"Not for flour, but for science."
(Folk wisdom)
Calculations using chemical reaction equations
Classification of chemical reactions. Reactions of connection, decomposition, substitution, double exchange, redox reactions. Equations of chemical reactions. Selection of stoichiometric coefficients in reaction equations. Calculations using reaction equations. Determination of the amount of substance and mass of reagents and products. Determination of the volume of gaseous reagents and products. Theoretical and practical yield of the reaction product. Degree of purity of chemicals.
Examples of solving typical problems
Task 1. During X-ray examination of the human body, so-called radiocontrast agents are used. So, before scanning the stomach, the patient is given a suspension of insoluble barium sulfate, which does not transmit x-rays, to drink. What quantities of barium oxide and sulfuric acid are required to produce 100 barium sulfate?
Solution.
BaO + H 2 SO 4 = BaSO 4 + H 2 O
m(BaSO 4) = 100 g; M(BaSO 4) = 233 g/mol
n(BaO) = ?
n(H 2 SO 4) = ?
In accordance with the coefficients of the reaction equation, which in our case are all equal to 1, to obtain a given amount of BaSO 4 it is required:
n(BaO) = n(BaSO 4) = m(BaSO 4) / M(BaSO 4) = 100: 233
[g: (g/mol)] = 0.43 mol
n(H2SO4) = n(BaSO 4) = m(BaSO 4) / M(BaSO 4) = 100: 233
[g: (g/mol)] = 0.43 mol
Answer. To obtain 100 g of barium sulfate, 0.43 mol of barium oxide and 0.43 mol of sulfuric acid are required.
Task 2. Before liquid laboratory waste containing hydrochloric acid is poured down the drain, it is necessary to neutralize it with alkali (for example, sodium hydroxide) or soda (sodium carbonate). Determine the masses of NaOH and Na 2 CO 3 required to neutralize waste containing 0.45 mol HCl. What volume of gas (at normal conditions) will be released when the specified amount of waste is neutralized with soda?
Solution. Let us write the reaction equations and the conditions of the problem in formula form:
(1) HCl + NaOH = NaCl + H 2 O
(2) 2HCl + Na 2 CO 3 = 2NaCl + H 2 O + CO 2
n(HCl) = 0.45 mol; M(NaOH) = 40 g/mol;
M(Na 2 CO 3) = 106 g/mol; V M = 22.4 l/mol (n.s.)
n(NaOH) = ? m(NaOH) = ?
n(Na 2 CO 3) = ? m(Na 2 CO 3) = ?
V(CO2) = ? (Well.)
To neutralize a given amount of HCl in accordance with reaction equations (1) and (2), it is required:
n(NaOH) = n(HCl) = 0.45 mol;
m(NaOH) = n(NaOH). M(NaOH) = 0.45. 40
[mol. g/mol] = 18 g
n(Na 2 CO 3) = n
m(Na 2 CO 3) = n(Na 2 CO 3) / M(Na 2 CO 3) = 0.225. 106
[mol. g/mol] = 23.85 g
To calculate the volume of carbon dioxide released during neutralization according to reaction (2), an additional equation is used that relates the amount of a gaseous substance, its volume and molar volume:
n(CO2) = n(HCl) / 2 = 0.45: 2 [mol] = 0.225 mol;
V(CO2) = n(CO2). V M = 0.225. 22.4 [mol. l/mol] = 5.04 l
Answer. 18 g NaOH; 23.85 g Na 2 CO 3; 5.04 l CO 2
Task 3. Antoine-Laurent Lavoisier discovered the nature of combustion of various substances in oxygen after his famous twelve-day experiment. In this experiment, he first heated a sample of mercury in a sealed retort for a long time, and later (and at a higher temperature) he heated mercury(II) oxide formed at the first stage of the experiment. This released oxygen, and Lavoisier became, together with Joseph Priestley and Karl Scheele, the discoverer of this important chemical element. Calculate the amount and volume of oxygen (at normal conditions) collected during the decomposition of 108.5 g of HgO.
Solution. Let us write the reaction equation and the problem condition in formula form:
2HgO = 2Hg + O2
m(HgO) = 108.5 g; M(HgO) = 217 g/mol
V M = 22.4 l/mol (n.s.)
V(O2) = ? (Well.)
Amount of oxygen n(O 2), which is released during the decomposition of mercury(II) oxide, is:
n(O2) = 1/2 n(HgO) = 1/2 m(HgO) / M(HgO) = 108.5 / (217.2)
[g: (g/mol)] = 0.25 mol,
and its volume at ground level - V(O2) = n(O2). V M = 0.25. 22.4
[mol. l/mol] = 5.6 l
Answer. 0.25 mol, or 5.6 liters (at standard conditions) of oxygen.
