Using the chemical reaction equation, we determine the chemical quantity. Calculations using chemical equations

Methodological lesson development (2 hours)

Calculations using chemical equations

Khapugina Polina Ivanovna,
chemistry teacher GBOU secondary school 277
Kirovsky district of St. Petersburg

Lesson objectives: Teach eighth-graders to make calculations using chemical equations: find the quantity, mass and volume of reaction products from the quantity, volume and mass of the starting substances.

During the classes:

Before moving on to studying a new topic, we need to remember the quantities and formulas you already know. Remember the types of chemical reactions. And you already know how to create chemical equations and equalize them. Let's test your knowledge by doing the following: assessment test.(the contents of the test can be viewed on my personal website in the files and photos folder)

Teacher's explanation:

1. Before learning how to make calculations using chemical equations, you need to once again remember the formulas we already know for finding the amount of a substance, mass and volume of substances; in one step you can test yourself after the work you have done. To do this, let's turn to the following presentation, which will help restore our knowledge in memory. Presentation No. 1.

Write down the formulas you already know in your notebook:

n- amount of substance (mol)

m- mass of substance (g)

M - molar mass (g/mol)

V - gas volume (l)

V m - molar volume = 22.4 l/mol

n = m/M ; m = n.M

n = V/Vm; V = n.V m

2. Now, we must understand that a chemical equation shows not only the qualitative (transformation of substances) side of the process, but also the quantitative side of it. To do this, let's turn to the following Presentations#2(the presentation can be viewed on my personal website in the files and photos folder)

Write in your notebook:

The coefficient in the reaction equation indicates the number of particles, and the number of particles in turn determines the number of moles!

2H₂ + O₂ = 2H₂ ABOUT

Number of particles 2 molecules 1 molecules 2 molecules

Quantity ↓ ↓ ↓

substances,n2 mole 1 mole 2 mole

2 Fe(OH)3 = Fe2 O3 + 3 H2 O

↓ ↓ ↓

n= 2mol 1mol 3mol

3. The next stage that we must examine is the ability to create a proportion using the reaction equation, as well as solve it. To do this, let's turn to the following Presentations 3.(the presentation can be viewed on my personal website in the files and photos folder)

Write in your notebook:

Known by the condition: 2 mol X mol (numerator)

4P+5O₂ = 2P₂ O₅

Known by the equation: 4 mol 5 mol 2 mol (denominator)

Let's compose and solve the proportion:

2 mol X mol

_______ = _______

4 mol 2 mol

X mole = 2 mol. 2 mol= 1 mol

4 mol

X =n(P₂ O₅ )= 1 mol

4. Let's move on to solving problems using equations of chemical reactions. In order to solve a calculation problem in chemistry, you need to use the following algorithm - take five steps. Presentation 4. (the presentation can be viewed on my personal website in the files and photos folder) Textbook p. 101

In your notebook:

Students are given a ready-made solution algorithm to paste into their notebooks.

Algorithm for solving calculation problems using chemical reaction equations:

1. Create an equation for a chemical reaction (i.e., be sure to set the coefficients!)

2. Above the corresponding formulas in the equation, write down quantitative data about substances with units of measurement that are known or can be calculated based on the conditions of the problem, and the desired value X also with units of measurement.

3. Under these formulas, write down the corresponding quantitative quantities specified by the equation itself, also with units of measurement.

4. Compose and solve a proportion.

5. Formalize the answer.

5. Let's solve the problem.

Calculate the mass of water that is formed as a result of the interaction of 0.5 mol of aluminum oxide with sulfuric acid when heated.

Read the problem.

Write down the problem statement. (Given, find.)

In your notebook: (students write down the solution according to the teacher’s explanations according to Presentations 5 ) (the presentation can be viewed on my personal website in the files and photos folder)

Given:

n(Al₂O₃)=0.5 mol

_________________

Find:

Solution:

n=0.5 mol X mol

Al₂ O₃ +3 H₂ SO₄ = Al₂ (SO₄ ) ₃ +3 H₂ O

n= 1 mol 3 mol

M = 102 g/mol 18 g/mol

Calculation of molecular weight:

Мr(Al₂O₃)= 2.27+3.16= 54+48=102

Мr(H₂O)= 2.1+16=18

Compose and solve a proportion.

