A problem to solve in chemistry. Solving typical problems in chemistry

Municipal budgetary educational institution

"Average comprehensive school № 37

with in-depth study of individual subjects "

Vyborg, Leningrad region

"Solution of calculation problems advanced level difficulties"

(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 the check USE results in chemistry 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"Solving problems of increased complexity."

Tasks - special kind 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 weight

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

To pass 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) \u003d 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 (g)

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. Received solid dissolved in 15 ml of sulfuric acid solution (mass fraction of acid 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 nitric acid solution. 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 (AT a(O H) 2)

ω (FROM a FROM O 3)

ω (mg FROM 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

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

1 mol 1mol

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

1 mol 1mol

mg FROM 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(AT a(O H) 2) = 1.71 g

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

m(carbonates) = 3.05 - 1.71 = 1.34 g

Let m(FROM a FROM O 3) = x, then m(FROM a FROM O 3) = 1,34 – x

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

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

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

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

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

ω (mg FROM O 3) =28%

Answer: ω (AT a(O H) 2) = 56%, ω (FROM a FROM O 3) = 16%, ω (mg FROM 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 (NO 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

Probably every student technical university at least once wondered how to solve problems in chemistry. As practice shows, most students consider this science complex and incomprehensible, often they simply do not believe in their strength and give up without revealing their potential.

In fact, chemistry is only a problem from a psychological point of view. Having overcome yourself, realizing your capabilities, you can easily master the basics of this subject and move on to more complex issues. So, we learn to solve problems in chemistry quickly, correctly and easily, and also get the maximum pleasure from the result.

Why you should not be afraid to delve into science

Chemistry is not a collection of incomprehensible formulas, symbols and substances. It is a science closely related to environment. Without realizing it, we face it at every step. When cooking, damp cleaning the house, washing, walking in the fresh air, we constantly use chemical knowledge.

Following this logic, when you understand how to learn how to solve problems in chemistry, you can make your life much easier. But people who come across science while studying or working in production cannot do without special knowledge and skills. Workers in the medical field need chemistry no less, since any person in this profession must know how a particular drug affects the patient's body.

Chemistry is a science that is constantly present in our life, it is interconnected with a person, it is a part of it. Therefore, any student, whether he realizes it or not, is able to master this branch of knowledge.

Fundamentals of Chemistry

Before thinking about how to learn how to solve problems in chemistry, it is important to understand that without basic knowledge you can't do it. The basics of any science is the foundation of its comprehension. Even experienced professionals use this framework when solving the most complex problems, perhaps without realizing it.

So, check out the list of information you'll need:

• The valence of elements is a factor with the participation of which any problems are solved. Formulas of substances, equations will not be made correctly without this knowledge. You can find out what valency is in any chemistry textbook, since this is the basic concept that any student must master in the first lesson.
• The periodic table is familiar to almost every person. Learn to use it properly and you won't have to keep a lot of information in your head.
• Learn to identify what substance you are dealing with. The liquid, solid and gaseous state of the object with which you have to work can tell a lot.

After obtaining the above knowledge, many people will have much less questions about how to solve problems in chemistry. But if you still can't believe in yourself, read on.

Step-by-step instructions for solving any problem

After reading the previous information, many may have the opinion that it is extremely easy to solve problems in chemistry. The formulas you need to know may be really simple, but to master the science you will need to muster all your patience, diligence and perseverance. From the first time, few people manage to achieve their goal.

Over time, with perseverance, you can solve absolutely any problem. The process usually consists of the following steps:

• Making a brief condition of the problem.
• Drawing up a reaction equation.
• Arrangement of coefficients in the equation.
• Equation solution.

Experienced chemistry teachers assure that in order to freely solve any type of problem, you need to practice on 15 similar tasks on your own. After that, you will freely master the given topic.

A little about theory

It is impossible to think about how to solve problems in chemistry without mastering the theoretical material to the required extent. No matter how dry, useless and uninteresting it may seem, this is the basis of your skills. Theory is applied always and in all sciences. Without its existence, the practice is meaningless. Study the school curriculum in chemistry sequentially, step by step, without skipping even, as it seems to you, insignificant information, in order to eventually notice a breakthrough in your knowledge.

How to solve problems in chemistry: time for learning

Often, students who have mastered a certain type of task move on, forgetting that consolidating and repeating knowledge is a process no less important than obtaining it. Each topic should be fixed if you are counting on a long-term result. Otherwise, you will forget all the information very quickly. Therefore, do not be lazy, devote more time to each question.

Finally, do not forget about motivation - the engine of progress. Do you want to become an excellent chemist and surprise others with a huge store of knowledge? Act, try, decide, and you will succeed. Then you will be consulted on all chemical issues.

Chemistry is the science of substances, their properties and transformations. .
That is, if nothing happens to the substances around us, then this does not apply to chemistry. But what does "nothing happens" mean? If a thunderstorm suddenly caught us in the field, and we all got wet, as they say, “to the skin”, then is this not a transformation: after all, the clothes were dry, but became wet.

If, for example, you take an iron nail, process it with a file, and then assemble iron filings (Fe) , then this is also not a transformation: there was a nail - it became powder. But if after that to assemble the device and hold obtaining oxygen (O 2): heat up potassium permanganate(KMpo 4) and collect oxygen in a test tube, and then place these iron filings heated “to red” in it, then they will flare up with a bright flame and, after combustion, will turn into a brown powder. And this is also a transformation. So where is the chemistry? Despite the fact that in these examples the shape (iron nail) and the state of clothing (dry, wet) change, these are not transformations. The fact is that the nail itself, as it was a substance (iron), remained so, despite its different form, and our clothes soaked up the water from the rain, and then it evaporated into the atmosphere. The water itself has not changed. So what are transformations in terms of chemistry?