Task 4. The most important problem in the industrial production of fertilizers is the production of so-called “fixed nitrogen”. Currently, it is solved by synthesizing ammonia from nitrogen and hydrogen. What volume of ammonia (at standard conditions) can be obtained in this process if the volume of the initial hydrogen is 300 l and the practical yield (z) is 43%?
Solution. Let us write the reaction equation and the problem condition in formula form:
N2 + 3H2 = 2NH3
V(H 2) = 300 l; z(NH 3) = 43% = 0.43
V(NH 3) = ? (Well.)
Ammonia volume V(NH 3), which can be obtained in accordance with the problem condition, is:
V(NH 3) pract = V(NH 3) theor. z(NH 3) = 2/3. V(H2) . z(NH 3) =
2/3. 300. 0.45 [l] = 86 l
Answer. 86 l (at no.) of ammonia.
If the typical tasks given here are completely clear to you, proceed to solve them.
When solving computational chemical problems, it is necessary to be able to perform calculations using the equation of a chemical reaction. The lesson is devoted to studying the algorithm for calculating the mass (volume, quantity) of one of the reaction participants from the known mass (volume, quantity) of another reaction participant.
Topic: Substances and their transformations
Lesson:Calculations using the chemical reaction equation
Consider the reaction equation for the formation of water from simple substances:
2H 2 + O 2 = 2H 2 O
We can say that two molecules of water are formed from two molecules of hydrogen and one molecule of oxygen. On the other hand, the same entry says that for the formation of every two moles of water, you need to take two moles of hydrogen and one mole of oxygen.
The molar ratio of reaction participants helps to make calculations important for chemical synthesis. Let's look at examples of such calculations.
TASK 1. Let us determine the mass of water formed as a result of the combustion of hydrogen in 3.2 g of oxygen.
To solve this problem, you first need to create an equation for a chemical reaction and write down the given conditions of the problem over it.
If we knew the amount of oxygen that reacted, we could determine the amount of water. And then, we would calculate the mass of water, knowing its amount of substance and. To find the amount of oxygen, you need to divide the mass of oxygen by its molar mass.
Molar mass is numerically equal to relative mass. For oxygen, this value is 32. Let’s substitute it into the formula: the amount of oxygen substance is equal to the ratio of 3.2 g to 32 g/mol. It turned out to be 0.1 mol.
To find the amount of water substance, let’s leave the proportion using the molar ratio of the reaction participants:
For every 0.1 mole of oxygen there is an unknown amount of water, and for every 1 mole of oxygen there are 2 moles of water.
Hence the amount of water substance is 0.2 mol.
To determine the mass of water, you need to multiply the found value of the amount of water by its molar mass, i.e. multiply 0.2 mol by 18 g/mol, we get 3.6 g of water.
Rice. 1. Recording a brief condition and solution to Problem 1
In addition to the mass, you can calculate the volume of the gaseous reaction participant (at normal conditions) using a formula known to you, according to which the volume of gas at normal conditions. equal to the product of the amount of gas substance and the molar volume. Let's look at an example of solving a problem.
TASK 2. Let's calculate the volume of oxygen (at normal conditions) released during the decomposition of 27 g of water.
Let us write down the reaction equation and the given conditions of the problem. To find the volume of oxygen released, you must first find the amount of water substance through the mass, then, using the reaction equation, determine the amount of oxygen substance, after which you can calculate its volume at ground level.
The amount of water substance is equal to the ratio of the mass of water to its molar mass. We get a value of 1.5 mol.
Let's make a proportion: from 1.5 moles of water an unknown amount of oxygen is formed, from 2 moles of water 1 mole of oxygen is formed. Hence the amount of oxygen is 0.75 mol. Let's calculate the volume of oxygen at normal conditions. It is equal to the product of the amount of oxygen and the molar volume. The molar volume of any gaseous substance at ambient conditions. equal to 22.4 l/mol. Substituting the numerical values into the formula, we obtain a volume of oxygen equal to 16.8 liters.
Rice. 2. Recording a brief condition and solution to Problem 2
Knowing the algorithm for solving such problems, it is possible to calculate the mass, volume or amount of substance of one of the reaction participants from the mass, volume or amount of substance of another reaction participant.
1. Collection of problems and exercises in chemistry: 8th grade: for textbooks. P.A. Orzhekovsky and others. “Chemistry. 8th grade” / P.A. Orzhekovsky, N.A. Titov, F.F. Hegel. - M.: AST: Astrel, 2006. (p.40-48)
2. Ushakova O.V. Chemistry workbook: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry. 8th grade” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; under. ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006. (p. 73-75)
3. Chemistry. 8th grade. Textbook for general education institutions / P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova. - M.: Astrel, 2013. (§23)
4. Chemistry: 8th grade: textbook. for general education institutions / P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. M.: AST: Astrel, 2005. (§29)
5. Chemistry: inorganic. chemistry: textbook. for 8th grade general education establishment /G.E. Rudzitis, F.G. Feldman. - M.: Education, OJSC “Moscow Textbooks”, 2009. (p.45-47)
6. Encyclopedia for children. Volume 17. Chemistry / Chapter. ed.V.A. Volodin, Ved. scientific ed. I. Leenson. - M.: Avanta+, 2003.
