Calculations using chemical equations lesson. Calculations using equations of chemical reactions. Topic: Substances and their transformations
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
Let us 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 produce important chemical synthesis calculations. 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. Molar volume of any gaseous substance at no. 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. Single collection of digital educational resources ().
1) p. 73-75 No. 2, 3, 5 from Workbook in chemistry: 8th grade: to the textbook 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
Exercise. How many liters of oxygen (n.o.) will react during the combustion of 4.8 g of magnesium?
Periodic Law (PL) and Periodic System (PS)
elements of D. I. Mendeleev
The discovery of PZ and the construction of PS were the pinnacle of the development of chemistry in the 19th century (1869). DI. Mendeleev arranged all the elements known at that time (63) in ascending order atomic masses and at the same time discovered a connection between properties chemical elements with their atomic masses, which consisted in the fact that at certain intervals the properties of the elements were repeated. D. I. Mendeleev formulated periodic law So: The properties of simple substances, as well as the forms and properties of compounds of elements, are periodically dependent on the magnitude of the atomic masses of the elements.
Despite the enormous significance of such a conclusion, PZ and PS represented only a brilliant empirical (experimental) generalization of facts, and their physical meaning remained unclear for a long time. The reason for this is that in the 19th century there was a complete lack of understanding about the complexity of the structure of the atom.
Three PS options are most often used:
1. Short period;
2. Semi-long (all elements of the 4th and 5th periods are extended in one line of 18 elements;
3. Long-period (all s, p, d and f elements are drawn into one line.
The short form of PS consists of 7 periods and 8 groups.
A period is a horizontal series that begins with an alkali metal(except the first period) and ends with an inert element(except for the seventh period).
The first, second and third periods consist of one row and are called small. The fourth, fifth and sixth periods consist of two rows and are called large. Total in periodic table 10 rows. The top row is even, the bottom row is odd. Even rows contain only metals and the properties of the elements change little from left to right. An even series of a large period ends with three elements similar in properties: triads. The odd-numbered rows contain metals and non-metals; in them, from left to right, there is a gradual transition from metallic to non-metallic properties.
In the sixth period of postlanthanum La (No. 57) there are 14 elements with similar properties (No. 58 - 71): lanthanides. All of them are reactive metals, they react with water, and they have a strong horizontal analogy.
In the seventh period after actinium Ac (No. 89), 14 elements (No. 90 - 103) similar to actinium are similarly located: actinides. The nuclei of their atoms are extremely unstable, that is, they are radioactive.
Each group consists of two subgroups: main and secondary.
Subgroups that include elements of small and large periods are called main (A). Subgroups that include elements of only large periods are called secondary (B). Subgroups combine the elements that are most similar to each other.
The elements of one group are characterized by the following patterns:
1. All elements, except noble gases, form oxygen compounds.
2. The highest valence and highest positive oxidation state usually corresponds to the group number. Exceptions: 1) in group 8, only ruthenium Ru and Os have a valency of VIII; Cu +1, Cu +2; O–2; F –1.
3. Elements of the main subgroup from groups IV to VIII form volatile compounds with hydrogen. Their valence in these compounds is equal to the difference between the number 8 and the group number. For example, N is in V group and its valency is 8 – 5 = 3 in the NH 3 compound.
Atomic structure
In the 19th century believed that an atom is an indivisible particle that does not change during chemical reactions. At the end of the 19th and beginning of the 20th centuries. X-ray radiation was discovered (by the German scientist K. Roentgen, 1895), radioactivity (by the French scientist A. Becquerel, 1896), and the electron (by the English scientist J. Thomson, 1897). Mass m(e)=9.109×10 –28 g and negative charge q(e)=1.602×10 –19 Cl. The value of the electron charge is taken as a unit of elementary electric charge.
In 1903, J. Thomson proposed a model of the structure of the atom, according to which the positive charge is uniformly distributed throughout the volume of the atom and is neutralized by electrons interspersed in it. Developing these ideas, E. Rutherford in 1911. proposed a planetary model of the structure of the atom. According to this theory, at the center of the atom there is a positively charged nucleus, around which electrons move. The collection of electrons in an atom is called its electronic shell. In 1913, the English scientist D. Moseley discovered that the value positive charge the nucleus of an atom is equal to the serial number of the element in D. I. Mendeleev’s periodic table of elements. The atom is electrically neutral, therefore the number of electrons in the electron shell of an atom is equal to the charge of the nucleus Z or the atomic number of the element in the periodic table.
