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Question No. 21 of the OGE exam materials in chemistry is a problem on the equation of a chemical reaction. The Specification of control measuring materials for the main state exam in chemistry in 2018 specifies the following skills to be tested and methods of action when performing this task: « Calculation of the mass fraction of solute in a solution. Calculation of the amount of a substance, mass or volume of a substance from the amount of substance, mass or volume of one of the reactants or products of a reaction.” Analysis of demonstration works and open bank tasks made it possible to identify three types of tasks used in examination papers. In preparation for the OGE, I solve examples of each type of problem with students and offer similar tasks selected from an open bank for independent solution. When solving problems on chemical reaction equations, I use the algorithm presented in the 8th grade chemistry textbook by O.S. Gabrielyan.

1 type

The mass of a solution of the product or one of the starting materials of the reaction is given. Calculate the mass (volume) of the starting substance or reaction product.

1 action: We calculate the mass of the product or one of the starting materials of the reaction.

Action 2: We calculate the mass or volume of the starting substance using the algorithm.

Example task: TO solution aluminum chloride weighing 53.2 g and a mass fraction of 5%, an excess of silver nitrate solution was added. Calculate the mass of the sediment formed.

Analysis of the solution

1. TO solution aluminum sulfate weighing 34.2 g and a mass fraction of 10%, an excess of barium nitrate solution was added. Calculate the mass of the sediment formed.
2. Carbon dioxide was passed through a solution of calcium hydroxide. 324 g formed solution calcium bicarbonate with a mass fraction of 1%. Calculate the volume of reacted gas.

2nd view

The mass of a solution of a substance or reaction product is given. Calculate the mass fraction of the substance or reaction product.

1 action: Using the algorithm, we calculate the mass of the starting substance (product) of the reaction. We do not pay attention to the mass of its solution.

Action 2: We know the mass of the starting substance (product) - we found it in the first step. We know the mass of the solution - it is given in the condition. Finding the mass fraction.

Example task: 73 g solution hydrochloric acid was mixed with a portion of calcium carbonate. In this case, 0.896 liters of gas were released. Calculate the mass fraction of the original solution of hydrochloric acid.

Analysis of the solution

2. ω = m(in-va)/m(solution) · 100%

ω = 2.92/73 100= 4%

Problems for independent solution.

1. To 200 g solution calcium chloride, sodium carbonate solution was added until precipitation stopped. The mass of the sediment was 12.0 g. Calculate the mass fraction of calcium chloride in the original solution. (Take the relative atomic mass of chlorine to be 35.5)
2. After passing 4.4 g of carbon dioxide through 320 g solution potassium hydroxide to obtain a solution of medium salt. Calculate the mass fraction of alkali in the solution

Type 3

The mass fraction of the solution of the starting substance is given. Determine the mass of the starting substance.

1 Action. Using the algorithm, find the mass of the starting substance.

2 Action. We know the mass of the starting substance (from the first action). We know the mass fraction (from the condition). Find the mass of the solution.

Sample task: An excess of barium chloride solution was added to a solution of potassium carbonate with a mass fraction of 6%. As a result, a precipitate weighing 9.85 g was formed. Determine the mass of the initial potassium carbonate solution.

Analysis of the solution

2. ω = m(in-va)/m(solution) · 100%

m(solution) = 6.9/6 ▪100% = 115 g.

Problems to solve independently

1. After passing 11.2 liters (N.S.) of ammonia through a 10% solution of sulfuric acid, a solution of medium salt was obtained. Determine the mass of the original sulfuric acid solution.
2. When 4.48 liters of carbon dioxide (n.o.) were passed through a solution of barium hydroxide with a mass fraction of 12%, barium carbonate was formed. Calculate the mass of the initial barium hydroxide solution.

Algorithm for solving problems using equations of chemical reactions

1. Brief description of the problem conditions.
2. Writing the equation of a chemical reaction.
3. Writing known and unknown quantities over the formulas of substances.
4. Record under the formulas of substances the quantities, molar masses and masses (or molar volumes and volumes) of substances.
5. Drawing up and solving proportions.
6. Recording the task response.

Part 1 contains 19 tasks with a short answer, including 15 tasks of a basic level of complexity (the serial numbers of these tasks: 1, 2, 3, 4, ...15) and 4 tasks of an increased level of complexity (the serial numbers of these tasks: 16, 17, 18, 19). Despite all their differences, the tasks in this part are similar in that the answer to each of them is written briefly in the form of one number or a sequence of numbers (two or three). The sequence of numbers is written on the answer form without spaces or other additional characters.

