The simplest reaction equation is:

Fe + S => FeS

The reaction equation must be able not only to write, but also to read. This equation in its simplest version reads as follows: an iron molecule interacts with a sulfur molecule, one molecule of iron sulfide is obtained.

The most difficult thing in writing a reaction equation is to write formulas for the reaction products, i.e. formed substances. There is only one rule here: the formulas of molecules are built strictly according to the valency of their constituent elements.

In addition, when compiling reaction equations, one must remember the law of conservation of the mass of substances: all atoms of the molecules of the initial substances must be part of the molecules of the reaction products. Not a single atom should disappear or suddenly appear. Therefore, sometimes, having written all the formulas in the reaction equation, it is necessary to equalize the number of atoms in each part of the equation - to arrange the coefficients. Here is an example:C + O 2 => CO 2

Here, each element has the same number of atoms on both the right and left sides of the equation. The equation is ready.

Cu+O 2 => CuO

And here there are more oxygen atoms on the left side of the equation than on the right. It takes so many copper oxide molecules to getCuO , so that they contain the same number of oxygen atoms, i.e. 2. Therefore, before the formulaСuО set factor 2:

Cu+O2 => 2 CuO

Now the number of copper atoms is not the same. On the left side of the equation, before the sign of copper, we put the coefficient 2:

2 Cu+O2 => 2 CuO

Count whether the atoms of each element are equal in the left and right sides of the equation. If so, then the reaction equation is correct.

One more example: Al+O 2 = Al 2 O 3

And here the atoms of each element different amount before and after the reaction. We begin to equalize with gas - with oxygen molecules:

1) Left 2 oxygen atoms, and on the right 3. We are looking for the least common multiple of these two numbers. This smallest number, which is divisible by both 2 and 3, i.e. 6. Before the formulas of oxygen and aluminum oxideAl 2 O 3 we set such coefficients that the total number of oxygen atoms in these molecules is 6:

Al+ 3O2= 2Al2O 3

2) We count the number of aluminum atoms: 1 atom on the left, and 2 atoms on the right in two molecules, i.e. 4. Before the sign of aluminum on the left side of the equation, we put the coefficient 4:

4 Al + 3O 2 => 2 Al2O3

3) Once again, we count all the atoms before and after the reaction: 4 aluminum atoms and 6 oxygen atoms each.

Everything is in order, the reaction equation is correct. And if the reaction proceeds when heated, then a sign is additionally placed above the arrow t .

The equation of a chemical reaction is a record of the course of a chemical reaction using chemical formulas and coefficients.

L Is it easy to arrange the coefficients in chemical equations?

Here are my children and they have grown up to chemistry (I am a class teacher in the 8th "B" class). Chemistry is most often given to children at the first lesson, but on Thursday I don’t have the first lesson, and I asked for a lesson with Valentina Ivanovna “to look at the children” and check the diaries. The topic fascinated me, at school I loved chemistry, and I did not check the diaries. Once again, I was convinced that students most often experience difficulties due to the fact that they do not see interdisciplinary connections. In this chemistry lesson, students had to write chemical equations, knowing the valency chemical substances. And many students had difficulty in determining the numerical coefficients. The next chemistry lesson on Saturday was spent together with Valentina Ivanovna.

Exercise 1.

Write the following sentences in the form of chemical equations:

A) “During the firing of calcium carbonate, calcium oxide and carbon monoxide (IV) are formed”; b) "When phosphorus (V) oxide reacts with water, phosphoric acid is obtained."

Solution:

A) CaCO 3 \u003d CaO + CO 2 - the reaction is endothermic. There were no difficulties with this task, since it was not necessary to look for numerical coefficients. Initially, in the left and right parts of the equality, one calcium atom, one carbon atom and three oxygen atoms each.

B) P 2 O 5 + 3H 2 O \u003d 2H 3 PO 4 - the reaction is exothermic. Problems arose with the second equation, without numerical coefficients the correct equality did not turn out: P 2 O 5 + H 2 O → H 3 PO 4. Obviously, to compile the correct equality, you need to select numerical coefficients. If you select, you can start with phosphorus: there are two atoms on the left, and one on the right, so we put a numerical factor equal to two in front of the nitric acid formula and then we get: P 2 O 5 + H 2 O → 2H 3 PO 4. But now it remains to equalize the number of oxygen and hydrogen atoms: there are two hydrogen atoms on the left, and six atoms on the right, so we put a numerical coefficient equal to three in front of the water formula and then we get: P 2 O 5 + 3H 2 O → 2H 3 PO 4. Now it is easy to make sure that in each part of the equation there are equal amounts of phosphorus atoms and hydrogen atoms and oxygen atoms, therefore, we have obtained the correct chemical reaction equation: P 2 O 5 + 3H 2 O \u003d 2H 3 PO 4.