Additional web resources
2. Unified collection of digital educational resources ().
Homework
1) p. 73-75 No. 2, 3, 5 from the Workbook in Chemistry: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry. 8th grade” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; under. ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006.
2) p. 135 No. 3,4 from the textbook P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova “Chemistry: 8th grade,” 2013
Stoichiometry- quantitative relationships between reacting substances.
If reagents enter into a chemical interaction in strictly defined quantities, and as a result of the reaction substances are formed, the amount of which can be calculated, then such reactions are called stoichiometric.
Laws of stoichiometry:
The coefficients in chemical equations before the formulas of chemical compounds are called stoichiometric.
All calculations using chemical equations are based on the use of stoichiometric coefficients and are associated with finding quantities of a substance (number of moles).
The amount of substance in the reaction equation (number of moles) = the coefficient in front of the corresponding molecule.
N A=6.02×10 23 mol -1.
η - ratio of the actual mass of the product m p to a theoretically possible m t, expressed in fractions of a unit or as a percentage.
If the yield of reaction products is not indicated in the condition, then in the calculations it is taken equal to 100% (quantitative yield).
Calculation scheme using chemical reaction equations:
- Write an equation for a chemical reaction.
- Above the chemical formulas of substances write known and unknown quantities with units of measurement.
- Under the chemical formulas of substances with known and unknowns, write down the corresponding values of these quantities found from the reaction equation.
- Compose and solve a proportion.
Example. Calculate the mass and amount of magnesium oxide formed during complete combustion of 24 g of magnesium.
|
Given: m(Mg) = 24 g Find: ν (MgO) m (MgO) |
Solution: 1. Let's create an equation for a chemical reaction: 2Mg + O 2 = 2MgO. 2. Under the formulas of substances we indicate the amount of substance (number of moles) that corresponds to the stoichiometric coefficients: 2Mg + O2 = 2MgO 2 mole 2 mole 3. Determine the molar mass of magnesium: Relative atomic mass of magnesium Ar (Mg) = 24. Because the molar mass value is equal to the relative atomic or molecular mass, then M (Mg)= 24 g/mol. 4. Using the mass of the substance specified in the condition, we calculate the amount of the substance:
5. Above the chemical formula of magnesium oxide MgO, the mass of which is unknown, we set xmole, above the magnesium formula Mg we write its molar mass: 1 mole xmole 2Mg + O2 = 2MgO 2 mole 2 mole
According to the rules for solving proportions:
Amount of magnesium oxide ν (MgO)= 1 mol. 7. Calculate the molar mass of magnesium oxide: M (Mg)=24 g/mol, M(O)=16 g/mol. M(MgO)= 24 + 16 = 40 g/mol. We calculate the mass of magnesium oxide: m (MgO) = ν (MgO) × M (MgO) = 1 mol × 40 g/mol = 40 g. Answer: ν (MgO) = 1 mol; m (MgO) = 40 g. |
Calculations using chemical equations (stoichiometric calculations) are based on the law of conservation of mass of substances. In real chemical processes, due to incomplete reactions and losses, the mass of products is usually less than theoretically calculated. Reaction output (ŋ) is the ratio of the actual mass of the product (m practical) to the theoretically possible (m theoretical), expressed in fractions of a unit or as a percentage:
ŋ= (m practical / m theoretical) 100%.
If the yield of reaction products is not specified in the problem conditions, it is taken as 100% in the calculations (quantitative yield).
Example 1. How many g of copper are formed when 8 g of copper oxide is reduced with hydrogen if the reaction yield is 82% of theoretical?
Solution: 1. Calculate the theoretical yield of copper using the reaction equation:
CuO + H2 = Cu + H2O
80 g (1 mol) CuO upon reduction can form 64 g (1 mol) Cu; 8 g CuO upon reduction can form X g Cu
2. Let’s determine how many grams of copper are formed at 82% product yield:
6.4 g – 100% yield (theoretical)
X g –– 82%
X = (8 82) / 100 = 5.25 g
Example 2. Determine the yield of the reaction for producing tungsten by the aluminothermy method if 12.72 g of metal was obtained from 33.14 g of ore concentrate containing WO 3 and non-reducing impurities (mass fraction of impurities 0.3).