From the point of view of chemistry, transformations are such phenomena that are accompanied by a change in the composition of a substance. Let's take the same nail as an example. It does not matter what form it took after being filed, but after being collected from it iron filings placed in an atmosphere of oxygen - it turned into iron oxide(Fe 2 O 3 ) . So, has something really changed? Yes, it has. There was a nail substance, but under the influence of oxygen a new substance was formed - element oxide gland. molecular equation this transformation can be represented by the following chemical symbols:

4Fe + 3O 2 = 2Fe 2 O 3 (1)

For a person uninitiated in chemistry, questions immediately arise. What is the "molecular equation", what is Fe? Why are there numbers "4", "3", "2"? What are the small numbers "2" and "3" in the formula Fe 2 O 3? This means that the time has come to sort things out in order.

Signs chemical elements.

Despite the fact that they begin to study chemistry in the 8th grade, and some even earlier, many people know the great Russian chemist D. I. Mendeleev. And of course, his famous "Periodic Table of Chemical Elements". Otherwise, more simply, it is called the "Mendeleev's Table".

In this table, in the appropriate order, the elements are located. To date, about 120 of them are known. The names of many elements have been known to us for a long time. These are: iron, aluminum, oxygen, carbon, gold, silicon. Previously, we used these words without hesitation, identifying them with objects: an iron bolt, aluminum wire, oxygen in the atmosphere, a golden ring, etc. etc. But in fact, all these substances (bolt, wire, ring) consist of their respective elements. The whole paradox is that the element cannot be touched, picked up. How so? They are in the periodic table, but you can’t take them! Yes exactly. A chemical element is an abstract (that is, abstract) concept, and is used in chemistry, however, as in other sciences, for calculations, drawing up equations, and solving problems. Each element differs from the other in that it is characterized by its own electronic configuration of an atom. The number of protons in the nucleus of an atom is equal to the number of electrons in its orbitals. For example, hydrogen is element #1. Its atom consists of 1 proton and 1 electron. Helium is element number 2. Its atom consists of 2 protons and 2 electrons. Lithium is element number 3. Its atom consists of 3 protons and 3 electrons. Darmstadtium - element number 110. Its atom consists of 110 protons and 110 electrons.

Each element is denoted by a certain symbol, Latin letters, and has a certain reading in translation from Latin. For example, hydrogen has the symbol "N", read as "hydrogenium" or "ash". Silicon has the symbol "Si" read as "silicium". Mercury has a symbol "Hg" and is read as "hydrargyrum". And so on. All these designations can be found in any chemistry textbook for the 8th grade. For us now, the main thing is to understand that when compiling chemical equations, it is necessary to operate with the indicated symbols of the elements.

Simple and complex substances.

Denoting various substances with single symbols of chemical elements (Hg mercury, Fe iron, Cu copper, Zn zinc, Al aluminum) we essentially denote simple substances, that is, substances consisting of atoms of the same type (containing the same number of protons and neutrons in an atom). For example, if iron and sulfur substances interact, then the equation will take the following form:

Fe + S = FeS (2)

Simple substances include metals (Ba, K, Na, Mg, Ag), as well as non-metals (S, P, Si, Cl 2, N 2, O 2, H 2). And you should pay attention
special attention to the fact that all metals are denoted by single symbols: K, Ba, Ca, Al, V, Mg, etc., and non-metals - either by simple symbols: C, S, P or may have different indices that indicate their molecular structure: H 2 , Cl 2 , O 2 , J 2 , P 4 , S 8 . In the future, this will be very great importance when writing equations. It is not at all difficult to guess that complex substances are substances formed from atoms. different kind, for example,

one). Oxides:
aluminium oxide Al 2 O 3,

sodium oxide Na 2 O
copper oxide CuO,
zinc oxide ZnO
titanium oxide Ti2O3,
carbon monoxide or carbon monoxide (+2) CO
sulfur oxide (+6) SO 3

2). Reasons:
iron hydroxide(+3) Fe (OH) 3,
copper hydroxide Cu(OH)2,
potassium hydroxide or potassium alkali KOH,
sodium hydroxide NaOH.

3). Acids:
hydrochloric acid HCl
sulfurous acid H2SO3,
Nitric acid HNO3

four). Salts:
sodium thiosulfate Na 2 S 2 O 3,
sodium sulfate or Glauber's salt Na 2 SO 4,
calcium carbonate or limestone CaCO 3,
copper chloride CuCl 2

5). organic matter:
sodium acetate CH 3 COOHa,
methane CH 4,
acetylene C 2 H 2,
glucose C 6 H 12 O 6

Finally, after we figured out the structure various substances, you can start compiling chemical equations.

Chemical equation.

The word “equation” itself is derived from the word “equalize”, i.e. divide something into equal parts. In mathematics, equations are almost the very essence of this science. For example, you can give such a simple equation in which the left and right sides will be equal to "2":

40: (9 + 11) = (50 x 2): (80 - 30);

And in chemical equations, the same principle: the left and right sides of the equation must correspond to the same number of atoms, the elements participating in them. Or, if an ionic equation is given, then in it number of particles must also meet this requirement. A chemical equation is a conditional record of a chemical reaction using chemical formulas and mathematical symbols. A chemical equation inherently reflects a particular chemical reaction, that is, the process of interaction of substances, during which new substances arise. For example, it is necessary write a molecular equation reactions that take part barium chloride BaCl 2 and sulphuric acid H 2 SO 4. As a result of this reaction, an insoluble precipitate is formed - barium sulfate BaSO 4 and hydrochloric acid Hcl:

ВаСl 2 + H 2 SO 4 = BaSO 4 + 2НCl (3)

First of all, it is necessary to understand that the large number “2” in front of the HCl substance is called the coefficient, and the small numbers “2”, “4” under the formulas ВаСl 2, H 2 SO 4, BaSO 4 are called indices. Both the coefficients and indices in chemical equations play the role of factors, not terms. In order to correctly write a chemical equation, it is necessary arrange the coefficients in the reaction equation. Now let's start counting the atoms of the elements on the left and right sides of the equation. On the left side of the equation: the substance BaCl 2 contains 1 barium atom (Ba), 2 chlorine atoms (Cl). In the substance H 2 SO 4: 2 hydrogen atoms (H), 1 sulfur atom (S) and 4 oxygen atoms (O). On the right side of the equation: in the BaSO 4 substance there is 1 barium atom (Ba) 1 sulfur atom (S) and 4 oxygen atoms (O), in the HCl substance: 1 hydrogen atom (H) and 1 chlorine atom (Cl). Whence it follows that on the right side of the equation the number of hydrogen and chlorine atoms is half that on the left side. Therefore, before the HCl formula on the right side of the equation, it is necessary to put the coefficient "2". If we now add the number of atoms of the elements involved in this reaction, both on the left and on the right, we get the following balance:

In both parts of the equation, the number of atoms of the elements participating in the reaction are equal, therefore it is correct.