Second way: algebraic. Let's assume that we put three coefficients in the equation a, b, c , resulting in the correct chemical reaction equation: A P2O5+ V H 2 O = With H3PO4. Since the equation uses atoms of three types, we will compose a system of three linear equations with three unknowns a, in And With .

Substances that were used in the chemical reaction: P - phosphorus; O 2 - oxygen; P 2 O 5 - phosphorus oxide (V).

C) Fe 2 (SO 4) 3 + KOH → Fe (OH) 3 + K 2 SO 4.

Solution: ) Fe 2 (SO 4) 3 + 6KOH \u003d 2Fe (OH) 3 + 3K 2 SO 4. We decided by selection: we equalized the number of iron atoms (2); equalized the number of sulfur atoms (3); equalized the number of potassium atoms (6); equalize the number of oxygen atoms.

Substances that were used in the chemical reaction: Fe 2 (SO 4) 3 - iron sulfate (III); KOH, potassium hydroxide; Fe (OH) 3 - iron (III) hydroxide; K 2 SO 4 - potassium sulfate.

D) CuOH → Cu 2 O + H 2 O.

Solution: 2CuOH \u003d Cu 2 O + H 2 O. The problem of determining the numerical coefficients was solved by compiling a system of equations:

Substances that were used in the chemical reaction: CuOH - copper (I) hydroxide; Cu 2 O - copper (I) oxide; H 2 O - water.

E) CS 2 + O 2 → CO 2 + SO 2.

Solution: CS 2 + 3O 2 \u003d CO 2 + 2SO 2. Decided by the selection of coefficients: equalized the number of sulfur atoms (2); equalized the number of oxygen atoms (3).

Substances that were used in the chemical reaction: CS 2 - sulfur sulfide (IV); O2-
Substances that were used in the chemical reaction: FeS 2 - pyrites; O 2 - oxygen; Fe 2 O 3 - iron oxide (III); SO 2 - sulfur oxide (IV).
Exercise 3

(It was proposed to be solved as an independent work).

Condition:

Write down the equations of chemical reactions according to the following schemes:

A) phosphoric acid + sodium hydroxide → sodium phosphate + water;

B) sodium oxide + water → sodium hydroxide;

C) iron oxide (II) + aluminum → aluminum oxide + iron;

D) copper (II) hydroxide → copper (II) oxide + water.

Answer:

A) 2H 3 PO 4 + 6NaOH \u003d 2Na 3 PO 4 + 6H 2 O;

B) Na 2 O + H 2 O \u003d 2NaOH;

C) 3FeO + 2Al = Al 2 O 3 + 3Fe;

D) Cu (OH) 2 \u003d CuO + H 2 O.

In 10 minutes, 85% of the students completed the task with excellent marks, which pleasantly surprised Valentina Ivanovna.

Today we will talk about how to arrange coefficients in chemical equations. This question is of interest not only to high school students. educational institutions, but also guys who are just getting acquainted with the basic elements of a complex and interesting science. If you understand at the first stage, in the future there will be no problems with solving problems. Let's get it right from the start.

What is an equation

By it it is customary to mean a conditional record of a chemical reaction occurring between the selected reagents. For such a process, indices, coefficients, formulas are used.

Compilation algorithm

How to write chemical equations? Examples of any interactions can be written by summing up the original compounds. The equal sign indicates that there is an interaction between the reacting substances. Next, a formula for products by valence (oxidation state) is compiled.

How to record a reaction

For example, if you need to write down chemical equations that confirm the properties of methane, choose the following options:

  • halogenation (radical interaction with element VIIA of the periodic table of D. I. Mendeleev);
  • combustion in atmospheric oxygen.

For the first case, on the left side we write starting materials, in the right - the received products. After checking the number of atoms of each chemical element we get the final record of the ongoing process. When methane burns in atmospheric oxygen, an exothermic process occurs, as a result of which carbon dioxide and water vapor are formed.

In order to correctly put the coefficients in chemical equations, the law of conservation of mass of substances is used. We start the adjustment process by determining the number of carbon atoms. Next, we carry out calculations for hydrogen and only after that we check the amount of oxygen.

OVR

Complex chemical equations can be equalized using the method of electronic balance or half-reactions. We offer a sequence of actions designed to arrange the coefficients in the reactions the following types:

  • decomposition;
  • substitution.

First, it is important to arrange the oxidation state of each element in the compound. When placing them, it is necessary to take into account some rules:

  1. For a simple substance, it is equal to zero.
  2. In a binary compound, their sum is 0.
  3. In a compound of three or more elements, the first shows a positive value, the extreme ion - negative meaning degree of oxidation. The central element is calculated mathematically, given that the sum should be 0.