Solution 1) Determine the mass (g) of WO 3 in 33.14 g of ore concentrate:
ω(WO 3)= 1.0 - 0.3 = 0.7
m(WO 3) = ω(WO 3) m ore = 0.7 33.14 = 23.2 g
2) Let us determine the theoretical yield of tungsten as a result of the reduction of 23.2 g of WO 3 with aluminum powder:
WO 3 + 2Al = Al 2 O 3 + W.
When 232 g (1 g-mol) WO 3 is reduced, 187 g (1 g-mol) W is formed, and from 23.2 g WO 3 – X g W
X = (23.2 187) / 232 = 18.7 g W
3) Let's calculate the practical yield of tungsten:
18.7 g W –– 100%
12.72 g W –– Y%
Y = (12.72 100) / 18.7 = 68%.
Example 3. How many grams of barium sulfate precipitate are formed when solutions containing 20.8 g of barium chloride and 8.0 g of sodium sulfate are combined?
Solution. Reaction equation:
BaCl 2 + Na 2 SO 4 = BaSO 4 + 2NaCl.
The amount of the reaction product is calculated using the original substance taken in deficiency.
1). It is first determined which of the two starting substances is in short supply.
Let us denote the amount of g Na 2 SO 4 –– X.
208 g (1 mol) BaCl 2 reacts with 132 g (1 mol) Na 2 SO 4; 20.8 g –– with X g
X = (20.8 132) / 208 = 13.2 g Na 2 SO 4.
We have established that the reaction with 20.8 g of BaCl 2 will require 13.2 g of Na 2 SO 4, and 18.0 g is given. Thus, sodium sulfate is taken into the reaction in excess, and further calculations should be carried out using BaCl 2 taken in short supply.
2). We determine the number of grams of precipitate BaSO 4. 208 g (1 mol) BaCl 2 forms 233 g (1 mol) BaSO 4; 20.8 g –– Y g
Y = (233 20.8) / 208 = 23.3 g.
Law of Constancy of Composition
It was first formulated by J. Proust (1808).
All individual chemical substances of molecular structure have a constant qualitative and quantitative composition and a specific chemical structure, regardless of the method of preparation.
From the law of constancy of composition it follows that chemical elements are combined in certain quantitative ratios.
For example, carbon and oxygen form compounds with different mass ratios of the elements carbon and oxygen. CO C: O = 3: 4 CO2 C: O = 3: 8 In no other way do carbon and oxygen combine. This means that CO and CO2 compounds have a constant composition, which is determined by the oxidation states of the carbon valency in the compounds. The valence of each element has certain values (there may be several of them, variable valence), therefore the composition of the compounds is certain.
All of the above applies to substances of molecular structure. Since molecules have a certain chemical formula (composition), the substance they form has a constant composition (obviously coinciding with the composition of each molecule). The exception is polymers (consisting of molecules of different lengths).
The situation is more complicated with substances of non-molecular structure. We are talking about substances in condensed (solid and liquid) states. Because NaCl - an ionic compound in the solid state (alternating Na+ and Cl-) in the gaseous state - represents individual NaCl molecules. It is impossible to isolate individual molecules in a drop of liquid or in a crystal. For example FeO
Fe 2+ O 2– Fe 2+ O 2–, etc. perfect crystal
The law of constant composition requires that the number of Fe2+ ions be exactly equal to the number of O2– ions. And these numbers are huge even for very small crystals (a cube, an edge of 0.001 mm is 5 × 1011). This is impossible for a real crystal. In a real crystal, regularity violations are inevitable. Iron(II) oxide may contain a variable amount of oxygen depending on the production conditions. The actual composition of the oxide is expressed by the formula Fe1 – xO, where 0.16 ³ x ³ 0.04. This is berthollide, a compound of variable composition, in contrast to daltonides with x = 0. With a non-stoichiometric composition of the ionic compound, electrical neutrality is ensured. Fe 3+ is present instead of the missing Fe 2+ ion
In an atomic (non-ionic) substance, some atoms may be absent, and some may replace each other. Such compounds are also classified as daltonides. The formula of the intermetallic compound of copper with zinc, which is a component of brass, existing in the composition range of 40 – 55 at% Zn can be written as follows: (Cu0.9 – 1.0Zn0.1 – 0)(Cu0 –.0.2Zn0 – 0 ,8) copper atoms can be replaced by zinc atoms and vice versa.
The law of constancy of composition is thus strictly observed for substances of molecular structure (exceptions are high molecular weight) and has limited application for non-molecular substances.
Mass fraction of element ω(E)– This is the proportion of one element in the total mass of a substance. Calculated as a percentage or in shares. Denoted by the Greek letter ω (omega). ω shows what part the mass of a given element is of the total mass of the substance:
ω(E) = (n Ar(E)) / Mr
where n is the number of atoms; Ar(E) - relative atomic mass of the element; Mr is the relative molecular mass of the substance.
