Solving chemical reaction equations causes difficulties for a considerable number of secondary school students, largely due to the wide variety of elements involved in them and the ambiguity of their interactions. But since the main part of the general chemistry course at school examines the interaction of substances based on their reaction equations, students must necessarily fill gaps in this area and learn to solve chemical equations in order to avoid problems with the subject in the future.
The equation of a chemical reaction is a symbolic notation that displays the interacting chemical elements, their quantitative ratio and the substances resulting from the interaction. These equations reflect the essence of the interaction of substances from the point of view of atomic-molecular or electronic interaction.
- At the very beginning of the school chemistry course, they are taught to solve equations based on the concept of valence of elements of the periodic table. Based on this simplification, let us consider the solution of a chemical equation using the example of the oxidation of aluminum with oxygen. Aluminum reacts with oxygen to form aluminum oxide. Having the specified initial data, we will draw up an equation diagram.
Al + O 2 → AlO
In this case, we have written down an approximate diagram of a chemical reaction, which only partially reflects its essence. The substances involved in the reaction are written on the left side of the diagram, and the result of their interaction is written on the right. Additionally, oxygen and other typical oxidizing agents are usually written to the right of metals and other reducing agents on both sides of the equation. The arrow shows the direction of the reaction.
- In order for this compiled reaction scheme to acquire a complete form and comply with the law of conservation of mass of substances, it is necessary:
- Place indices on the right side of the equation for the substance resulting from the interaction.
- Level the amount of elements participating in the reaction with the amount of the resulting substance in accordance with the law of conservation of mass of substances.
- Place indices on the right side of the equation for the substance resulting from the interaction.
- Let's start by suspending the subscripts in the chemical formula of the finished substance. Indices are set in accordance with the valence of chemical elements. Valence is the ability of atoms to form compounds with other atoms due to the combination of their unpaired electrons, when some atoms give up their electrons, while others add them to themselves at an external energy level. It is generally accepted that the valency of a chemical element is determined by its group (column) in the periodic table. However, in practice, the interaction of chemical elements is much more complex and varied. For example, the oxygen atom has a valence of Ⅱ in all reactions, despite the fact that it is in the sixth group in the periodic table.
- To help you navigate this diversity, we offer you the following small reference assistant that will help you determine the valence of a chemical element. Select the element you are interested in and you will see the possible values of its valence. Rare valencies for the selected element are indicated in brackets.
- Let's return to our example. Let us write down its valence on the right side of the reaction diagram above each element.
For aluminum Al the valence will be equal to Ⅲ, and for the oxygen molecule O 2 the valence will be equal to Ⅱ. Find the least common multiple of these numbers. It will be equal to six. We divide the least common multiple by the valence of each element and get the indices. For aluminum, divide six by valence to obtain an index of 2, for oxygen 6/2 = 3. The chemical formula of aluminum oxide obtained as a result of the reaction will take the form Al 2 O 3.
Al + O 2 → Al 2 O 3
- After obtaining the correct formula of the finished substance, it is necessary to check and, in most cases, equalize the right and left parts of the diagram according to the law of conservation of mass, since the reaction products are formed from the same atoms that were originally part of the starting substances participating in the reaction.
- Law of conservation of mass states that the number of atoms that entered into the reaction must be equal to the number of atoms resulting from the interaction. In our scheme, the interaction involves one aluminum atom and two oxygen atoms. As a result of the reaction, we obtain two aluminum atoms and three oxygen atoms. Obviously, the diagram must be leveled using coefficients for elements and matter in order for the law of conservation of mass to be observed.
- Equalization is also performed by finding the least common multiple, which is located between the elements with the largest indices. In our example, this will be oxygen with an index on the right side equal to 3 and on the left side equal to 2. The least common multiple in this case will also be equal to 6. Now we divide the least common multiple by the value of the largest index on the left and right sides of the equation and get the following indices for oxygen.
Al + 3∙O 2 → 2∙Al 2 O 3
- Now all that remains is to equalize the aluminum on the right side. To do this, put a coefficient of 4 on the left side.