0.5 mol = X mole

1 mole 3 mole

X mole =n(H₂ O) = 0.5 mol. 3 mol= 1.5 mol

1 mole

Let's find the mass of water.

m(H₂ O) = n(H₂ O) . M(H₂ O)

m(H₂ O) = 1.5 mol.18 g/mol = 27 g

Answer:m(H₂ O)=27 g

6. Solve problems yourself.

Two students are called to the board for assessment.

1. Determine the volume of chlorine (no.) required to obtain 634 g of aluminum chloride according to the equation: 2Al + 3Cl 2 = 2AlCl 3. Answer: 159.6 l

2. Calculate the amount of substance and mass of lithium required for the reaction with oxygen weighing 128 g according to the equation: 4Li + O 2 = 2Li 2 O Answer: 16 mol, 112 g

7. Homework.

Task.

Find the mass of zinc oxide, which is formed by the interaction of 13 g of zinc with oxygen.

Calculations using chemical equations

Zakirova Olisya Telmanovna – chemistry teacher.

9th grade

Lesson objectives: Develop knowledge about how to solve problems using chemical equations: find the quantity, mass and volume of reaction products from the quantity, mass or volume of starting substances, develop the ability to analyze, compare, highlight the main thing, draw up an action plan, draw conclusions, skills in working with the text of a problem, ability choose a way to solve a learning problem, the ability to compose equations of chemical reactions. Foster independence in decision making and the ability to evaluate results.

Course and content of the lesson

I Organizational moment. Hello guys. Today, a very interesting lesson on problem solving.

II. Updating knowledge, skills and abilities.

In order to know and understand chemistry, you must not only assimilate the material, but also be able to apply the acquired knowledge. You know the signs indicating the occurrence of chemical reactions, you know how to write equations for chemical reactions.

1. List the signs of chemical transformations:

2.Match the match.

4Al + 3O 2 = 2Al 2 O 3 a) equations of compound reactions

MgCO 3 = MgO + CO 2 b) equations of substitution reactions

2HgO = 2Hg + O 2: c) equations of decomposition reactions

2Na + S=Na 2 S

Zn + Br 2 = ZnBr2

Zn + 2HCl = ZnCl 2 + H 2

Fe + CuSO 4 = FeSO 4 + Cu

III.Assimilation of new knowledge.In order to solve a calculation problem in chemistry, you can use the following algorithm - take five steps:

1. Write an equation

2. Above the formulas of substances, write known and unknown quantities with the corresponding units of measurement.
(only for pure substances, that is, those that do not contain substances). If, according to the conditions of the problem, substances containing impurities enter into a reaction, then the content of the pure substance must first be determined.

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.

Let's start solving problems using an algorithm

Calculating the volume of a substance from the known mass of another substance participating in the reaction

V(0 2)=?l(n.s.)

M(H 2 O) = 18 g/mol

Vm=22.4 l/mol Let's write 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

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

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

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 the found value of the amount of the substance, and under the formulas of the substances we write the stoichiometric ratios displayed by the 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 molTherefore 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

IV .Reinforcement of the studied material. What volume of air (n.a.) will be required to interact with 270 gr. containing 20% ​​impurities? What amount of aluminum oxide will be produced?

V.Homework:

Level 1Calculate the volume of oxygen (no.) released as a result of the decomposition of a portion of water weighing 9 g.

Level 2 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?

Level 3 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?

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

ALGORITHM FOR SOLVING PROBLEMS N m V n NxNx mxmx VxVx nxnx = 1. Write the reaction equation Using a known amount of a substance from one participant in the reaction, calculate the amount of the desired substance. If the amount of a substance is unknown, first find it using a known mass, or volume, or number of molecules. Using the found amount of the substance, find the desired characteristic of the desired reaction participant (mass, volume, or number of molecules).