Chemical equation and chemical reactions

As we have already found out, chemical equations are a reflection of chemical reactions. Chemical reactions are such phenomena in the process of which the transformation of one substance into another occurs. Among their diversity, two main types can be distinguished:

one). Connection reactions
2). decomposition reactions.

The overwhelming majority of chemical reactions belong to addition reactions, since changes in its composition can rarely occur with a single substance if it is not subjected to external influences (dissolution, heating, light). Nothing characterizes a chemical phenomenon, or reaction, as much as the changes that occur when two or more substances interact. Such phenomena can occur spontaneously and be accompanied by an increase or decrease in temperature, light effects, color changes, sedimentation, release of gaseous products, noise.

For clarity, we present several equations that reflect the processes of compound reactions, during which we obtain sodium chloride(NaCl), zinc chloride(ZnCl 2), silver chloride precipitate(AgCl), aluminum chloride(AlCl 3)

Cl 2 + 2Nа = 2NaCl (4)

CuCl 2 + Zn \u003d ZnCl 2 + Cu (5)

AgNO 3 + KCl \u003d AgCl + 2KNO 3 (6)

3HCl + Al(OH) 3 \u003d AlCl 3 + 3H 2 O (7)

Among the reactions of the compound, the following should be especially noted : substitution (5), exchange (6), and as a special case of the exchange reaction, the reaction neutralization (7).

Substitution reactions include those in which atoms of a simple substance replace the atoms of one of the elements in a complex substance. In example (5), zinc atoms replace copper atoms from the CuCl 2 solution, while zinc passes into the soluble ZnCl 2 salt, and copper is released from the solution in the metallic state.

Exchange reactions are those reactions in which two complex substances exchange their constituents. In the case of reaction (6), the soluble salts of AgNO 3 and KCl, when both solutions are drained, form an insoluble precipitate of the AgCl salt. At the same time, they exchange their constituent parts - cations and anions. Potassium cations K + are attached to NO 3 anions, and silver cations Ag + - to Cl - anions.

A special, particular case of exchange reactions is the neutralization reaction. Neutralization reactions are reactions in which acids react with bases to form salt and water. In example (7), hydrochloric acid HCl reacts with base Al(OH) 3 to form AlCl 3 salt and water. In this case, aluminum cations Al 3+ from the base are exchanged with Cl anions - from the acid. As a result, it happens hydrochloric acid neutralization.

Decomposition reactions include those in which two or more new simple or complex substances, but of a simpler composition, are formed from one complex one. As reactions, one can cite those in the process of which 1) decompose. potassium nitrate(KNO 3) with the formation of potassium nitrite (KNO 2) and oxygen (O 2); 2). Potassium permanganate(KMnO 4): potassium manganate is formed (K 2 MnO 4), manganese oxide(MnO 2) and oxygen (O 2); 3). calcium carbonate or marble; in the process are formed carbonicgas(CO 2) and calcium oxide(Cao)

2KNO 3 \u003d 2KNO 2 + O 2 (8)
2KMnO 4 \u003d K 2 MnO 4 + MnO 2 + O 2 (9)
CaCO 3 \u003d CaO + CO 2 (10)

In reaction (8), one complex and one simple substance is formed from a complex substance. In reaction (9) there are two complex and one simple. In reaction (10) there are two complex substances, but simpler in composition

All classes of complex substances undergo decomposition:

one). Oxides: silver oxide 2Ag 2 O = 4Ag + O 2 (11)

2). Hydroxides: iron hydroxide 2Fe(OH) 3 = Fe 2 O 3 + 3H 2 O (12)

3). Acids: sulphuric acid H 2 SO 4 \u003d SO 3 + H 2 O (13)

four). Salts: calcium carbonate CaCO 3 \u003d CaO + CO 2 (14)

5). organic matter: alcoholic fermentation of glucose

C 6 H 12 O 6 \u003d 2C 2 H 5 OH + 2CO 2 (15)

According to another classification, all chemical reactions can be divided into two types: reactions that take place with the release of heat, they are called exothermic, and reactions that go with the absorption of heat - endothermic. The criterion for such processes is thermal effect of the reaction. As a rule, exothermic reactions include oxidation reactions, i.e. interactions with oxygen methane combustion:

CH 4 + 2O 2 \u003d CO 2 + 2H 2 O + Q (16)

and to endothermic reactions - decomposition reactions, already given above (11) - (15). The Q sign at the end of the equation indicates whether heat is released during the reaction (+Q) or absorbed (-Q):

CaCO 3 \u003d CaO + CO 2 - Q (17)

You can also consider all chemical reactions according to the type of change in the degree of oxidation of the elements involved in their transformations. For example, in reaction (17), the elements participating in it do not change their oxidation states:

Ca +2 C +4 O 3 -2 \u003d Ca +2 O -2 + C +4 O 2 -2 (18)

And in reaction (16), the elements change their oxidation states:

2Mg 0 + O 2 0 \u003d 2Mg +2 O -2

These types of reactions are redox . They will be considered separately. To formulate equations for reactions of this type, it is necessary to use half-reaction method and apply electronic balance equation.