Knowing the quantitative elemental composition of a compound, it is possible to establish its simplest molecular formula. To establish the simplest molecular formula:
1) Designate the formula of the compound A x B y C z
2) Calculate the ratio X: Y: Z through the mass fractions of elements:
ω (A) = (x Ar(A)) / Mr(A x B y C z)
ω (B) = (y Ar(B)) / Mr(A x B y C z)
ω (C) = (z Ar(C)) / Mr(A x B y C z)
X = (ω (A) Mr) / Ar(A)
Y = (ω (B) Mr) / Ar(B)
Z = (ω (C) Mr) / Ar(C)
x: y: z = (ω (A) / Ar(A)) : (ω (B) / Ar(B)) : (ω (C) / Ar(C))
3) The resulting numbers are divided by the smallest to obtain the integers X, Y, Z.
4) Write down the formula of the compound.
Law of Multiples
(D. Dalton, 1803)
If two chemical elements give several compounds, then the weight fractions of the same element in these compounds that fall on the same weight fraction of the second element are related to each other as small integers.
N 2 O N 2 O 3 NO 2 (N 2 O 4) N 2 O 5
The number of oxygen atoms in the molecules of these compounds per two nitrogen atoms is in the ratio 1: 3: 4: 5.
Law of volumetric relations
(Gay-Lussac, 1808)
“The volumes of gases entering into chemical reactions and the volumes of gases formed as a result of the reaction are related to each other as small whole numbers.”
Consequence. Stoichiometric coefficients in the equations of chemical reactions for molecules of gaseous substances show in what volume ratios gaseous substances react or are obtained.
Examples.
a) 2CO + O 2 = 2CO 2
When two volumes of carbon (II) oxide are oxidized by one volume of oxygen, 2 volumes of carbon dioxide are formed, i.e. the volume of the initial reaction mixture is reduced by 1 volume.
b) During the synthesis of ammonia from the elements:
N2 + 3H2 = 2NH3
One volume of nitrogen reacts with three volumes of hydrogen; In this case, 2 volumes of ammonia are formed - the volume of the initial gaseous reaction mass will decrease by 2 times.
“A mole is equal to the amount of substance in a system containing the same number of structural elements as there are atoms in carbon - 12 (12 C) weighing 0.012 kg (exactly). When using a mole, the structural elements must be specified and may be atoms, molecules, ions, electrons and other particles or specified groups of particles." We are not talking about carbon in general, but its isotope 12 C, as with the introduction of the atomic mass unit. Since 12 g of carbon 12 C contains 6.02 × 10 23 atoms, we can say that a mole is the amount of substance containing 6.02 × 10 23 of its structural elements (atoms or groups of atoms, molecules, groups of ions (Na 2 SO 4), complex groups, etc.). The number N A = 6.02 × 10 23 is named Avogadro's constant. The molar mass of a substance is the mass of one mole. Its usual unit is g/mol, symbol M.
Recall that relative molecular mass (M r) is the ratio of the mass of one molecule to the mass of an atomic mass unit, which is equal to 1/N A g.
Let the relative molecular mass of a substance be equal to M r. Let's calculate its molecular weight M.
Mass of one molecule: m = M r a.m.u. = M r × g
Mass of one mole (N A molecules): M = m N A = M r × = M r. We see that the numerical molar mass in grams coincides with the relative molecular mass. This is a consequence of the choice of a certain atomic mass unit (1/12 of the mass of the carbon isotope 12 C).
SECTION I. GENERAL CHEMISTRY
4. Chemical reaction
Examples of solving typical problems
II.Calculations using chemical reaction equations
Problem 7. What volume of hydrogen (n.s.) will be spent on the reduction of 0.4 mol of chromium(III) oxide?
Given:
Solution
Let's write the reaction equation:
1. From the written equation it is clear that
2. To find the volume of hydrogen, we use the formula
Answer: V (H 2 ) = 26.88 l.
Problem 8. What mass of aluminum reacted with chloride acid if 2688 ml (n.s.) of hydrogen was released?
Given:
Solution
Let's write the reaction equation:
Let's make a proportion: 54 g of aluminum corresponds to 67.2 liters of hydrogen, and x g of aluminum corresponds to 2.688 liters of hydrogen:
Answer: m (A l) = 2.16 g.
Problem 9. What volume of oxygen must be used to burn 120 m 3 of a mixture of nitrogen and carbon(II) oxide, if the volume fraction of nitrogen in the mixture is 40%?
Given:
Solution
1. In the initial mixture, only carbon(II) oxide burns, the volume fraction of which is:
2. According to the formula 
Let's calculate the volume of carbon(II) oxide in the mixture:
3. Let's write down the reaction equation and, using the law of volumetric relations, carry out the calculation:
Answer: V (O 2 ) = 3 6 m 3.