4∙Al + 3∙O 2 = 2∙Al 2 O 3
- After arranging the coefficients, the equation of a chemical reaction corresponds to the law of conservation of mass and an equal sign can be placed between its left and right sides. The coefficients placed in the equation indicate the number of molecules of substances participating in the reaction and resulting from it, or the ratio of these substances in moles.
Chemistry test Chemical equations grade 8 with answers. The test contains 2 parts. Part 1 contains 15 basic level tasks. Part 2 contains 3 advanced level tasks.
Part 1
1. Are the following statements true?
A. The mass of the reactants is equal to the mass of the reaction products.
B. A chemical equation is a conventional notation of a chemical reaction using chemical formulas and mathematical symbols.
1) only A is correct
2) only B is correct
3) both judgments are correct
4) both judgments are incorrect
2. During a chemical reaction, the number of atoms of a certain element
1) only increases
2) only decreases
3) does not change
3. During a chemical reaction, the number of reactant molecules
1) only increases
2) only decreases
3) does not change
4) can either increase or decrease
4. During a chemical reaction, the number of molecules of reaction products
1) only increases
2) only decreases
3) does not change
4) can either increase or decrease
5.
CH 4 + O 2 → CO 2 + H 2 O.
1) 5
2) 6
3) 7
4) 8
6.
Write down the reaction equation according to the diagram:
FeS + O 2 → Fe 2 O 3 + SO 2.
1) 13
2) 15
3) 17
4) 19
7.
Write down the reaction equation according to the diagram:
Na 2 O + H 2 O → NaOH.
1) 4
2) 5
3) 6
4) 7
8.
Write down the reaction equation according to the diagram:
H 2 O + N 2 O 5 → HNO 3.
Give your answer as the sum of the coefficients in the reaction equation.
1) 7
2) 6
3) 5
4) 4
9.
Write down the reaction equation according to the diagram:
NaOH + N 2 O 3 → NaNO 2 + H 2 O.
Give your answer as the sum of the coefficients in the reaction equation.
1) 7
2) 6
3) 5
4) 4
10.
Create a reaction equation according to the scheme: Al 2 O 3 + HCI → AlCl 3 + H 2 O.
Give your answer as the sum of the coefficients in the reaction equation.
1) 10
2) 11
3) 12
4) 14
11.
Write down the reaction equation according to the diagram:
Fe(OH) 3 + H 2 SO 4 → Fe 2 (SO 4) 3 + H 2 O. Give your answer as the sum of the coefficients in the reaction equation.
1) 12
2) 13
3) 14
4) 15
12. Write down the reaction equation according to the diagram:
copper (II) hydroxide + hydrochloric acid → copper (II) chloride + water.
Give your answer as the sum of the coefficients in the reaction equation.
1) 7
2) 6
3) 5
4) 4
13. Write down the reaction equation according to the diagram:
aluminum hydroxide → aluminum oxide + water.
Give your answer as the sum of the coefficients in the reaction equation.
1) 4
2) 5
3) 6
4) 7
14. Write down the reaction equation according to the diagram:
iron (III) oxide + hydrogen → iron + water.
Give your answer as the sum of the coefficients in the reaction equation.
1) 6
2) 7
3) 8
4) 9
15. Write down the reaction equation according to the diagram:
calcium carbonate + hydrochloric acid → calcium chloride + water + carbon monoxide (IV).
Give your answer as the sum of the coefficients in the reaction equation.
1) 6
2) 7
3) 8
4) 9
Part 2
1. Establish a correspondence between the starting substances and the products of the corresponding chemical reactions. Give your answer as a sequence of numbers corresponding to the letters in the alphabet.
Starting materials
A) H 2 + O 2 →
B) C 2 H 6 + O 2 →
B) Al(OH) 3 + H 2 SO 4 →
D) Ca(NO 3) 2 + Na 3 PO 4 →
Reaction products
1) CO 2 + H 2 O
2) H 2 O
3) Ca 3 (PO 4) 2 + NaNO 3
4) Al 2 (SO 4) 3 + H 2 O
2. Establish a correspondence between the reaction scheme and the sum of the coefficients in the reaction equation. Give your answer as a sequence of numbers corresponding to the letters in the alphabet.