In 1932, Soviet scientists D. D. Ivanenko and E. N. Gapon and, independently of them, the German scientist W. Heisenberg created proton-neutron theory of nuclear structure. Proton p is a particle with a mass equal to 1 a. eat.
(1.66 ×10 –24 g), and charge + 1. Neutron n is an electrically neutral particle with a mass close to the mass of a proton. Protons and neutrons are called nucleons.
The charge of the nucleus of an atom is determined by the number of protons. Hence, the number of protons in the nucleus of an atom is also equal to the atomic number of the element in the periodic table. Total number protons and neutrons is called mass number (A). It is equal to the value of the relative atomic mass rounded to the nearest whole number.
Exercise. What is the nuclear charge and how many electrons, protons, neutrons are there in a zinc atom?
Z=+30, p=30, e=30, n = 65–30 = 35.
Isotopes
Varieties of atoms of the same element that have the same nuclear charges but different mass numbers (the same number of protons and different numbers of neutrons) are called isotopes. Chemical properties All isotopes of one element are the same.
Each isotope is characterized by two quantities: mass number (indicated at the top left of chemical symbol) and serial number (placed below to the left of the chemical symbol) and is indicated by the symbol of the corresponding element. For example, the element hydrogen has three isotopes. N – protium (1 p); D(H) - deuterium (1p, 1n); T(H) - tritium (1 p, 2 n).
Detailed outline of the lesson “Calculations according to chemical equations».
Textbook: O.S. Gabrielyan.
Class: 8
Lesson topic: Calculations using chemical equations.
Lesson type: combined.
Educational objectives: introduce calculations using chemical equations; to develop students’ knowledge of calculations using chemical equations; begin to develop skills in composing chemical equations and calculating equations.
Educational tasks: continue the formation of a natural science worldview, an idea of the individual and the whole.
Developmental tasks: continue to develop the ability to observe, analyze, explain, and draw conclusions.
Teaching methods: verbal (explanation and story of the teacher), verbal - visual (explanation using notes on the blackboard).
Equipment: chalkboard, table of D.I. Mendeleev.
During the classes:
1. Organizational moment (2-5 min.)
Hello guys, have a seat. Today in the lesson you and I will have to learn how to carry out calculations using chemical equations.
2. Test of knowledge and skills (10 – 15 min.)
In previous classes we went through equations chemical reactions, let's remember what a chemical equation is? (A chemical equation is a conditional representation of a chemical reaction using chemical formulas and mathematical symbols).
On the basis of what law are chemical reactions written? (Law of conservation of mass of substances).
What does it sound like? (The mass of substances that entered into a chemical reaction is equal to the mass of substances resulting from it).
3. Explanation of new material (20 – 30 min.)
Using a chemical equation, you can determine which substances reacted and which were formed, and you can also use a chemical equation to calculate the mass, volume and quantity of reacting substances.
For calculations, it is very important to choose the units of mass, volume and quantity of a substance that correspond to each other. Let's open the textbooks on page 146 and find table No. 7. Using this table, let us consider the ratio of some units of physical and chemical quantities.
In order to solve calculation problems in chemistry, you can use an algorithm. The algorithm for solving problems is given in the textbook on page 147.
Using the problem solving algorithm, let's solve the following problem:
Task: Calculate the volume of hydrogen (n.s.) that will be required to react with 230 kg of iron (III) oxide. Calculate the amount of water that is formed in this case.
Given: Solution:
m(Fe 2 O 3) = 230 kg 1. Write the equation of the chemical reaction:
V(H 2) - ?
n(H 2 O) - ? 2. Write down the known and unknown numerical values above the formulas of the substances in the equation.
