O = 2Na + 3H 2

2 mole 2 mole 3 mole

Al 2 O 3 + 2NaOH + 3H 2 O = 2 Na

1mol 2mol

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

n ( Al) = 0.1 mol m (Al) = 2.7 g

ω (Al) = 2.7 / 9 = 0.3 (30%)

ω(Al 2 O 3 ) = 70%

m (Al 2 O 3 ) = 9 – 2.7 = 6.3 ( G )

n(Al 2 O 3 ) = 6,3 / 102 = 0,06 ( mole )

n 1 (NaOH) = 0.1 mole

n 2 (NaOH) = 0.12 mole

n common . (NaOH) = 0.22 mole

m R - ra (NaOH) = 0.22∙ 40 /0.4 = 22 ( G )

V R - ra (NaOH) = 22 / 1.4 = 16 ( ml )

Answer : ω(Al) = 30%, ω(Al 2 O 3 ) = 70%, V R - ra (NaOH) = 16 ml

9) An alloy of aluminum and copper weighing 2 g was treated with a solution of sodium hydroxide, with a mass fraction of alkali 40% (ρ =1.4 g/ml). The undissolved precipitate was filtered off, washed and treated with a solution nitric acid. The resulting mixture was evaporated to dryness, the residue was calcined. The mass of the resulting product was 0.8 g. Determine the mass fraction of metals in the alloy and the volume of the spent sodium hydroxide solution.

Find:

ω (Cu); ω (Al)

V r-ra ( NaOH)

Given:

m(mixture)=2 g

ω (NaOH)=40%

M( Al)=27 g/mol

M( Cu)=64 g/mol

M( NaOH)=40 g/mol

Alkali dissolves only aluminum.

2Al + 2NaOH + 6H 2 O = 2 Na + 3 H 2

2mol 2mol 3mol

Copper is an undissolved residue.

3Cu + 8HNO 3 = 3Cu(NO 3 ) 2 +4H 2 O + 2 NO

3 mole 3 mole

2Cu(NO 3 ) 2 = 2 CuO + 4NO 2 +O 2

2mol 2mol

n (CuO) = 0.8 / 80 = 0.01 (mol)

n (CuO) = n (Cu(NO 3 ) 2 ) = n(Cu) = 0.1 mole

m(Cu) = 0.64 G

ω (Cu) = 0.64 / 2 = 0.32 (32%)

ω(Al) = 68%

m(Al) = 9 - 0.64 = 1.36(g)

n ( Al) = 1.36 / 27 = 0.05 (mol)

n ( NaOH) = 0.05 mol

m r-ra ( NaOH) = 0,05 ∙ 40 / 0.4 = 5 (g)

V r-ra ( NaOH) = 5 / 1.43 = 3.5 (ml)

Answer: ω (Cu) = 32%, ω (Al) = 68%, V r-ra ( NaOH) = 3.5 ml

10) A mixture of potassium, copper and silver nitrates was calcined, weighing 18.36 g. The volume of released gases was 4.32 l (n.o.). The solid residue was treated with water, after which its mass decreased by 3.4 g. Find the mass fractions of nitrates in the initial mixture.

Find:

ω (KNO 3 )

ω (Cu(NO 3 ) 2 )

ω (AgNO 3)

Given:

m(blends) = 18.36 g

∆m(hard. rest.)=3.4 g

V (CO 2) = 4.32 l

M(K NO 2) \u003d 85 g / mol

M(K NO 3) =101 g/mol

2 K NO 3 = 2 K NO 2 + O 2 (1)

2 mol 2 mol 1mol

2 Cu(NO 3 ) 2 = 2 CuO + 4 NO 2 +O 2 (2)

2 mol 2 mol 4 mol 1 mol

2 AgNO 3 = 2 Ag + 2 NO 2 + O 2 (3)

2 mol 2 mol 2 mol 1 mol

CuO + 2H 2 O= interaction not possible

Ag+ 2H 2 O= interaction not possible

TO NO 2 + 2H 2 O= salt dissolution

The change in the mass of the solid residue occurred due to the dissolution of the salt, therefore:

m(TO NO 2) = 3.4 g

n(K NO 2) = 3.4 / 85 = 0.04 (mol)

n(K NO 3) = 0.04 (mol)

m(TO NO 3) = 0,04∙ 101 = 4.04 (g)