Part 2, depending on the CMM model, contains 3 or 4 tasks of a high level of complexity, with a detailed answer. The difference between exam models 1 and 2 lies in the content and approaches to completing the last tasks of the exam options:

Examination model 1 contains task 22, which involves performing a “thought experiment”;

Examination model 2 contains tasks 22 and 23, which involve performing laboratory work (a real chemical experiment).

Scale for converting points to grades:

"2"– from 0 to 8

"3"– from 9 to 17

"4"– from 18 to 26

"5"– from 27 to 34

System for assessing the performance of individual tasks and the examination work as a whole

Correct completion of each of tasks 1–15 is scored 1 point. Correct completion of each of tasks 16–19 is assessed with a maximum of 2 points. Tasks 16 and 17 are considered completed correctly if two answer options are correctly selected in each of them. For an incomplete answer - one of two answers is correctly named or three answers are named, two of which are correct - 1 point is given. The remaining answer options are considered incorrect and are scored 0 points. Tasks 18 and 19 are considered completed correctly if three correspondences are correctly established. An answer in which two out of three matches are established is considered partially correct; it is worth 1 point. The remaining options are considered an incorrect answer and are scored 0 points.

The tasks of Part 2 (20–23) are checked by a subject commission. Maximum score for a correctly completed task: for tasks 20 and 21 - 3 points each; in model 1 for task 22 – 5 points; in model 2 for task 22 - 4 points, for task 23 - 5 points.

To complete the examination work in accordance with model 1, 120 minutes are allotted; according to model 2 – 140 minutes

Novikova Galina Petrovna
Job title: teacher of chemistry and biology
Educational institution: MBOU Secondary School Yaroslavka
Locality: Yaroslavka village, Duvansky district, Republic of Belarus
Name of material: Master Class
Subject:"Solving tasks of the expanded part of the OGE in chemistry"
Publication date: 10.01.2018
Chapter: complete education

Master class “Solving tasks of the expanded part of the OGE in chemistry”

Prepared

republican

promotion

Petrovna,

biology

Yaroslavka village

municipal district Duvansky district of the Republic of Bashkortostan

A master class is an effective form of sharing experience with colleagues. IN

to completing examination tasks with a detailed answer.

examination

option

contains

high

level of difficulty, with a detailed answer – 20, 21, 22.

Tasks with extended answers are the most difficult in the exam room.

work. These assignments test mastery of the following content elements:

receiving

chemical

properties

various

inorganic

connections;

oxidative-

restorative

relationship

various

amount of substance, molar volume and molar mass of substance; massive

fraction of solute.

Completing tasks of this type presupposes the presence of formation

complex skills:

make up

electronic

the equation

oxidative-

recovery reaction;

explain the conditionality of the properties and methods of obtaining their substances

composition and structure;

carry out combined calculations using chemical equations.

execution tasks 20 necessary based on the reaction scheme,

presented in its condition, create an electronic balance and equation

ORR, determine the oxidizing agent and reducing agent.

Examples of solutions to task 20. The task is worth 3 points.

4 – oxidation process

1- recovery process

Molecular equation

Iodine in the oxidation state -1 (i.e. KI) is a reducing agent, and sulfur in

oxidation state +6 (i.e. H

) is an oxidizing agent.

5 - recovery process

1 – oxidation process

Molecular equation

oxidation

is

oxidizing agent

oxidation 0 is a reducing agent.

1 – oxidation process

6 – recovery process

Molecular equation

Sulfur in oxidation state 0 is a reducing agent, and nitrogen in degree

) is an oxidizing agent.

Task 21 involves performing two types of calculations: calculation

mass fraction of solute in solution and calculation of the amount

substance, mass or volume of substance by amount of substance, mass or

the volume of one of the reactants or products of the reaction.

To solve the problem, we use the following algorithm:

Algorithm

solutions

tasks

calculation

chemical

equations of mass (volume, quantity) of starting substances and products

reactions.

1.Read the problem and write a short condition for it.

2.Make an equation for the chemical reaction taking place.

3. Calculate the molar masses of the known and desired substances.

4. Calculate the amount of the substance that is known according to the conditions of the problem.

To do this, you can use the following formulas:

n = m: M or n =V: V

chemical

equation

write down

quantitative

ratio

between the substance specified by the problem conditions and the desired substance.

The quantitative ratio is determined by the coefficients recorded

before the formulas of the corresponding substances in a chemical equation.

quantitative

ratio

define

quantity

what you're looking for

substances.