Problem 10. Calculate the volume of the gas mixture that is formed as a result of the thermal decomposition of 75.2 g of cuprum(II) nitrate.
Given:
Solution
Let's write the reaction equation:
1. Let's calculate the amount of cuprum(II) nitrate. M (Cu (NO 3 ) 2) = 188 g/mol:
2. We calculate the amount of gas substances that are formed according to the reaction equations:
3. Let's calculate the volume of the gas mixture. V M = 22.4 l/mol:
Answer: V (mixture) = 22.4 l.
Problem 11. What volume of sulfur(I V ) oxide can be obtained by roasting 2.425 tons of zinc blende, the mass fraction of zinc sulfide in which is 80%?
Given:
Solution
1. Let's calculate the mass ZnS in zinc blende:
2. Let's create a reaction equation, using which we calculate the volume SO2. M (ZnS) = 97 g/mol, V M = 22.4 l/mol:
Answer: V (SO 2 ) = 448 m 3 .
Problem 12. Calculate the volume of oxygen that can be obtained with complete thermal decomposition of 34 g of dihydrogen peroxide solution with a mass fraction of H 2 O 2 30%.
Given:
Solution
1. Let's calculate the mass of dihydrogen peroxide in solution. M(H 2 O 2 ) = 34 g/mol:
2. Let's create a reaction equation and carry out calculations based on them. V M = 22.4 l/mol:
Answer: V (O 2 ) = 3.36 l.
Problem 13. What mass of technical aluminum with a mass fraction of impurities of 3% must be used to extract 2.5 mol of iron from iron scale?
Given:
Solution
1. Let's write the reaction equation and calculate the mass of pure aluminum that needs to be used for the reaction:
2. Since aluminum contains 3% impurities, then
3. From the formula 
Let's calculate the mass of technical aluminum (that is, with impurities):
Answer: m (A l) Tech. = 61.9 g.
Problem 14. As a result of heating 107.2 g of a mixture of potassium sulfate and potassium nitrate, 0.1 mol of gas was released. Calculate the mass of potassium sulfate in the original mixture of salts.
Given:
Solution
1. Potassium sulfate is a thermally stable substance. Consequently, only potassium nitrate decomposes when heated. Let's write down the reaction, put the proportion, determine the amount of the substance potassium nitrate that was dissolved:
2. Let's calculate the mass of 0.2 mol of potassium nitrate. M (KNO 3 ) = 101 g/mol:
3. Let's calculate the mass of potassium sulfate in the initial mixture:
Answer: m(K 2 SO 4) = 87 g.
Problem 15. With complete thermal decomposition of 0.8 mol of aluminum nitrate, 35.7 g of solid residue was obtained. Calculate the relative yield of the substance (%) contained in the solid residue.
Given:
Solution
1. Let us write down the equation for the decomposition reaction of aluminum nitrate. Let's make a proportion, determine the amount of substance n (A l 2 O 3):
2. Let's calculate the mass of the formed oxide. M(A l 2 O 3 ) = 102 g/mol:
3. Let's calculate the relative output A l 2 O 3 according to the formula:
Answer: η (A l 2 O 3 ) = 87.5%.
Problem 16. 0.4 mol of ferum(III) hydroxide was heated until complete decomposition. The resulting oxide was reduced with hydrogen to obtain 19.04 g of iron. Calculate the relative iron yield (%).
Given:
Solution
1. Let's write down the reaction equations:
2. Using the equations, we draw up a stoichiometric scheme and, using the proportion, we determine the theoretical yield of iron n(Fe)t attack. :
3. Let’s calculate the mass of iron that could theoretically be obtained based on the reactions carried out(M(Fe) = 56 g/mol):
4. Calculate the relative yield of iron:
Answer: η (Fe) = 85%.
Problem 17. When 23.4 g of potassium is dissolved in water, 5.6 liters of gas (n.o.) are obtained. Calculate the relative yield of this gas (%).
Given:
Solution
1. Let’s write down the reaction equation and calculate the volume of hydrogen, which theoretically, i.e. in accordance with the reaction equation, can be obtained from a given mass of potassium:
Let's make a proportion:
2. Let's calculate the relative yield of hydrogen:
Answer: η (H 2) = 83.3%.
Problem 18. When burning 0.0168 m 3 of acetylene, 55 g of carbon(I) were obtained V ) oxide. Calculate the relative yield of carbon dioxide (%).
Given:
Solution
1. Write down the equation for the combustion reaction of acetylene, compose the proportion and calculate the mass of carbon (And V ) oxide, which can be obtained theoretically. V M = 22.4 l/mol, M(CO 2) = 44 g/mol:
2. Let's calculate the relative yield of carbon (And V) oxide:
Answer: η (CO 2 ) = 83.3%.