ω (KNO 3) = 4,04 / 18,36 = 0,22 (22%)

n 1 (O 2) = 0.02 (mol)

n total (gases) = 4.32 / 22.4 = 0.19 (mol)

n 2+3 (gases) = 0.17 (mol)

m(mixtures without K NO 3) \u003d 18.36 - 4.04 \u003d 14.32 (g)

Let m (Cu(NO 3 ) 2 ) = x, Then m (AgNO 3 ) = 14.32 – x.

n (Cu(NO 3 ) 2 ) = x / 188,

n (AgNO 3) = (14,32 – x) / 170

n 2 (gases) = 2.5x / 188,

n 3 (gases) = 1.5 ∙ (14.32 - x) / 170,

2.5x/188 + 1.5 ∙ (14.32 - x) / 170 \u003d 0.17

X = 9.75, m (Cu(NO 3 ) 2 ) = 9,75 G

ω (Cu(NO 3 ) 2 ) = 9,75 / 18,36 = 0,531 (53,1%)

ω (AgNO 3 ) = 24,09%

Answer : ω (KNO 3 ) = 22%, ω (Cu(NO 3 ) 2 ) = 53.1%, ω (AgNO 3 ) = 24,09%.

11) A mixture of barium hydroxide, calcium and magnesium carbonates weighing 3.05 g was calcined to remove volatile substances. The mass of the solid residue was 2.21 g. Volatile products were brought to normal conditions, and the gas was passed through a solution of potassium hydroxide, the mass of which increased by 0.66 g. Find the mass fractions of substances in the initial mixture.

ω (IN a(O H) 2)

ω (WITH a WITH O 3)

ω (mg WITH O 3)

m(mixture) = 3.05 g

m(solid rest) = 2.21 g

∆ m(KOH) = 0.66 g

M ( H 2 O) =18 g/mol

M (CO 2) \u003d 44 g / mol

M (B a(O H) 2) \u003d 171 g / mol

M (CaCO 2) \u003d 100 g / mol

M ( mg CO 2) \u003d 84 g / mol

IN a(O H) 2 = H 2 O+ V aO

1 mol 1mol

WITH a WITH O 3 \u003d CO 2 + C aO

1 mol 1mol

mg WITH O 3 \u003d CO 2 + MgO

1 mol 1mol

The mass of KOH increased due to the mass of absorbed CO 2

KOH + CO 2 →…

According to the law of conservation of mass of substances

m (H 2 O) \u003d 3.05 - 2.21 - 0.66 \u003d 0.18 g

n ( H 2 O) = 0.01 mol

n (B a(O H) 2) = 0.01 mol

m(IN a(O H) 2) = 1.71 g

ω (IN a(O H) 2) = 1.71 / 3.05 = 0.56 (56%)

m(carbonates) = 3.05 - 1.71 = 1.34 g

Let m(WITH a WITH O 3) = x, Then m(WITH a WITH O 3) = 1,34 – x

n 1 (C O 2) = n (C a WITH O 3) = x /100

n 2 (C O 2) = n ( mg WITH O 3) = (1,34 - x)/84

x /100 + (1,34 - x)/84 = 0,015

x = 0,05, m(WITH a WITH O 3) = 0.05 g

ω (WITH a WITH O 3) = 0,05/3,05 = 0,16 (16%)

ω (mg WITH O 3) =28%

Answer: ω (IN a(O H) 2) = 56%, ω (WITH a WITH O 3) = 16%, ω (mg WITH O 3) =28%

2.5 An unknown substance enters the reaction o / is formed during the reaction.

1) When a hydrogen compound of a monovalent metal interacted with 100 g of water, a solution was obtained with a mass fraction of a substance of 2.38%. The mass of the solution turned out to be 0.2 g less than the sum of the masses of water and the initial hydrogen compound. Determine which connection was taken.