Calculate the amount of aluminum required to produce 1.5 mol of hydrogen when reacting with hydrochloric acid. Given: n(H2) = 1.5 mol n(Al) – ? Solution: x mol 1.5 mol 2Al + 6HCl = 2AlCl 3 + 3H 2 2 mol 3 mol We make up the proportion: x mol 1.5 mol = 2 mol 3 mol 2 1.5 x = 3 x = 1 (mol) Answer : n(Al) = 1 mol A PS

Given: n(Al 2 S 3) = 2.5 mol n(S) – ? Solution: x mol 2.5 mol 2Al + 3S = Al 2 S 3 3 mol 1 mol x = n(S) = 3 n(Al 2 S 3) = = 3 2.5 mol = 7.5 mol Answer: n(S) = 7.5 mol Determine the amount of sulfur required to obtain 2.5 mol of aluminum sulfide. PS A

Given: m(Cu(OH) 2) = 14.7 g m(CuO) – ? M(Cu(OH) 2) = 64+(16+1) 2 = 98 g/mol M(CuO) = = 80 g/mol Solution: 14.7 g x mol Cu(OH) 2 = CuO + H 2 O 1 mol 1 mol m(Cu(OH) 2) n(Cu(OH) 2) = M(Cu(OH) 2) 14.7 g n(Cu(OH) 2) = = 0.15 mol 98 g/ mol x = n(CuO) = n(Cu(OH) 2) = 0.15 mol m(CuO) = n(CuO) M(CuO) = 0.15 mol 80 g/mol = 12 g Answer: m(CuO) = 12 g Calculate the mass of copper (II) oxide formed during the decomposition of 14.7 g of copper (II) hydroxide. PS A 0.15 mol

Given: m(Zn)=13 g m(ZnCl 2) – ? M(Zn) = 65 g/mol M(ZnCl 2 =65 +35.5 2 = 136 g/mol Solution: 0.2 mol x mol Zn + 2HCl = ZnCl 2 + H 2 1 mol 1 mol m(Zn ) n(Zn) = M(Zn) 13 g n(Zn) = = 0.2 mol 65 g/mol x = n(ZnCl 2) = n(Zn) = 0.2 mol m(ZnCl 2) = n (ZnCl 2) M(ZnCl 2) = 0.2 mol 136 g/mol = 27.2 g Answer: m(ZnCl 2) = 27.2 g Calculate the mass of salt that is formed when 13 g of zinc reacts with hydrochloric acid acid. PS A

Given: m(MgO) = 6 g V(O2) – ? M(MgO) = = 40 g/mol Vm = 22.4 l/mol Solution: 0.15 mol x mol 2MgO = 2Mg + O 2 2 mol 1 mol m(MgO) n(MgO) = M(MgO) 6 g n(MgO) = = 0.15 mol 40 g/mol x = n(O 2) = ½ n(MgO) = 1/2 0, 15 mol = 0.075 mol V(O 2) = n(O 2 )·Vm = 0.075 mol·22.4 l/mol = 1.68 l Answer: V(O 2) = 1.68 l What volume of oxygen (no.) is formed during the decomposition of 6 g of magnesium oxide. PS A

Given: m(Cu)=32 g V(H 2) – ? M(Cu) = 64 g/mol Vm = 22.4 l/mol Solution: x mol 0.5 mol H 2 + CuO = H 2 O + Cu 1 mol 1 mol m(Cu) n(Cu) = M( Cu) 32 g n(Cu)= = 0.5 mol 64 g/mol x = n(H 2) = n(Cu) = 0.5 mol V(H 2) = n(H 2) Vm = 0 .5 mol·22.4 l/mol = 11.2 l Answer: V(H 2) = 11.2 l Calculate how much hydrogen must react with copper (II) oxide to form 32 g of copper. PS A

INDEPENDENT WORK: OPTION 1: Calculate the mass of copper that is formed when 4 g of copper (II) oxide is reduced with excess hydrogen. CuO + H 2 = Cu + H 2 O OPTION 2: 20 g of sodium hydroxide reacted with sulfuric acid. Calculate the mass of salt formed. 2NaOH + H 2 SO 4 = Na 2 SO 4 + 2H 2 O