Next, those atoms or ions are selected for which the oxidation state has changed. The plus and minus signs show the number of electrons (accepted, given away). Next, the smallest multiple is determined between them. When dividing the NOC by these numbers, the numbers are obtained. This algorithm will be the answer to the question of how to arrange the coefficients in chemical equations.

First example

Let's say the task is given: "Arrange the coefficients in the reaction, fill in the gaps, determine the oxidizing agent and reducing agent." Such examples are offered to school graduates who have chosen chemistry as their exam.

KMnO 4 + H 2 SO 4 + KBr = MnSO 4 + Br 2 +…+…

Let's try to understand how to arrange the coefficients in the chemical equations offered to future engineers and physicians. After arranging the oxidation states of the elements in the starting materials and available products, we find that the manganese ion acts as an oxidizing agent, and the bromide ion demonstrates reducing properties.

We conclude that the missing substances do not participate in the redox process. One of the missing products is water, and the second will be potassium sulfate. After compiling the electronic balance, the final step will be the setting of the coefficients in the equation.

Second example

Let's give another example to understand how to arrange the coefficients in the chemical equations of the redox type.

Let's say we have the following schema:

P + HNO 3 \u003d NO 2 + ... + ...

Phosphorus, which by convention is a simple substance, exhibits reducing properties, increasing the oxidation state to +5. Therefore, one of the missing substances will be phosphoric acid H 3 PO 4. OVR assumes the presence of a reducing agent, which will be nitrogen. It goes into nitric oxide (4), forming NO 2

In order to put the coefficients in this reaction, we will make an electronic balance.

P 0 gives back 5e = P +5

N +5 takes e = N +4

Given that nitric acid and nitric oxide (4) must be preceded by a factor of 5, we get the finished reaction:

P + 5HNO 3 \u003d 5NO 2 + H 2 O + H 3 PO 4

Stereochemical coefficients in chemistry allow solving various computational problems.

Third example

Given that the placement of coefficients causes difficulties for many high school students, it is necessary to work out the sequence of actions using specific examples. We offer another example of a task, the implementation of which requires mastery of the method of arranging the coefficients in the redox reaction.

H 2 S + HMnO 4 \u003d S + MnO 2 +…

The peculiarity of the proposed task is that it is necessary to supplement the missing reaction product, and only after that you can proceed to setting the coefficients.

After arranging the oxidation states of each element in the compounds, it can be concluded that manganese, which lowers the valence, exhibits oxidizing properties. Sulfur demonstrates the reducing ability in the proposed reaction, being reduced to a simple substance. After compiling the electronic balance, we will only have to place the coefficients in the proposed process scheme. And the deed is done.

Fourth example

A chemical equation is called a complete process when the law of conservation of mass of substances is fully observed in it. How to check this pattern? The number of atoms of the same type that have entered into the reaction must correspond to their number in the interaction products. Only in this case it will be possible to talk about the usefulness of the recorded chemical interaction, the possibility of its application for calculations, solving computational problems different levels difficulties. Here is a variant of the task, which involves the arrangement of the missing stereochemical coefficients in the reaction:

Si + ... + HF = H 2 SiF 6 + NO + ...

The complexity of the task is that both the starting substances and the products of the interaction are omitted. After setting all the elements of the oxidation states, we see that the silicon atom exhibits reducing properties in the proposed task. Nitrogen (II) is present among the reaction products, one of the starting compounds is nitric acid. Logically, we determine that the missing product of the reaction is water. The final step will be the arrangement of the obtained stereochemical coefficients in the reaction.

3Si + 4HNO 3 + 18HF = 3H 2 SiF 6 + 4NO + 8 H 2 O

An example of an equation problem

It is necessary to determine the volume of a 10% solution of hydrogen chloride, the density of which is 1.05 g / ml, necessary for the complete neutralization of calcium hydroxide formed during the hydrolysis of its carbide. It is known that the gas released during hydrolysis occupies a volume of 8.96 liters (n.a.).

CaC 2 + 2H 2 O \u003d Ca (OH) 2 + C 2 H 2

Calcium hydroxide interacts with hydrogen chloride, complete neutralization occurs:

Ca (OH) 2 + 2HCl \u003d CaCl 2 + 2H 2 O

We calculate the mass of acid that is required for this process. Determine the volume of the hydrogen chloride solution. All calculations for the problem are carried out taking into account stereochemical coefficients, which confirms their importance.

Finally

An analysis of the results of the unified state exam in chemistry indicates that tasks related to setting stereochemical coefficients in equations, compiling an electronic balance, determining an oxidizing agent and a reducing agent cause serious difficulties for modern graduates. general education schools. Unfortunately, the degree of independence of modern graduates is practically minimal, so high school students do not work out the theoretical base proposed by the teacher.