Reaction equations
A) Fe 3 O 4 + Al → Al 2 O 3 + Fe
B) P 2 O 5 + H 2 O → H 3 PO 4
B) Al + O 2 → Al 2 O 3
D) Fe(OH) 3 → Fe 2 O 3 + H 2 O
Sum of odds
1) 6
2) 9
3) 12
4) 18
5) 24
3. The law of conservation of mass of matter is part of the more general law of conservation of matter. Types of matter (energy and matter) are interconnected according to Einstein's formula: ΔE = Δm ⋅ s 2 (where the speed of light c = 3 ⋅ 10 8 m/s). If during the reaction, for example, ΔE = 90 kJ = 9 ⋅ 10 4 J of energy was released, then the mass of the system decreased by the amount: Δm = ΔE/s 2 = 9 ⋅ 10 4 /(3 ⋅ 10 8) 2 = 10 -12 kg = 10 -9 g. This value is less than the accuracy of analytical balances (10 -6 g). Therefore, changes in mass during chemical reactions can be neglected. Calculate the amount of energy released ΔE in kJ if the mass of the system during the reaction decreased by 2.5 ⋅ 10 -9 g. In your answer, write down the value ΔE without indicating the units of measurement.
Answers to the chemistry test Chemical equations grade 8
Part 1
1-3
2-3
3-2
4-1
5-2
6-3
7-1
8-4
9-2
10-3
11-1
12-2
13-3
14-4
15-1
Part 2
1-2143
2-5121
3-225
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Attention! Slide previews are for informational purposes only and may not represent all the features of the presentation. If you are interested in this work, please download the full version.
The purpose of the lesson: help students develop knowledge of a chemical equation as a conditional recording of a chemical reaction using chemical formulas.
Tasks:
Educational:
- systematize previously studied material;
- teach the ability to compose equations of chemical reactions.
Educational:
- develop communication skills (work in pairs, ability to listen and hear).
Educational:
- develop educational and organizational skills aimed at accomplishing the task;
- develop analytical thinking skills.
Lesson type: combined.
Equipment: computer, multimedia projector, screen, assessment sheets, reflection card, “set of chemical symbols”, notebook with printed base, reagents: sodium hydroxide, iron(III) chloride, alcohol lamp, holder, matches, Whatman paper, multi-colored chemical symbols.
Lesson presentation (Appendix 3)
Lesson structure.
I. Organizing time.
II. Updating knowledge and skills.
III. Motivation and goal setting.
IV. Learning new material:
4.1 combustion reaction of aluminum in oxygen;
4.2 decomposition reaction of iron (III) hydroxide;
4.3 algorithm for arranging coefficients;
4.4 minutes of relaxation;
4.5 set the coefficients;
V. Consolidation of acquired knowledge.
VI. Summing up the lesson and grading.
VII. Homework.
VIII. Final words from the teacher.
During the classes
Chemical nature of a complex particle
determined by the nature of elementary
components,
their number and
chemical structure.
D.I.Mendeleev
Teacher. Hello guys. Sit down.
Please note: you have a printed notebook on your desk. (Appendix 2), in which you will work today, and a score sheet in which you will record your achievements, sign it.
Updating knowledge and skills.
Teacher. We got acquainted with physical and chemical phenomena, chemical reactions and signs of their occurrence. We studied the law of conservation of mass of substances.
Let's test your knowledge. I suggest you open your printed notebooks and complete task 1. You are given 5 minutes to complete the task.
Test on the topic “Physical and chemical phenomena. Law of conservation of mass of substances.”
1. How do chemical reactions differ from physical phenomena?
- Change in shape and state of aggregation of a substance.
- Formation of new substances.
- Change of location.
2. What are the signs of a chemical reaction?
- Precipitate formation, color change, gas evolution.
3. In accordance with what law are equations of chemical reactions drawn up?
- The law of constancy of the composition of matter.
- Law of conservation of mass of matter.