Find:

Given:

m (H 2 O) = 100 g

ω (Me Oh) = 2,38%

∆ m(solution) = 0.2 g

M ( H 2 O) = 18 g/mol

Men + H 2 O= Me Oh+ H 2

1 mole 1 mole 1 mole

0.1 mol 0.1 mol 0.1 mol

The mass of the final solution decreased by the mass of hydrogen gas.

n (H 2) \u003d 0.2 / 2 \u003d 0.1 (mol)

n ( H 2 O) proreact. = 0.1 mol

m (H 2 O) proreag = 1.8 g

m (H 2 O in solution) = 100 - 1.8 = 98.2 (g)

ω (Me Oh) = m(Me Oh) / m(r-ra g/mol

Let m(Me Oh) = x

0.0238 = x / (98.2 + x)

x = 2,4, m(Me O H) = 2.4 g

n(Me O H) = 0.1 mol

M (Me O H) \u003d 2.4 / 0.1 \u003d 24 (g / mol)

M (Me) = 7 g/mol

Me - Li

Answer: Li N.

2) When 260 g of an unknown metal is dissolved in highly dilute nitric acid, two salts are formed: Me (NABOUT 3 ) 2 AndX. When heatedXwith calcium hydroxide, gas is released, which with phosphoric acid forms 66 g of ammonium hydroorthophosphate. Determine the metal and salt formulaX.

Find:

Given:

m(Me) = 260 g

m ((NH 4) 2 HPO 4) = 66 g

M (( NH 4) 2 HPO 4) =132 g/mol

Answer: Zn, salt - NH 4 NO 3.

4Me + 10HNO 3 = 4Me(NO 3 ) 2 +NH 4 NO 3 + 3H 2 O

4 mole 1 mole

2NH 4 NO 3 +Ca(OH) 2 = Ca(NO 3 ) 2 +2NH 3 + 2H 2 O

2 mole 2 mole

2NH 3 + H 3 PO 4 = (NH 4 ) 2 HPO 4

2 mol 1mol

n ((NH 4) 2 HPO 4) = 66/132 = 0.5 (mol)

n (N H 3) = n (NH 4 NO 3) = 1 mol

n (Me) = 4mol

M (Me) = 260/4 = 65 g/mol

Me - Zn

3) In 198.2 ml of aluminum sulfate solution (ρ = 1 g/ml) lowered a plate of an unknown divalent metal. After some time, the mass of the plate decreased by 1.8 g, and the concentration of the formed salt was 18%. Define metal.

Find:

ω 2 (NaOH)

Given:

V solution = 198.2 ml

ρ (solution) = 1 g/ml

ω 1 (salt) = 18%

∆m(p-ra) \u003d 1.8 g

M ( Al) =27 g/mol

Al 2 (SO 4 ) 3 + 3Me = 2Al+ 3MeSO 4

3 mole 2 mole 3 mole

m(r-ra to r-tion) = 198.2 (g)

m(p-ra after p-tion) \u003d 198.2 + 1.8 \u003d 200 (g)

m (MeSO 4) in-va \u003d 200 ∙ 0.18 = 36 (g)

Let M (Me) = x, then M ( MeSO 4) = x + 96

n ( MeSO 4) = 36 / (x + 96)

n (Me) \u003d 36 / (x + 96)

m(Me) = 36 x/ (x + 96)

n ( Al) = 24 / (x + 96),

m (Al) = 24 ∙ 27/(x+96)

m(Me) ─ m (Al) = ∆m(r-ra)

36x/ (x + 96) ─ 24 ∙ 27 / (x + 96) = 1.8

x \u003d 24, M (Me) \u003d 24 g / mol

Metal - mg

Answer: mg.