After bringing various types of chemical reactions, you can proceed to the principle of compiling chemical equations, in other words, the selection of coefficients in their left and right parts.

Mechanisms for compiling chemical equations.

Whatever type this or that chemical reaction belongs to, its record (chemical equation) must correspond to the condition of equality of the number of atoms before the reaction and after the reaction.

There are equations (17) that do not require adjustment, i.e. placement of coefficients. But in most cases, as in examples (3), (7), (15), it is necessary to take actions aimed at equalizing the left and right sides of the equation. What principles should be followed in such cases? Is there any system in the selection of coefficients? There is, and not one. These systems include:

one). Selection of coefficients according to given formulas.

2). Compilation according to the valencies of the reactants.

3). Compilation according to the oxidation states of the reactants.

In the first case, it is assumed that we know the formulas of the reactants both before and after the reaction. For example, given the following equation:

N 2 + O 2 →N 2 O 3 (19)

It is generally accepted that until the equality between the atoms of the elements before and after the reaction is established, the equal sign (=) is not put in the equation, but is replaced by an arrow (→). Now let's get down to the actual balancing. On the left side of the equation there are 2 nitrogen atoms (N 2) and two oxygen atoms (O 2), and on the right side there are two nitrogen atoms (N 2) and three oxygen atoms (O 3). It is not necessary to equalize it by the number of nitrogen atoms, but by oxygen it is necessary to achieve equality, since two atoms participated before the reaction, and after the reaction there were three atoms. Let's make the following diagram:

before reaction after reaction
O 2 O 3

Let's define the smallest multiple between the given numbers of atoms, it will be "6".

O 2 O 3
\ 6 /

Divide this number on the left side of the oxygen equation by "2". We get the number "3", put it in the equation to be solved:

N 2 + 3O 2 →N 2 O 3

We also divide the number "6" for the right side of the equation by "3". We get the number "2", just put it in the equation to be solved:

N 2 + 3O 2 → 2N 2 O 3

The number of oxygen atoms in both the left and right parts of the equation became equal, respectively, 6 atoms:

But the number of nitrogen atoms in both sides of the equation will not match:

On the left side there are two atoms, on the right side there are four atoms. Therefore, in order to achieve equality, it is necessary to double the amount of nitrogen on the left side of the equation, putting the coefficient "2":

Thus, the equality for nitrogen is observed and, in general, the equation will take the form:

2N 2 + 3O 2 → 2N 2 O 3

Now in the equation, instead of an arrow, you can put an equal sign:

2N 2 + 3O 2 \u003d 2N 2 O 3 (20)

Let's take another example. The following reaction equation is given:

P + Cl 2 → PCl 5

On the left side of the equation there is 1 phosphorus atom (P) and two chlorine atoms (Cl 2), and on the right side there is one phosphorus atom (P) and five oxygen atoms (Cl 5). It is not necessary to equalize it by the number of phosphorus atoms, but for chlorine it is necessary to achieve equality, since two atoms participated before the reaction, and after the reaction there were five atoms. Let's make the following diagram:

before reaction after reaction
Cl 2 Cl 5

Let's define the smallest multiple between the given numbers of atoms, it will be "10".

Cl 2 Cl 5
\ 10 /

Divide this number on the left side of the equation for chlorine by "2". We get the number "5", put it in the equation to be solved:

Р + 5Cl 2 → РCl 5

We also divide the number "10" for the right side of the equation by "5". We get the number "2", just put it in the equation to be solved:

Р + 5Cl 2 → 2РCl 5

The number of chlorine atoms in both the left and right parts of the equation became equal, respectively, 10 atoms:

But the number of phosphorus atoms in both sides of the equation will not match:

Therefore, in order to achieve equality, it is necessary to double the amount of phosphorus on the left side of the equation, putting the coefficient "2":

Thus, the equality for phosphorus is observed and, in general, the equation will take the form:

2Р + 5Cl 2 = 2РCl 5 (21)

When writing equations by valency must be given definition of valency and set values ​​for the most famous elements. Valence is one of the previously used concepts, currently not used in a number of school programs. But with its help it is easier to explain the principles of compiling equations of chemical reactions. By valency is meant number chemical bonds, which one or another atom can form with another, or other atoms . Valence has no sign (+ or -) and is indicated by Roman numerals, usually above the symbols of chemical elements, for example:

Where do these values ​​come from? How to apply them in the preparation of chemical equations? The numerical values ​​of the valencies of the elements coincide with their group number Periodic system chemical elements D. I. Mendeleev (Table 1).

For other elements valency values may have other values, but never greater than the number of the group in which they are located. Moreover, for even numbers of groups (IV and VI), the valences of elements take only even values, and for odd ones, they can have both even and odd values ​​(Table.2).

Of course, there are exceptions to the valency values ​​for some elements, but in each specific case, these points are usually specified. Now consider general principle compiling chemical equations for given valences for certain elements. Most often, this method is acceptable in the case of compiling equations of chemical reactions of the compound simple substances, for example, when interacting with oxygen ( oxidation reactions). Suppose you want to display the oxidation reaction aluminum. But recall that metals are denoted by single atoms (Al), and non-metals that are in a gaseous state - with indices "2" - (O 2). First, we write the general scheme of the reaction:

Al + O 2 → AlO

At this stage, it is not yet known what the correct spelling should be for alumina. And it is precisely at this stage that knowledge of the valencies of the elements will come to our aid. For aluminum and oxygen, we put them above the proposed formula for this oxide:

IIIII
Al O

After that, "cross"-on-"cross" these symbols of the elements will put the corresponding indices below:

IIIII
Al 2 O 3

Composition of a chemical compound Al 2 O 3 determined. The further scheme of the reaction equation will take the form:

Al + O 2 → Al 2 O 3

It remains only to equalize the left and right parts of it. We proceed in the same way as in the case of formulating equation (19). We equalize the number of oxygen atoms, resorting to finding the smallest multiple:

before reaction after reaction

O 2 O 3
\ 6 /

Divide this number on the left side of the oxygen equation by "2". We get the number "3", put it in the equation to be solved. We also divide the number "6" for the right side of the equation by "3". We get the number "2", just put it in the equation to be solved:

Al + 3O 2 → 2Al 2 O 3

In order to achieve equality for aluminum, it is necessary to adjust its amount on the left side of the equation by setting the coefficient "4":

4Al + 3O 2 → 2Al 2 O 3

Thus, the equality for aluminum and oxygen is observed and, in general, the equation will take the final form:

4Al + 3O 2 \u003d 2Al 2 O 3 (22)

Using the valency method, it is possible to predict which substance is formed in the course of a chemical reaction, what its formula will look like. Suppose nitrogen and hydrogen with the corresponding valences III and I entered into the reaction of the compound. Let's write the general reaction scheme:

N 2 + H 2 → NH

For nitrogen and hydrogen, we put down the valencies over the proposed formula of this compound:

As before, "cross"-on-"cross" for these element symbols, we put the corresponding indices below:

III I
N H 3

The further scheme of the reaction equation will take the form:

N 2 + H 2 → NH 3

Equalizing in the already known way, through the smallest multiple for hydrogen, equal to "6", we obtain the desired coefficients, and the equation as a whole:

N 2 + 3H 2 \u003d 2NH 3 (23)

When compiling equations for oxidation states reacting substances, it must be recalled that the degree of oxidation of an element is the number of electrons received or given away in the process of a chemical reaction. The oxidation state in compounds basically, numerically coincides with the values ​​of the element's valences. But they differ in sign. For example, for hydrogen, the valence is I, and the oxidation state is (+1) or (-1). For oxygen, the valence is II, and the oxidation state is (-2). For nitrogen, the valencies are I, II, III, IV, V, and the oxidation states are (-3), (+1), (+2), (+3), (+4), (+5), etc. . The oxidation states of the elements most commonly used in equations are shown in Table 3.

In the case of compound reactions, the principle of compiling equations in terms of oxidation states is the same as in compiling in terms of valencies. For example, let's give the reaction equation for the oxidation of chlorine with oxygen, in which chlorine forms a compound with an oxidation state of +7. Let's write the proposed equation:

Cl 2 + O 2 → ClO

We put the oxidation states of the corresponding atoms over the proposed ClO compound:

As in the previous cases, we establish that the desired compound formula will take the form:

7 -2
Cl 2 O 7

The reaction equation will take the following form:

Cl 2 + O 2 → Cl 2 O 7

Equalizing for oxygen, finding the smallest multiple between two and seven, equal to "14", we finally establish the equality:

2Cl 2 + 7O 2 \u003d 2Cl 2 O 7 (24)

A slightly different method must be used with oxidation states when compiling exchange, neutralization, and substitution reactions. In some cases, it is difficult to find out: what compounds are formed during the interaction of complex substances?

How do you know what happens in a reaction?

Indeed, how do you know: what reaction products can arise in the course of a particular reaction? For example, what is formed when barium nitrate and potassium sulfate react?

Ba (NO 3) 2 + K 2 SO 4 →?

Maybe VAC 2 (NO 3) 2 + SO 4? Or Ba + NO 3 SO 4 + K 2? Or something else? Of course, during this reaction, compounds are formed: BaSO 4 and KNO 3. And how is this known? And how to write formulas of substances? Let's start with what is most often overlooked: the very concept of "exchange reaction". This means that in these reactions, the substances change with each other in constituent parts. Since the exchange reactions are mostly carried out between bases, acids or salts, the parts with which they will change are metal cations (Na +, Mg 2+, Al 3+, Ca 2+, Cr 3+), H + ions or OH -, anions - acid residues, (Cl -, NO 3 2-, SO 3 2-, SO 4 2-, CO 3 2-, PO 4 3-). In general, the exchange reaction can be given in the following notation:

Kt1An1 + Kt2An1 = Kt1An2 + Kt2An1 (25)

Where Kt1 and Kt2 are the metal cations (1) and (2), and An1 and An2 are the anions (1) and (2) corresponding to them. In this case, it must be taken into account that in compounds before and after the reaction, cations are always established in the first place, and anions in the second. Therefore, if it reacts potassium chloride and silver nitrate, both in solution

KCl + AgNO 3 →

then in the process of it substances KNO 3 and AgCl are formed and the corresponding equation will take the form:

KCl + AgNO 3 \u003d KNO 3 + AgCl (26)

In neutralization reactions, protons from acids (H +) will combine with hydroxyl anions (OH -) to form water (H 2 O):

HCl + KOH \u003d KCl + H 2 O (27)

The oxidation states of metal cations and the charges of anions of acid residues are indicated in the table of the solubility of substances (acids, salts and bases in water). Metal cations are shown horizontally, and anions of acid residues are shown vertically.

Based on this, when compiling the equation for the exchange reaction, it is first necessary to establish the oxidation states of the particles receiving in this chemical process in its left part. For example, you need to write an equation for the interaction between calcium chloride and sodium carbonate. Let's draw up the initial scheme for this reaction:

CaCl + NaCO 3 →

Ca 2+ Cl - + Na + CO 3 2- →

Having performed the already known “cross”-to-“cross” action, we determine the real formulas of the starting substances:

CaCl 2 + Na 2 CO 3 →

Based on the principle of exchange of cations and anions (25), we establish the preliminary formulas of the substances formed during the reaction:

CaCl 2 + Na 2 CO 3 → CaCO 3 + NaCl

We put down the corresponding charges over their cations and anions:

Ca 2+ CO 3 2- + Na + Cl -

Substance formulas are written correctly, in accordance with the charges of cations and anions. Let's make a complete equation by equating the left and right parts of it in terms of sodium and chlorine:

CaCl 2 + Na 2 CO 3 \u003d CaCO 3 + 2NaCl (28)

As another example, here is the equation for the neutralization reaction between barium hydroxide and phosphoric acid:

VaON + NPO 4 →

We put the corresponding charges over cations and anions:

Ba 2+ OH - + H + RO 4 3- →

Let's define the real formulas of the starting substances:

Va (OH) 2 + H 3 RO 4 →

Based on the principle of exchange of cations and anions (25), we establish the preliminary formulas of the substances formed during the reaction, taking into account that in the exchange reaction, one of the substances must necessarily be water:

Ba (OH) 2 + H 3 RO 4 → Ba 2+ RO 4 3- + H 2 O

Let's determine the correct record of the formula of the salt formed during the reaction:

Ba (OH) 2 + H 3 RO 4 → Ba 3 (RO 4) 2 + H 2 O

Equate the left side of the equation for barium:

3VA (OH) 2 + H 3 RO 4 → Ba 3 (RO 4) 2 + H 2 O

Since on the right side of the equation the residue of phosphoric acid is taken twice, (PO 4) 2, then on the left it is also necessary to double its amount:

3VA (OH) 2 + 2H 3 RO 4 → Ba 3 (RO 4) 2 + H 2 O

It remains to match the number of hydrogen and oxygen atoms on the right side of the water. Since the total number of hydrogen atoms on the left is 12, on the right it must also correspond to twelve, therefore, before the water formula, it is necessary put a coefficient"6" (since there are already 2 hydrogen atoms in the water molecule). For oxygen, equality is also observed: on the left 14 and on the right 14. So, the equation has the correct form of writing:

3Ва (ОН) 2 + 2Н 3 РО 4 → Ва 3 (РО 4) 2 + 6Н 2 O (29)

Possibility of chemical reactions

The world is made up of a great variety of substances. The number of variants of chemical reactions between them is also incalculable. But can we, having written this or that equation on paper, assert that a chemical reaction will correspond to it? There is a misconception that if the right arrange odds in the equation, then it will be feasible in practice. For example, if we take sulfuric acid solution and drop into it zinc, then we can observe the process of hydrogen evolution:

Zn + H 2 SO 4 \u003d ZnSO 4 + H 2 (30)

But if copper is lowered into the same solution, then the process of gas evolution will not be observed. The reaction is not feasible.

Cu + H 2 SO 4 ≠

If concentrated sulfuric acid is taken, it will react with copper:

Cu + 2H 2 SO 4 \u003d CuSO 4 + SO 2 + 2H 2 O (31)

In reaction (23) between nitrogen and hydrogen gases, thermodynamic balance, those. how many molecules ammonia NH 3 is formed per unit time, the same number of them will decompose back into nitrogen and hydrogen. Shift in chemical equilibrium can be achieved by increasing the pressure and decreasing the temperature

N 2 + 3H 2 \u003d 2NH 3

If you take potassium hydroxide solution and pour on it sodium sulfate solution, then no changes will be observed, the reaction will not be feasible:

KOH + Na 2 SO 4 ≠

Sodium chloride solution when interacting with bromine, it will not form bromine, despite the fact that this reaction can be attributed to a substitution reaction:

NaCl + Br 2 ≠

What are the reasons for such discrepancies? The fact is that it is not enough just to correctly define compound formulas, it is necessary to know the specifics of the interaction of metals with acids, to skillfully use the table of solubility of substances, to know the rules of substitution in the series of activity of metals and halogens. This article outlines only the most basic principles of how arrange the coefficients in the reaction equations, how write molecular equations, how determine the composition of a chemical compound.

Chemistry, as a science, is extremely diverse and multifaceted. This article reflects only a small part of the processes taking place in the real world. Types, thermochemical equations, electrolysis, organic synthesis processes and much, much more. But more on that in future articles.

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Solving school problems in chemistry can present some difficulties for schoolchildren, so we lay out a number of examples of solving the main types of problems in school chemistry with a detailed analysis.

To solve problems in chemistry, you need to know a number of formulas indicated in the table below. Properly using this simple set, you can solve almost any problem from the course of chemistry.

Substance Calculations	Share calculations	Reaction Product Yield Calculations
ν=m/M, ν=V/V M , ν=N/N A , ν=PV/RT	ω=m h / m about, φ \u003d V h / V about, χ=ν h / ν about	η = m pr. /m theor. , η = V pr. / V theor. , η = ν ex. / ν theor.
ν is the amount of substance (mol); ν h - the amount of substance private (mol); ν about - the amount of substance total (mol); m is the mass (g); m h - quotient mass (g); m about - total weight (g); V - volume (l); V M - volume 1 mol (l); V h - private volume (l); V about - total volume (l); N is the number of particles (atoms, molecules, ions); N A - Avogadro's number (the number of particles in 1 mol of a substance) N A \u003d 6.02 × 10 23; Q is the amount of electricity (C); F is the Faraday constant (F » 96500 C); P - pressure (Pa) (1 atm "10 5 Pa); R is the universal gas constant R » 8.31 J/(mol×K); T is the absolute temperature (K); ω is the mass fraction; φ is the volume fraction; χ is the mole fraction; η is the yield of the reaction product; m pr., V pr., ν pr. - mass, volume, amount of substance practical; m theor.,V theor., ν theor. - mass, volume, amount of substance theoretical.

Calculating the mass of a certain amount of a substance

Exercise:

Determine the mass of 5 moles of water (H 2 O).

Solution:

1. Calculate the molar mass of a substance using the periodic table of D. I. Mendeleev. The masses of all atoms are rounded up to units, chlorine - up to 35.5.
   M(H 2 O)=2×1+16=18 g/mol
2. Find the mass of water using the formula:
   m \u003d ν × M (H 2 O) \u003d 5 mol × 18 g / mol \u003d 90 g
3. Record response:
   Answer: The mass of 5 moles of water is 90 g.

Solute Mass Fraction Calculation

Exercise:

Calculate the mass fraction of salt (NaCl) in the solution obtained by dissolving 25 g of salt in 475 g of water.