- Periodic law.
- Law of dynamics.
- The law of universal gravitation.
4. The law of conservation of mass of matter discovered:
- DI. Mendeleev.
- C. Darwin.
- M.V. Lomonosov.
- I. Newton.
- A.I. Butlerov.
5. A chemical equation is called:
- Conventional notation of a chemical reaction.
Teacher. You've done the job. I suggest you check it out. Exchange notebooks and check each other. Attention to the screen. For each correct answer - 1 point. Enter the total number of points on the evaluation sheets.
Motivation and goal setting.
Teacher. Using this knowledge, today we will draw up equations of chemical reactions, revealing the problem “Is the law of conservation of mass of substances the basis for drawing up equations of chemical reactions”
Learning new material.
Teacher. We are accustomed to thinking that an equation is a mathematical example where there is an unknown, and this unknown needs to be calculated. But in chemical equations there is usually nothing unknown: everything is simply written down in them using formulas: which substances react and which are obtained during this reaction. Let's see the experience.
(Reaction of sulfur and iron compound.) Appendix 3
Teacher. From the point of view of the mass of substances, the reaction equation for the compound of iron and sulfur is understood as follows
Iron + sulfur → iron (II) sulfide (task 2 tpo)
But in chemistry, words are reflected by chemical signs. Write this equation using chemical symbols.
Fe + S → FeS
(One student writes on the board, the rest in TVET.)
Teacher. Now read it.
Students. An iron molecule interacts with a sulfur molecule to produce one molecule of iron (II) sulfide.
Teacher. In this reaction, we see that the amount of starting substances is equal to the amount of substances in the reaction product.
We must always remember that when composing reaction equations, not a single atom should be lost or unexpectedly appear. Therefore, sometimes, having written all the formulas in the reaction equation, you have to equalize the number of atoms in each part of the equation - set the coefficients. Let's see another experiment
(Combustion of aluminum in oxygen.) Appendix 4
Teacher. Let's write the equation of a chemical reaction (task 3 in TPO)
Al + O 2 → Al +3 O -2
To write the oxide formula correctly, remember that
Students. Oxygen in oxides has an oxidation state of -2, aluminum is a chemical element with a constant oxidation state of +3. LCM = 6
Al + O 2 → Al 2 O 3
Teacher. We see that 1 aluminum atom enters into the reaction, two aluminum atoms are formed. Two oxygen atoms enter, three oxygen atoms are formed.
Simple and beautiful, but disrespectful to the law of conservation of mass of substances - it is different before and after the reaction.
Therefore, we need to arrange the coefficients in this chemical reaction equation. To do this, let's find the LCM for oxygen.
Students. LCM = 6
Teacher. We put coefficients in front of the formulas for oxygen and aluminum oxide so that the number of oxygen atoms on the left and right is equal to 6.
Al + 3 O 2 → 2 Al 2 O 3
Teacher. Now we find that as a result of the reaction, four aluminum atoms are formed. Therefore, in front of the aluminum atom on the left side we put a coefficient of 4
Al + 3O 2 → 2Al 2 O 3Let us once again count all the atoms before and after the reaction. We bet equal.
4Al + 3O 2 _ = 2 Al 2 O 3
Teacher. Let's look at another example
(The teacher demonstrates an experiment on the decomposition of iron (III) hydroxide.)
Fe(OH) 3 → Fe 2 O 3 + H 2 O
Teacher. Let's arrange the coefficients. One iron atom reacts and two iron atoms are formed. Therefore, before the formula of iron hydroxide (3) we put a coefficient of 2.
Fe(OH) 3 → Fe 2 O 3 + H 2 OTeacher. We find that 6 hydrogen atoms enter into the reaction (2x3), 2 hydrogen atoms are formed.
Students. NOC =6. 6/2 = 3. Therefore, we set the coefficient of 3 for the water formula
2Fe(OH) 3 → Fe 2 O 3 + 3 H 2 O
Teacher. We count oxygen.