Calculate

what you're looking for

substances.

use the following formulas: m = n

Examples of problem solving. The task is worth 3 points.

Task 1.

solution

mass

fractions of NaOH 6% missed

carbon dioxide to form sodium carbonate. Calculate volume (no.s.)

reacting gas.

m(NaOH solution)=40g 2NaOH + CO

2mol 1mol

w(NaOH)=6% or 0.06 M=40g/mol

Find: 1) Calculate the mass and amount of NaOH:

a) m(NaOH)=40g

)- ? b) n(NaOH)=2.4g:40g/mol=0.06 mol

2) Determine the volume of CO

: a) according to the reaction equation

)=2:1, which means n(CO

)=22.4 l/mol

0.03 mol=0.672l

Answer: V(CO

Task 2. Calculate the mass of sediment that is formed during the action

excess barium chloride solution per 68.4 g of 8% sulfate solution

aluminum

Given: Solution:

)=8% or 0.08 1 mol 3 mol

Find: M=342g/mol M=233g/mol

)-? 1) Calculate the mass and amount of the substance

)= 5.472g: 342g/mol=0.016 mol

2) Calculate the mass of BaSO sediment

a) according to the reaction equation n(Al

)=1:3, which means n(BaSO

)=233g/mol

mole=11.184g

Answer: m(BaSO

Task 3. Calculate the mass of sediment that is formed when passing

hydrogen sulfide through 75.2 g of a 5% solution of copper (II) nitrate.

Given: Solution:

m (solution Cu(NO

)=5% or 0.05 1 mol 1 mol

Find: M=188g/mol M=96g/mol

m(CuS) -? 1) Calculate the mass and amount of the substance

)=3.76g:188g/mol=0.02 mol

2) Calculate the mass of sediment: a) using the equation

R e a k c i i

):n(CuS) = 1: 1, value n(CuS) =

b) m(CuS)=96 g/mol

0.02 mol=1.92g

Answer: m(CuS)= 1.92g

Task 22 is practice-oriented and in model 1 has

character

mental

experiment.

oriented

check

the following

to plan

carrying out

experiment

proposed

describe

signs

flow

chemical

reactions; write molecular and abbreviated ionic equations of these

reactions. The task is worth 5 points.

Examples of solving problems.

substances:

NaOH solutions,

two stages Zn(OH)

Describe the signs of the reactions being carried out. For reaction

ion exchange write the abbreviated ionic equation for the reaction.

Experiment scheme:

Ion exchange reaction

2 NaOH = Zn(OH)

Ion exchange reaction

Signs of reactions:

For the first reaction – the formation of a white dense precipitate;

For the second reaction - the formation of a white gelatinous precipitate.

substances: Al,

BaCl solutions

Using water and the necessary substances only from this list, you will get

two stages AlCl

Describe the signs of the reactions being carried out. For the 2nd reaction

write the abbreviated ionic equation for the reaction.

Experiment scheme:

Signs of reactions:

For the first reaction – the formation of a white precipitate;

For the second reaction - dissolution of the precipitate, formation of a transparent

solution.

Abbreviated ionic equation for reaction 2:

In conclusion, I would like to wish great and painstaking work on

preparation

state

final

certifications,

turned out to be

definitely successful!

In this section, I systematize the analysis of problems from the OGE in chemistry. Similar to the section, you will find detailed analyzes with instructions for solving typical problems in chemistry in the 9th grade OGE. Before analyzing each block of typical problems, I provide theoretical information, without which solving this task is impossible. There is only as much theory as is enough to know to successfully complete the task on the one hand. On the other hand, I tried to describe the theoretical material in an interesting and understandable language. I am sure that after completing the training using my materials, you will not only successfully pass the OGE in chemistry, but also fall in love with this subject.

General information about the exam

OGE in chemistry consists of three parts.

In the first part 15 tasks with one answer- this is the first level and the tasks in it are not difficult, provided, of course, you have basic knowledge of chemistry. These tasks do not require calculations, with the exception of task 15.

The second part consists of four questions- in the first two - 16 and 17, you need to choose two correct answers, and in 18 and 19, correlate the values ​​or statements from the right column with the left.

The third part is problem solving. At 20 you need to equalize the reaction and determine the coefficients, and at 21 you need to solve the calculation problem.

Fourth part - practical, is not difficult, but you need to be careful and careful, as always when working with chemistry.

Total amount given for work 140 minutes.

Below are typical variants of tasks, accompanied by the theory necessary for the solution. All tasks are thematic - opposite each task a topic is indicated for general understanding.