Problem 19. As a result of catalytic oxidation of 5.8 mol of ammonia, 0.112 m 3 of nitrogen(II) oxide was obtained. Calculate the relative yield of the resulting oxide (%).
Given:
Solution
1. Let’s write down the equation for the reaction of the catalytic oxidation of ammonia, compose the proportion and calculate the volume of nitrogen (And V ) oxide, which can theoretically be obtained ( V M = 22.4 l/mol):
2. Calculate the relative yield of nitrogen(II) oxide:
Answer: η(NO) = 86.2%.
Problem 20. 1.2 mol of nitrogen (I) was passed through an excess of potassium hydroxide solution V ) oxide. We obtained 0.55 mol of potassium nitrate. Calculate the relative yield of the resulting salt (%).
Given:
Solution
1. Let's write down the equation of the chemical reaction, make up the proportion and calculate the mass of potassium nitrate, which can theoretically be obtained:
2. Let’s calculate the relative yield of potassium nitrate:
Answer: η(KNO 3 ) = 91.7%.
Problem 2 1 . What mass of ammonium sulfate can be extracted from 56 liters of ammonia if the relative yield of salt is 90%.
Given:
Solution
1. Write down the reaction equation and compose the proportion and calculate the mass of salt that can theoretically be obtained from 56 liters NH3. V M = 22.4 l/mol M((NH 4) 2 S O 4 ) = 132 g/mol:
2. Let’s calculate the mass of salt that can be obtained practically:
Answer: m ((NH 4 ) 2 S O 4 ) = 148.5 g.
Problem 22. Chlorine completely oxidized 1.4 mol of iron. What mass of salt was obtained if its yield was 95%?
Given:
Solution
1. Let's write down the reaction equation and calculate the mass of salt that can be obtained theoretically. M (FeCl 3 ) = 162.5 g/mol:
2. Calculate the mass FeCl3, which we practically received:
Answer: m (FeCl 3) pr. ≈ 216
Problem 23. To a solution containing 0.15 mol of potassium orthophosphate, a solution containing 0.6 mol of argentum(I) nitrate was added. Determine the mass of the sediment that has formed.
Given:
Solution
1. Let us write the reaction equation ( M (Ag 3 P O 4) = 419 g/mol):
It shows that for a reaction with 0.15 mol K 3 PO 4, 0.45 mol (0.15 · 3 = 0.45) argentum(I) nitrate is needed. Since, according to the conditions of the problem, the amount of substance AgN B 3 is 0.6 mol, it is this salt that is taken in excess, that is, part of it remains unused. Potassium orthophosphate will react completely, and therefore the yield of products is calculated by its quantity.
2. We make up the proportion:
Answer: m (Ag 3 P O 4). = 62.85 g.
Problem 24. 16.2 g of aluminum were placed in a solution containing 58.4 g of hydrogen chloride. What volume of gas (no.s.) was released?
Given:
Solution
1. Let's calculate the amount of aluminum and hydrogen chloride. M(A l) = 27 g/mol, M(HC l) = 36.5 g/mol:
2. Let's write the reaction equation and establish the substance that is taken in excess:
Let's calculate the amount of aluminum substance that can be dissolved in a given amount of hydrochloric acid:
Consequently, aluminum is taken in excess: the amount of its substance (0.6 mol) is more than necessary. The volume of hydrogen is calculated by the amount of hydrogen chloride.
3. Let's calculate the volume of hydrogen that is released. V M = 22.4 l/mol:
Answer: V (H 2 ) = 17.92 l.
Problem 25. A mixture that contained 0.4 liters of acetylene and 1200 ml of oxygen was brought to reaction conditions. What volume carbon(I V ) oxide formed?
Given:
Solution
Let's write the reaction equation:
According to the law of volumetric relations, it follows from the above equation that for every 2 volumes of C 2 H 2 5 volumes are consumed O2 with the formation of 4 volumes of carbon (I V ) oxide. Therefore, first we will determine the substance that is in excess - we will check whether there is enough oxygen to burn acetylene:
Since, according to the conditions of the task for burning acetylene, 1.2 liters were taken, and 1 liter is needed, we conclude that oxygen was taken in excess, and the volume of carbon (I V ) oxide is calculated by the volume of acetylene, using the law of volumetric gas ratios:
Answer: V (CO 2 ) = 0.8 l.
Problem 26. A mixture containing 80 ml of hydrogen sulfide and 120 ml O2 , led to the reaction conditions and obtained 70 ml of sulfur(I V ) oxide. Measurements of gas volumes were carried out under the same conditions. Calculate the relative yield of sulfurs(IV) oxide (%).