Students. Left – 2x3 =6; on the right – 3+3 = 6
Students. The number of oxygen atoms that entered into the reaction is equal to the number of oxygen atoms formed during the reaction. You can bet equally.
2Fe(OH) 3 = Fe 2 O 3 +3 H 2 O
Teacher. Now let's summarize everything that was said earlier and get acquainted with the algorithm for arranging coefficients in the equations of chemical reactions.
- Count the number of atoms of each element on the right and left sides of the chemical reaction equation.
- Determine which element has a changing number of atoms and find the LCM.
- Divide the NOC into indices to obtain coefficients. Place them before the formulas.
- Recalculate the number of atoms and repeat the action if necessary.
- The last thing to check is the number of oxygen atoms.
Teacher. You've worked hard and you're probably tired. I suggest you relax, close your eyes and remember some pleasant moments in life. They are different for each of you. Now open your eyes and make circular movements with them, first clockwise, then counterclockwise. Now move your eyes intensively horizontally: right - left, and vertically: up - down.
Now let’s activate our mental activity and massage our earlobes.
Teacher. We continue to work.
In printed notebooks we will complete task 5. You will work in pairs. You need to place the coefficients in the equations of chemical reactions. You are given 10 minutes to complete the task.
- P + Cl 2 →PCl 5
- Na + S → Na 2 S
- HCl + Mg →MgCl 2 + H 2
- N 2 + H 2 →NH 3
- H 2 O → H 2 + O 2
Teacher. Let's check the completion of the task ( the teacher questions and displays the correct answers on the slide). For each correctly set coefficient - 1 point.
You completed the task. Well done!
Teacher. Now let's get back to our problem.
Guys, what do you think, is the law of conservation of mass of substances the basis for drawing up equations of chemical reactions?
Students. Yes, during the lesson we proved that the law of conservation of mass of substances is the basis for drawing up equations of chemical reactions.
Consolidation of knowledge.
Teacher. We have studied all the main issues. Now let's do a short test that will allow you to see how you have mastered the topic. You should only answer “yes” or “no”. You have 3 minutes to work.
Statements.
- In the reaction Ca + Cl 2 → CaCl 2, coefficients are not needed.(Yes)
- In the reaction Zn + HCl → ZnCl 2 + H 2, the coefficient for zinc is 2. (No)
- In the reaction Ca + O 2 → CaO, the coefficient for calcium oxide is 2.(Yes)
- In the reaction CH 4 → C + H 2 no coefficients are needed.(No)
- In the reaction CuO + H 2 → Cu + H 2 O, the coefficient for copper is 2. (No)
- In the reaction C + O 2 → CO, a coefficient of 2 must be assigned to both carbon monoxide (II) and carbon. (Yes)
- In the reaction CuCl 2 + Fe → Cu + FeCl 2 no coefficients are needed.(Yes)
Teacher. Let's check the progress of the work. For each correct answer - 1 point.
Lesson summary.
Teacher. You did a good job. Now calculate the total number of points scored for the lesson and give yourself a grade according to the rating that you see on the screen. Give me your evaluation sheets so that you can enter your grade into the journal.
Homework.
Teacher. Our lesson came to an end, during which we were able to prove that the law of conservation of mass of substances is the basis for composing reaction equations, and we learned how to compose chemical reaction equations. And as a final point, write down your homework
§ 27, ex. 1 – for those who received a rating of “3”
ex. 2 – for those who received a rating of “4”
ex. 3 – for those who received a rating“5”
Final words from the teacher.
Teacher. I thank you for the lesson. But before you leave the office, pay attention to the table (the teacher points to a piece of Whatman paper with an image of a table and multi-colored chemical symbols). You see chemical signs of different colors. Each color symbolizes your mood.. I suggest you create your own table of chemical elements (it will differ from D.I. Mendeleev’s PSHE) - a table of the mood of the lesson. To do this, you must go to the sheet of music, take one chemical element, according to the characteristic that you see on the screen, and attach it to a table cell. I will do this first by showing you how comfortable I am working with you.
F I felt comfortable in the lesson, I received answers to all my questions.
F I was bored in class, I didn’t learn anything new.