Solution:

1. Write down the formula for finding the mass fraction:
   ω (%) \u003d (m in-va / m solution) × 100%
2. Find the mass of the solution.
   m solution \u003d m (H 2 O) + m (NaCl) \u003d 475 + 25 \u003d 500 g
3. Calculate the mass fraction by substituting the values ​​into the formula.
   ω (NaCl) \u003d (m in-va / m solution) × 100% = (25/500)×100%=5%
4. Write down the answer.
   Answer: the mass fraction of NaCl is 5%

Calculation of the mass of a substance in a solution by its mass fraction

Exercise:

How many grams of sugar and water must be taken to obtain 200 g of a 5% solution?

Solution:

1. Write down the formula for determining the mass fraction of a solute.
   ω=m in-va /m r-ra → m in-va = m r-ra ×ω
2. Calculate the mass of salt.
   m in-va (salt) \u003d 200 × 0.05 \u003d 10 g
3. Determine the mass of water.
   m (H 2 O) \u003d m (solution) - m (salt) \u003d 200 - 10 \u003d 190 g
4. Write down the answer.
   Answer: you need to take 10 g of sugar and 190 g of water

Determination of the yield of the reaction product in% of the theoretically possible

Exercise:

Calculate the yield of ammonium nitrate (NH 4 NO 3) in% of the theoretically possible if 380 g of fertilizer was obtained by passing 85 g of ammonia (NH 3) into a solution of nitric acid (HNO 3).

Solution:

1. Write the equation of a chemical reaction and arrange the coefficients
   NH 3 + HNO 3 \u003d NH 4 NO 3
2. Write the data from the condition of the problem above the reaction equation.
   

   m = 85 g				m pr. = 380 g
   NH3	+	HNO3	=	NH4NO3

3. Under the formulas of substances, calculate the amount of the substance according to the coefficients as the product of the amount of the substance and the molar mass of the substance:
   
4. Practically obtained mass of ammonium nitrate is known (380 g). In order to determine the theoretical mass of ammonium nitrate, draw up a proportion
   85/17=x/380
   
5. Solve the equation, find x.
   x=400 g theoretical mass of ammonium nitrate
6. Determine the yield of the reaction product (%), referring the practical mass to the theoretical one and multiply by 100%
   η=m pr. /m theor. =(380/400)×100%=95%
7. Write down the answer.
   Answer: the yield of ammonium nitrate was 95%.

Calculation of the mass of the product from the known mass of the reagent containing a certain proportion of impurities

Exercise:

Calculate the mass of calcium oxide (CaO) obtained by firing 300 g of limestone (CaCO 3) containing 10% impurities.

Solution:

1. Write down the equation of the chemical reaction, put the coefficients.
   CaCO 3 \u003d CaO + CO 2
2. Calculate the mass of pure CaCO 3 contained in limestone.
   ω (pure) \u003d 100% - 10% \u003d 90% or 0.9;
   m (CaCO 3) \u003d 300 × 0.9 \u003d 270 g
3. The resulting mass of CaCO 3 is written over the formula CaCO 3 in the reaction equation. The desired mass of CaO is denoted by x.
   

   270 g		x r
   CaCO 3	=	Cao	+	CO 2

4. Under the formulas of substances in the equation, write the amount of the substance (according to the coefficients); the product of the quantities of substances by their molar mass (molecular mass of CaCO 3 \u003d 100 , CaO = 56 ).
   
5. Set up a proportion.
   270/100=x/56
6. Solve the equation.
   x = 151.2 g
7. Write down the answer.
   Answer: the mass of calcium oxide will be 151.2 g

Calculation of the mass of the reaction product, if the yield of the reaction product is known

Exercise:

How many g of ammonium nitrate (NH 4 NO 3) can be obtained by reacting 44.8 liters of ammonia (n.a.) with nitric acid, if it is known that the practical yield is 80% of the theoretically possible?

Solution:

1. Write down the equation of the chemical reaction, arrange the coefficients.
   NH 3 + HNO 3 \u003d NH 4 NO 3
2. Write these conditions of the problem above the reaction equation. The mass of ammonium nitrate is denoted by x.
   
3. Under the reaction equation write:
   a) the amount of substances according to the coefficients;
   b) the product of the molar volume of ammonia by the amount of substance; the product of the molar mass of NH 4 NO 3 by the amount of substance.
4. Set up a proportion.
   44.4/22.4=x/80
5. Solve the equation by finding x (theoretical mass of ammonium nitrate):
   x \u003d 160 g.
6. Find the practical mass of NH 4 NO 3 by multiplying the theoretical mass by the practical yield (in fractions of one)
   m (NH 4 NO 3) \u003d 160 × 0.8 \u003d 128 g
7. Write down the answer.
   Answer: the mass of ammonium nitrate will be 128 g.

Determining the mass of the product if one of the reagents is taken in excess

Exercise:

14 g of calcium oxide (CaO) was treated with a solution containing 37.8 g of nitric acid (HNO 3 ). Calculate the mass of the reaction product.

Solution:

1. Write the reaction equation, arrange the coefficients
   CaO + 2HNO 3 \u003d Ca (NO 3) 2 + H 2 O
2. Determine the mole of reagents using the formula: ν = m/M
   ν(CaO) = 14/56=0.25 mol;
   ν (HNO 3) \u003d 37.8 / 63 \u003d 0.6 mol.
   
3. Above the reaction equation, write the calculated amounts of the substance. Under the equation - the amount of substance according to stoichiometric coefficients.
   
4. Determine the substance taken in deficiency by comparing the ratios of the taken amounts of substances to stoichiometric coefficients.
   0,25/1 < 0,6/2
   Consequently, nitric acid is taken in deficiency. From it we will determine the mass of the product.
5. Under the formula of calcium nitrate (Ca (NO 3) 2) in the equation, put down:
   a) the amount of substance, according to the stoichiometric coefficient;
   b) the product of the molar mass by the amount of substance. Above the formula (Ca (NO 3) 2) - x g.