Given:
Solution
1. Let us write the equation for the combustion reaction of hydrogen sulfide:
2. Let's check whether there is enough oxygen to burn 80 ml of hydrogen sulfide:
Therefore, there will be enough oxygen, because it was taken in 120 ml in stoichiometric quantities. Excess of one from no substances. And therefore the volume SO 2 can be calculated using any of them:
3. Let us calculate the relative yield of sulfur(I V) oxide:
Answer: η (SO 2 ) = 87.5%.
Problem 27. When 0.28 g of an alkali metal was dissolved in water, 0.448 liters of hydrogen (n.s.) were released. Name the metaland indicate its proton number.
Given:
Solution
1. Let's write the reaction equation(V M = 22.4 l/mol):
Let's make a proportion and calculate the amount of metal substance:
2. Let's calculate the molar mass of the metal that reacted:
This is Lithium. The proton number of Lithium is 3.
Answer: Z(Me) = 3.
Problem 28. As a result of complete thermal decomposition of 42.8 g of hydroxide of a trivalent metal element, 32 g of a solid residue was obtained. Give the molar mass of the metal element.
Given:
Solution
1. Let's write the reaction equation in general form:
Since the only known substance of this reaction is water, we will carry out calculations based on the mass of water that is formed. Based on the law of conservation of mass of substances, we determine its mass:
2. Using the reaction equation, we will calculate the molar mass of the hydroxide of the metal element. Molarnusthe mass of Me(OH) 3 hydroxide will be denoted by x g/mol (M(H 2 O ) =18 g/mol):
3. Let's calculate the molar mass of the metal element:
This is Ferum.
Answer: M(Me) = 56 g/mol.
Problem 29. Cuprum(II) oxide was oxidized with 13.8 g of saturated monohydric alcohol and obtained 9.9 g of aldehyde, the relative yield of which was 75%. Name the alcohol and indicate its molar mass.
Given:
Solution
The most optimal option for writing the formula of a saturated monohydric alcohol to write the equation for the reaction of its oxidation is R - CH 2 OH, where R - alkyl substituent, the general formula of which is C n H 2 n +1 . This is due to the fact that it is the CH 2 OH groupchanges during the oxidation reaction, that is, it goes into the aldehyde group-CHO.
1. Let us write the reaction equation for the oxidation of alcohol to aldehyde in general form:
2. Let's calculate the theoretical mass of the aldehyde:
To further solve this problem, you can use 2 methods.
AND method (a mathematical method that involves performing a certain number of arithmetic operations).
Let us denote the molar mass of the alkyl substituent M(R) through x g/mol. Then:
Let's make a proportion and calculate the molar mass of the alkyl substituent:
So, the alkyl substituent is methyl-CH 3, and the alcohol is ethanol CH 3 -CH 2 -OH; M(C 2 H 5 OH) = 46 g/mol.
II method.
Let's calculate the difference in molar masses of organic products in accordance with the equation:
According to the condition Δ m р = 13.8 - 13.2 = 0.6 (g).
Let's make a proportion: if 1 mole reacts RCH2OH, then the mass difference is 2 g, and if in moles RCH2OH, then the mass difference is 0.6 g.
According to the formula 
Let's calculate the molar mass of alcohol:
So the result is the same.
Answer: M(C 2 H 5 OH) = 46 g/mol.
Problem 30 . Upon complete dehydration of 87.5 g of ferrum(III) nitrate crystalline hydrate, 1.5 mol of water vapor was obtained. Determine the formula of the starting substance.
Given:
Solution
1. Let's calculate the mass of 1.5 mol of water obtained as a result of the reaction. M(H 2 O ) =18 g/mol:
2. Based on the law of conservation of mass, we calculate the mass of salt that was obtained by heating the crystalline hydrate:
3. Let's calculate the amount of substance Fe(NO3)3. M (Fe (NO 3 ) 3 ) = 242 g/mol:
4. Let's calculate the ratio of the amounts of anhydrous salt and water:
For 0.25 moles of salt there are 1.5 moles of water per 1 mole of salt x mole:
Answer: crystal hydrate formula - Fe (NO 3) 3 6H 2 O.
Problem 31. Calculate the volume of oxygen required to burn 160 m 3 of a mixture of carbon(II) oxide, nitrogen and ethane, if the volume fractions of the mixture components are 50.0, 12.5 and 37.5%, respectively.
Given:
Solution
1. According to the formula 
Let's calculate the volumes of flammable components, namelycarbon(II) oxide and ethane (note that nitrogen does not burn):
2. Let's write the equations for the combustion reactions of CO and C 2 H 6:
3. Let's use the law of volumetric ratios of gases and calculate the volume of oxygen behind eachfrom the reaction equations:
4. Calculate the total volume of oxygen:
Answer: V (O 2) = 250 m 3.