Since the mass is given in kilograms, we find the volume in cubic meters, and the amount of substance in kilomoles. And then:
230kg x m 3 y kmol
Fe 2 O 3 + 3H 2 = 2Fe + 3H 2 O
where x is the volume of hydrogen V(H 2), y is the amount of water substance n(H 2 O).
3. a) Find the mass of 1 kmol Fe 2 O 3 specified by the chemical equation and write the resulting value under its formula:
Mr(Fe 2 O 3) = 56 * 2 + 16 * 3 = 160,
M(Fe 2 O 3) = 160 kg/kmol.
b) Find the volume of 3 kmol of hydrogen V = Vm*n specified by the equation, write the found value under the hydrogen formula: V(3H 2) = 22.4 m 3 /kmol * 3 kmol = 67.2 m 3.
c) Under the formula of water we indicate its quantity, given by the equation, - 3 kmol.
The equation becomes
230kg x m 3 y kmol
Fe 2 O 3 + 3H 2 = 2Fe + 3H 2 O
160kg 67.2 m 3 3 kmol
4. Let’s compose and solve the proportions:
a) 230 = x, x = 230*67.2 = 201.6 (m3) – volume of hydrogen V(H2)
b) 230 = y, y = 230*3 = 9 (kmol) – the amount of water substance n(H 2 O).
4.Primary consolidation of knowledge (10 – 12 min.)
Solve the problems (if possible, in several ways):
Task 1. 0.1 mol of zinc reacts with oxygen. How much oxygen is required? How much zinc oxide is formed?
Task 2. Zinc reacts with oxygen in an amount of 0.1 mol. Determine the mass of oxygen that reacts, as well as the mass of zinc oxide formed.
Task 3. Aluminum weighing 6.3 g reacts with oxygen. Determine the masses of oxygen and the resulting iron oxide if aluminum contains 20% impurities.
Task 4. What volume of hydrogen (n.s.) will be released when 2.7 g of 25% hydrochloric acid reacts with the amount of aluminum required by the reaction? What is this amount of substance?
Task 5. What volume carbon dioxide will be released when 60 kg of coal is burned?
Task 6. How many moles of calcium oxide are formed when 8 g of calcium containing 30% impurities is burned in oxygen?
5. Lesson summary (1 -3 min.)
Today in class we once again remembered writing chemical equations and learned how to carry out calculations using chemical equations.
6. Homework (1 – 4 min.)
§28, assignment in workbooks.
What mass of iron (III) oxide is formed when 0.6 mol of iron is burned in air?
Calculate the mass of aluminum sulfide formed when 5.4 g of aluminum powder is fused with sulfur. How many grams of iron (II) sulfide are formed when 11.2 g of iron powder is fused with sulfur?
Determine the mass of magnesium required to obtain 19 g of magnesium chloride (for example, by burning magnesium in chlorine).
How many liters of hydrogen chloride are formed when chlorine reacts with 5.5 liters of hydrogen?
What volume of hydrogen can react with 150 liters of oxygen?
What volume of carbon dioxide is formed when 8 liters of methane CH 4 are burned?
What volume of carbon dioxide is produced when 480 g of coal is burned?
How much oxygen will be released during decomposition? electric shock 100g of water?
What volume of nitrogen is formed during the explosion of 1 g of nitrogen iodide:
2NJ 3 = N 2 + 3J 2
How many grams of sulfur oxide (IV) are formed when 12.8 sulfur is burned?
What mass of magnesium oxide was formed when 6 g of magnesium shavings were burned in oxygen?
How many grams of water are produced when 9 g of hydrogen are burned in oxygen?
How many grams of aluminum must be taken to obtain 30.6 g of aluminum oxide?
How many grams of lithium must be burned in oxygen to produce 15 g of lithium oxide?
How many grams of sodium chloride are formed when 11.5 g of sodium is burned in chlorine?
How many moles of iron must be taken to obtain 32.5 g of iron (III) chloride?
How many grams of aluminum are needed to obtain 80.1 g of aluminum chloride?
How many moles of calcium oxide are formed when 8 g of calcium is burned in oxygen?
How many grams of aluminum chloride are produced when 10.8 g of aluminum foil is burned in chlorine?
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 mass fraction 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%.
Given:
Solution
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 (%).
