The concept of the yield of the product of a chemical reaction

All problems are solved on the basis of the law of conservation of the mass of substances: the mass of substances that have entered into a reaction is equal to the mass of substances that have been obtained as a result of the reaction.


The purpose of the lesson: the formation of the main concepts of stoichiometry and skills of solving problems on the "exit of a chemical reaction"

Educational tasks:

  • Control the degree of assimilation of basic concepts: methods of combating corrosion, metal activity.
  • To teach to solve problems on the "exit of a chemical reaction"

Developmental tasks:

  • Develop logical thinking in students

Educational tasks:

  • Foster students' interest in science;

Lesson type: combined.

Materials and equipment: Grade 9 chemistry textbook, whiteboard, projector, screen, laptop, presentation, measuring cup, potassium hydroxide solution, copper chloride solution, dressing gown and gloves.

Preparing students for active and conscious assimilation of the material.

The slide shows a furnace for producing calcium carbide. What do you think, will the mass of calcium carbide, which will be obtained in practice, differ from the mass that was planned for its production?

The topic of today's lesson is "The concept of the yield of a chemical reaction product." We will consider the basic concepts of the theory of the yield of the product of a chemical reaction, we will study the basic formulas that can be used to calculate the yield of the reaction product.

Explanation of the new material.

The explanation of the new material is carried out using the storytelling method.

Product yield is an important indicator of the efficiency of a chemical process. On its basis, in a real production process, a conclusion is made about the completeness of the conversion of raw materials, energy consumption, etc. In real conditions, the yield of any product is always less than the theoretical one.

The yield of the product of a chemical reaction is the ratio of the actually obtained product to the maximum possible, calculated from the reaction equation. The relative yield of the reaction product is denoted by the letter of the Greek alphabet ή (eta) and is expressed in fractions of a unit or percent.

It should be emphasized that the product yield can be calculated by the ratio of masses, volumes, chemical quantities of a substance:

Further, together with the students, several examples of solving problems for the yield of a chemical reaction product are analyzed:

Example 1: Problem 1. Calculate the mass of calcium carbide formed by the action of coal on calcium oxide weighing 16.8 g, if the yield is 80%.

Algorithm for solving problems.

Let's write down Given:

m (CaO) = 16.8 g

= 80% or 0.8

Find: m practice. (CaC2) =?

2. Let's write down UHR. Let's arrange the coefficients. CaO + 3C = CaC2 + CO

Above the formulas (from the given), we write the stoichiometric ratios displayed by the reaction equation.

16.8 g x (pr) g

CaO + 3C = CaC2 + CO

3. We find the molar masses of substances by PSCE

M (CaO) = 40 + 16 = 56 g / mol

M (CaC2) = 40 + 2 12 = 64g / mol

1 solution (through moths).

We find the amount of the reagent substance by the formulas:

n = m / M

n (CaO) = 16.8 (g) / 56 (g / mol) = 0.3 mol

Using UHR, we calculate the theoretical amount of matter (νtheor)

n (CaO) = n (CaC2) = 0.3 mol

theoretical mass (mtheor) of the reaction product

m = n * M; m (CaC2) = 0.3 mol * 64 g / mol = 19.2 g

solution option (through proportion).

4. We make up the proportion according to CCR

16.8g x (theoretical) g

56 g / mol = 64g / mol

Hence x = 16.8g * 64g / mol / 56g / mol = 19.2 g

6. Find the mass (volume) fraction of the product yield by the formula:

m practical (CaC2) = 0.8 19.2 g = 15.36 g

Answer: m practical (CaC2) = 15.36 g

Consolidation of the passed material.

Consolidation of the passed material: solving problems on the board, and individual cards with tasks of varying complexity.

Demonstration experience:

Pour a copper chloride solution into a measuring beaker, add an excess of potassium hydroxide solution. We observe precipitation of copper hydroxide II, in the form of a blue precipitate.

2 task. An excess of potassium hydroxide was added to a solution containing 27 g of copper (II) chloride. This precipitated a precipitate of copper (II) hydroxide, weighing 19 g, determine the yield of the reaction product (in percent).

D a n about: Solution:

m (CuCL2) = 27 g СuCl2 + 2KOH = Cu (OH) 2 + 2KCl

m pract. (Cu (OH) 2) = 19g

ŋ =?

M (CuCl2) = 64 + 35.5 * 2 = 135 g / mol

M (Cu (OH) 2) = 64 + 216 +2 = 98g / mol

ν (CuCl2) = 27 (g) / 135 (g / mol) = 0.2 mol

0.2 mol CuCl2 = 0.2 mol Cu (OH) 2

m theor (CuCl2) = 0.2 mol * 98 (g / mol) = 19.6 g

19 * 100 / 19.6 = 96.9%

Answer:  = 96.9%


Solve problems (handout)

 1. When sodium interacted with an amount of 0.5 mol of substance with water, hydrogen was obtained with a volume of 4.2 liters (n.u.). Calculate the practical gas yield (%).

2. Metallic chromium is obtained by reducing its oxide Cr2O3 with metallic aluminum. Calculate the mass of chromium that can be obtained by reducing its oxide with a mass of 228 g, if the practical yield of chromium is 95%.


Grading for work in the lesson with comments on the shortcomings and successes of students and the class as a whole.


Pupils write on a piece of paper the answers to the questions that are presented on the slide, then anonymously hand over these sheets to the teacher.