   0.25 mol		0.6 mol		x r
   CaO	+	2HNO 3	=	Ca(NO 3) 2	+	H2O
   1 mol		2 mol		1 mol
   				m = 1×164 g

6. Make a proportion
   0.25/1=x/164
7. Determine x
   x = 41 g
8. Write down the answer.
   Answer: the mass of salt (Ca (NO 3) 2) will be 41 g.

Calculations by thermochemical reaction equations

Exercise:

How much heat will be released when 200 g of copper (II) oxide (CuO) is dissolved in hydrochloric acid (aqueous HCl solution), if the thermochemical reaction equation:

CuO + 2HCl \u003d CuCl 2 + H 2 O + 63.6 kJ

Solution:

1. Write the data from the condition of the problem above the reaction equation
   
2. Under the copper oxide formula, write its amount (according to the coefficient); the product of the molar mass and the amount of the substance. Put x above the amount of heat in the reaction equation.
   

   200 g
   CuO	+	2HCl	=	CuCl 2	+	H2O	+	63.6 kJ
   1 mol
   m = 1×80 g

3. Set up a proportion.
   200/80=x/63.6
4. Calculate x.
   x=159 kJ
5. Write down the answer.
   Answer: when 200 g of CuO is dissolved in hydrochloric acid, 159 kJ of heat will be released.

Drawing up a thermochemical equation

Exercise:

When burning 6 g of magnesium, 152 kJ of heat is released. Write a thermochemical equation for the formation of magnesium oxide.

Solution:

1. Write an equation for a chemical reaction showing the release of heat. Arrange the coefficients.
   2Mg + O 2 \u003d 2MgO + Q
2. 
   

   6 g						152
   2Mg	+	O2	=	2MgO	+	Q

3. Under the formulas of substances write:
   a) the amount of substance (according to the coefficients);
   b) the product of the molar mass by the amount of substance. Place x under the heat of the reaction.
   
4. Set up a proportion.
   6/(2×24)=152/x
5. Calculate x (amount of heat, according to the equation)
   x=1216 kJ
6. Write down the thermochemical equation in the answer.
   Answer: 2Mg + O 2 = 2MgO + 1216 kJ

Calculation of gas volumes according to chemical equations

Exercise:

When ammonia (NH 3) is oxidized with oxygen in the presence of a catalyst, nitric oxide (II) and water are formed. What volume of oxygen will react with 20 liters of ammonia?

Solution:

1. Write the reaction equation and arrange the coefficients.
   4NH 3 + 5O 2 \u003d 4NO + 6H 2 O
2. Write the data from the condition of the problem above the reaction equation.
   

   20 l		x
   4NH3	+	5O2	=	4NO	+	6H2O

3. Under the reaction equation, write down the amounts of substances according to the coefficients.
   
4. Set up a proportion.
   20/4=x/5
5. Find x.
   x= 25 l
6. Write down the answer.
   Answer: 25 liters of oxygen.

Determination of the volume of a gaseous product from a known mass of a reagent containing impurities

Exercise:

What volume (n.c.) of carbon dioxide (CO 2) will be released when 50 g of marble (CaCO 3) containing 10% impurities in hydrochloric acid are dissolved?

Solution:

1. Write the equation of a chemical reaction, arrange the coefficients.
   CaCO 3 + 2HCl \u003d CaCl 2 + H 2 O + CO 2
2. Calculate the amount of pure CaCO 3 contained in 50 g of marble.
   ω (CaCO 3) \u003d 100% - 10% \u003d 90%
   To convert to fractions of one, divide by 100%.
   w (CaCO 3) \u003d 90% / 100% \u003d 0.9
   m (CaCO 3) \u003d m (marble) × w (CaCO 3) \u003d 50 × 0.9 \u003d 45 g
3. Write the resulting value over calcium carbonate in the reaction equation. Above CO 2 put x l.
   

   45 g								x
   CaCO3	+	2HCl	=	CaCl2	+	H2O	+	CO2

4. Under the formulas of substances write:
   a) the amount of substance, according to the coefficients;
   b) the product of the molar mass by the amount of substance, if we are talking about the mass of the substance, and the product of the molar volume by the amount of the substance, if we are talking about the volume of the substance.

   Calculation of the composition of the mixture according to the chemical reaction equation

   Exercise:

   The complete combustion of a mixture of methane and carbon monoxide (II) required the same volume of oxygen. Determine composition gas mixture in volume fractions.

   Solution:

   1. Write down the reaction equations, arrange the coefficients.
      CO + 1/2O 2 = CO 2
      CH 4 + 2O 2 \u003d CO 2 + 2H 2 O
   2. Designate the amount of carbon monoxide (CO) as x, and the amount of methane as y
      

   45 g								x
   CaCO3	+	2HCl	=

   X
   SO	+	1/2O 2	=	CO 2
   at
   CH 4	+	2O 2	=	CO 2	+	2H 2 O

5. Determine the amount of oxygen that will be consumed for combustion x moles of CO and y moles of CH 4.
   

   X		0.5 x
   SO	+	1/2O 2	=	CO 2
   at		2y
   CH 4	+	2O 2	=	CO 2	+	2H 2 O

6. Make a conclusion about the ratio of the amount of oxygen substance and gas mixture.
   The equality of the volumes of gases indicates the equality of the quantities of matter.
7. Write an equation.
   x + y = 0.5x + 2y
8. Simplify the equation.
   0.5 x = y
9. Take the amount of CO for 1 mol and determine the required amount of CH 4.
   If x=1 then y=0.5
10. Find the total amount of the substance.
    x + y = 1 + 0.5 = 1.5
11. Determine the volume fraction of carbon monoxide (CO) and methane in the mixture.
    φ(CO) \u003d 1 / 1.5 \u003d 2/3
    φ (CH 4) \u003d 0.5 / 1.5 \u003d 1/3
12. Write down the answer.
    Answer: the volume fraction of CO is 2/3, and CH 4 is 1/3.

Reference material:

periodic table

Solubility table

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