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%. V Problem 18. When burning 0.0168 m 3 of acetylene, 55 g of carbon(I) were obtained
Given:
Solution
) oxide. Calculate the relative yield of carbon dioxide (%). V 1. Write down the equation for the combustion reaction of acetylene, compose the proportion and calculate the mass of carbon (And) 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:
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
Given:
Solution
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. 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).
Given:
Solution
= 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?
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 120 ml of it was taken 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 2. Let's calculate the relative yield of carbon (And
Answer: η (SO 2 ) = 87.5%.
Problem 27. When dissolved in water 0.28 g alkali metal 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
Most best option writing the formula for a saturated monohydric alcohol to write the equation for its oxidation reaction is R - CH 2 OH, where R - alkyl substituent, general formula which 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:
For further solution There are 2 ways to do this task.
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.
Method II.
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. Establish 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 us 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.
Whatever you study, you
you study for yourself.
Petronius
Lesson objectives:
- introduce students to the basic ways of solving problems using chemical equations:
- find the quantity, mass and volume of reaction products from the quantity, mass or volume of starting substances,
- continue to develop skills in working with the text of a problem, the ability to reasonedly choose a method for solving an educational problem, and the ability to compose equations of chemical reactions.
- develop the ability to analyze, compare, highlight the main thing, draw up an action plan, and draw conclusions.
- cultivate tolerance towards others, independence in decision-making, and the ability to objectively evaluate the results of one’s work.
Forms of work: frontal, individual, pair, group.
Lesson type: combined with the use of ICT
I Organizational moment.
Hello guys. Today, we will learn how to solve problems using equations of chemical reactions. Slide 1 (see presentation).
Lesson objectives Slide 2.
II.Updating knowledge, skills and abilities.
Chemistry is a very interesting and at the same time complex science. In order to know and understand chemistry, you must not only assimilate the material, but also be able to apply the acquired knowledge. You learned what signs indicate the occurrence of chemical reactions, learned how to write equations for chemical reactions. I hope you have a good understanding of these topics and can answer my questions without difficulty.
Which phenomenon is not a sign of chemical transformations:
a) the appearance of sediment; c) change in volume;
b) gas release; d) the appearance of an odor. Slide 3
Please indicate in numbers:
a) equations of compound reactions
b) equations of substitution reactions
c) equations of decomposition reactions Slide 4
In order to learn how to solve problems, it is necessary to create an algorithm of actions, i.e. determine the sequence of actions.
Algorithm for calculations using chemical equations (on each student’s desk)
5. Write down the answer.
Let's start solving problems using an algorithm
Calculating the mass of a substance from the known mass of another substance participating in the reaction
Calculate the mass of oxygen released as a result of decomposition
portions of water weighing 9 g.
Let's find the molar mass of water and oxygen:
M(H 2 O) = 18 g/mol
M(O 2) = 32 g/mol Slide 6
Let's write the equation of the chemical reaction:
2H 2 O = 2H 2 + O 2
Above the formula in the reaction equation we write what we found
the value of the amount of a substance, and under the formulas of substances -
stoichiometric ratios displayed
chemical equation
0.5mol x mol
2H 2 O = 2H 2 + O 2
2mol 1mol
Let's calculate the amount of substance whose mass we want to find.
To do this, we create a proportion
0.5mol = hopmol
2mol 1mol
where x = 0.25 mol Slide 7
Therefore, n(O 2) = 0.25 mol
Find the mass of the substance that needs to be calculated
m(O 2)= n(O 2)*M(O 2)
m(O 2) = 0.25 mol 32 g/mol = 8 g
Let's write down the answer
Answer: m(O 2) = 8 g Slide 8
Calculating the volume of a substance from the known mass of another substance participating in the reaction
Calculate the volume of oxygen (no.) released as a result of the decomposition of a portion of water weighing 9 g.
V(0 2)=?l(n.s.)