4) During thermal decomposition of 6.4 g of salt in a vessel with a capacity of 1 l at 300.3 0 With a pressure of 1430 kPa. Determine the formula of salt if, during its decomposition, water and a gas poorly soluble in it are formed.

Find:

salt formula

Given:

m(salt) = 6.4 g

V(vessel) = 1 l

P = 1430 kPa

t=300.3 0 C

R= 8.31J/mol ∙ TO

n (gas) = PV/RT = 1430∙1 / 8,31∙ 573.3 = 0.3 (mol)

The condition of the problem corresponds to two equations:

NH 4 NO 2 = N 2 + 2 H 2 O ( gas)

1 mol 3 mol

NH 4 NO 3 = N 2 O + 2 H 2 O (gas)

1 mol 3 mol

n (salts) = 0.1 mol

M (salt) \u003d 6.4 / 0.1 \u003d 64 g / mol ( NH 4 NO 2)

Answer: NH 4 N

Literature.

1. N.E. Kuzmenko, V.V. Eremin, A.V. Popkov "Chemistry for high school students and university applicants", Moscow, "Drofa" 1999

2. G.P. Khomchenko, I.G. Khomchenko "Collection of problems in chemistry", Moscow "New Wave * Onyx" 2000

3. K.N. Zelenin, V.P. Sergutina, O.V., O.V. Solod "Manual in chemistry for those entering the Military - medical academy and other higher medical educational establishments»,

St. Petersburg, 1999

4. A guide for applicants to medical institutes "Problems in chemistry with solutions",

St. Petersburg Medical Institute named after I.P. Pavlov

5. FIPI "USE CHEMISTRY" 2009 - 2015

slide 2

“In order to avoid mistakes, one must gain experience; in order to gain experience, one must make mistakes.”

slide 3

C1. Using the electron balance method, write an equation for the reaction. Determine the oxidizing agent and reducing agent.

slide 4

Required Skills

Arrangement of oxidation states Ask yourself the main question: who donates electrons in this reaction, and who accepts them? Determine in which medium (acidic, neutral or alkaline) the reaction takes place. if we see acid in products, acid oxide means that it is definitely not an alkaline medium, and if metal hydroxide precipitates, it is definitely not acidic. Check that the reaction contains both an oxidizing agent and a reducing agent. If both substances can exhibit the properties of both a reducing agent and an oxidizing agent, it is necessary to consider which of them is a more active oxidizing agent. Then the second one will be the restorer.

slide 5

The sequence of the coefficients in the equation

First, put down the coefficients obtained from the electronic balance. If any substance acts both as a medium and as an oxidizing agent (reductant), it will need to be equalized later, when almost all the coefficients are placed. The penultimate equalizes hydrogen with oxygen, we only check

slide 6

Possible mistakes

Arrangement of oxidation states: a) oxidation states in hydrogen compounds non-metals: phosphine РН3 - the oxidation state of phosphorus is negative; b) c organic matter ah - check again whether the entire environment of the C atom is taken into account c) ammonia and ammonium salts - in them nitrogen always has an oxidation state of −3 c) oxygen salts and chlorine acids - in them chlorine can have an oxidation state of +1, +3, + 5, +7; d) double oxides: Fe3O4, Pb3O4 - in them, metals have two different oxidation states, usually only one of them is involved in electron transfer.

Slide 7

2. The choice of products without taking into account the transfer of electrons - that is, for example, in the reaction there is only an oxidizing agent without a reducing agent, or vice versa 3. Incorrect products from a chemical point of view: a substance that interacts with the environment cannot be obtained! a) in an acidic environment, metal oxide, base, ammonia cannot be obtained; b) in an alkaline environment, acid or acid oxide will not be obtained; c) an oxide, let alone a metal that reacts violently with water, is not formed in an aqueous solution.

Slide 8

Slide 9

Increasing the oxidation states of manganese