M(H 2 O) = 18 g/mol
Vm=22.4l/mol Slide 9
Let's write down the reaction equation. Let's arrange the coefficients
2H 2 O = 2H 2 + O 2
Above the formula in the reaction equation we write the found value of the amount of the substance, and under the formulas of the substances - the stoichiometric ratios displayed by the chemical equation
0.5 mol - x mol
2H 2 O = 2H 2 + O 2 Slide10
2mol - 1mol
Let's calculate the amount of substance whose mass we want to find. To do this, let's create a proportion
where x = 0.25 mol
Let's find the volume of the substance that needs to be calculated
V(0 2)=n(0 2) Vm
V(O 2) = 0.25 mol 22.4 l/mol = 5.6 l (no.)
Answer: 5.6 l Slide 11
III. Consolidation of the studied material.
Tasks for independent solution:
1. When reducing the oxides Fe 2 O 3 and SnO 2 with coal, 20 g of Fe and Sn were obtained. How many grams of each oxide were taken?
2.In which case is more water formed:
a) when reducing 10 g of copper (I) oxide (Cu 2 O) with hydrogen or
b) when reducing 10 g of copper(II) oxide (CuO) with hydrogen? Slide 12
Let's check the solution to Problem 1
M(Fe 2 O 3) = 160 g/mol
M(Fe)=56g/mol, 
m(Fe 2 O 3)=, m(Fe 2 O 3)= 0.18*160=28.6g
Answer: 28.6g
Slide 13
Let's check the solution to problem 2
M(CuO) = 80 g/mol
4. 
x mol = 0.07 mol,
n(H 2 O)=0.07 mol
m(H 2 O) = 0.07mol*18g/mol=1.26g
Slide 14
CuO + H 2 = Cu + H 2 O
n(CuO) = m/ M(CuO)
n(CuO) = 10g/ 80g/mol = 0.125 mol
0.125mol hops
CuO + H 2 = Cu + H 2 O
1mol 1mol
x mol = 0.125 mol, n(H 2 O) = 0.125 mol
m (H 2 O) = n * M (H 2 O);
m(H 2 O) = 0.125mol*18g/mol=2.25g
Answer: 2.25g Slide 15
Homework: study the textbook material p. 45-47, solve the problem
What is the mass of calcium oxide and what is the volume of carbon dioxide (n.s.)
can be obtained by decomposing calcium carbonate weighing 250 g?
CaCO 3 = CaO + CO Slide 16.
Literature
1. Gabrielyan O.S. Chemistry course program for grades 8-11 educational institutions. M. Bustard 2006
2. Gabrielyan O.S. Chemistry. 8th grade. Textbook for general education institutions. Bustard. M. 2005
3. Gorbuntsova S.V. Tests on the main sections of the school course. 8th - 9th grades. VAKO, Moscow, 2006.
4. Gorkovenko M.Yu. Lesson developments in chemistry. To the textbooks of O.S. Gabrielyan, L.S. Guzey, V.V. Sorokin, R.P. Surovtseva and G.E. Rudzitis, F.G. Feldman. 8th grade. VAKO, Moscow, 2004.
5. Gabrielyan O.S. Chemistry. Grade 8: Tests and tests. – M.: Bustard, 2003.
6. Radetsky A.M., Gorshkova V.P. Didactic material on chemistry for grades 8-9: A manual for teachers. – M.: Education, 2000
Application.
Calculations using chemical equations
Algorithm of actions.
In order to solve a calculation problem in chemistry, you can use the following algorithm - take five steps:
1. Write an equation for a chemical reaction.
2. Above the formulas of substances, write known and unknown quantities with the corresponding units of measurement (only for pure substances, without impurities). If, according to the conditions of the problem, substances containing impurities enter into a reaction, then you first need to determine the content of the pure substance.
3. Under the formulas of substances with known and unknowns, write down the corresponding values of these quantities found from the reaction equation.
4. Compose and solve a proportion.
5. Write down the answer.
The relationship between some physical and chemical quantities and their units
Mass (m) : g; kg; mg
Quantity of substances (n): mole; kmol; mmol
Molar mass (M): g/mol; kg/kmol; mg/mmol
Volume (V) : l; m 3 /kmol; ml
Molar volume (Vm) : l/mol; m 3 /kmol; ml/mmol
Number of particles (N): 6 1023 (Avagadro number – N A); 6 1026 ; 6 1020