Among the typical mistakes that schoolchildren make when placing coefficients in reactions different type, many mathematical errors. For example, not everyone knows how to find the least common multiple, divide and multiply numbers correctly. The reason for this phenomenon is the decrease in the number of hours devoted to educational schools to study this topic. With a basic program in chemistry, teachers do not have the opportunity to work out with their students issues related to the compilation of an electronic balance in the redox process.

COEFFICIENT ARRANGEMENT

The number of atoms of one element on the left side of the equation must be equal to the number of atoms of that element on the right side of the equation.

Task 1 (for groups).Determine the number of atoms of each chemical element involved in the reaction.

1. Calculate the number of atoms:

a) hydrogen: 8NH3, NaOH, 6NaOH, 2NaOH, H3PO4, 2H2SO4, 3H2S04, 8H2SO4;

6) oxygen: C02, 3C02, 2C02, 6CO, H2SO4, 5H2SO4, 4H2S04, HN03.

2. Calculate the number of atoms: a)hydrogen:

1) NaOH + HCl 2) CH4+H20 3) 2Na+H2

b) oxygen:

1) 2CO + 02 2) CO2 + 2H.O. 3)4NO2 + 2H2O + O2

Algorithm for arranging coefficients in the equations of chemical reactions

A1 + O2 → A12O3

A1-1 atom A1-2

O-2 atoms O-3

2. Among the elements with different number atoms in the left and right parts of the scheme, choose the one whose number of atoms is greater

O-2 atoms on the left

O-3 atoms on the right

3. Find the least common multiple (LCM) of the number of atoms of this element on the left side of the equation and the number of atoms of this element on the right side of the equation

LCM = 6

4. Divide the LCM by the number of atoms of this element on the left side of the equation, get the coefficient for the left side of the equation

6:2 = 3

Al + 3O 2 →Al 2 ABOUT 3

5. Divide the LCM by the number of atoms of this element on the right side of the equation, get the coefficient for the right side of the equation

6:3 = 2

A1+ O 2 →2А1 2 O3

6. If the set coefficient changed the number of atoms of another element, then repeat steps 3, 4, 5 again.

A1 + 3O 2 → →2А1 2 ABOUT 3

A1 -1 atom A1 - 4

LCM = 4

4:1=4 4:4=1

4A1 + ZO 2 → →2А1 2 ABOUT 3

. Primary test of knowledge acquisition (8-10 min .).

There are two oxygen atoms on the left side of the diagram, and one on the right side. The number of atoms must be aligned using coefficients.

1)2Mg+O2 →2MgO

2) CaCO3 + 2HCl → CaCl2 + H2 O + CO2

Task 2 Arrange the coefficients in the equations of chemical reactions (note that the coefficient changes the number of atoms of only one element):

1. Fe 2 O 3 + A l → A l 2 ABOUT 3 + Fe; Mg+N 2 → mg 3 N 2 ;

2 Al+S → Al 2 S 3 ; A1+ WITH → Al 4 C 3 ;

3. Al + Cr 2 O 3 → Cr + Al 2 O 3 ; Ca+P → Ca 3 P 2 ;

4. C + H 2 → CH 4 ; Ca + C → CaS 2 ;

5. Fe+O 2 → Fe 3 O 4 ; Si+Mg → mg 2 Si;

6/.Na + S → Na 2 S; CaO+ WITH → CaC 2 + CO;

7.Ca+N 2 → C a 3 N 2 ; Si + Cl 2 → SiCl 4 ;

8 Ag+S → Ag 2 S; H 2 + WITH l 2 → NS l;

9. N 2 +O 2 → NO; SO 2 + WITH → SO ;

10.HI → H 2 → + 1 2 ; mg+ NS l → MgCl 2 + H 2 ;

11. FeS+ NS 1 → FeCl 2 + H 2 S; Zn + HCl → ZnCl 2 + H 2 ;

12.Br 2 +KI → KBr+I 2 ; Si+HF (r) → SiF 4 + H 2 ;

1./HCl+Na 2 CO 3 → CO 2 +H 2 O + NaCl; KClO 3 +S → → KCl + SO 2 ;

14.Cl 2 +KBr → KCl + Br 2 ; SiO 2 + WITH → Si + CO;

15. SiO 2 + WITH → SiC+CO; Mg+SiO 2 → mg 2 Si+MgO

16 .

3. What does the "+" sign mean in the equation?

4. Why place the coefficients in the chemical equations

The teacher, being the main actor In the organisation cognitive activity students are constantly looking for ways to improve the effectiveness of learning. Organization effective learning is possible only with knowledge and skillful use of various forms of the pedagogical process.

1. Modern man must possess not only the sum of knowledge and skills, but also the ability to perceive the world as a single, complex, constantly evolving whole.

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Article on chemistry: "Arrangement of coefficients in chemical equations"

Compiled by: chemistry teacher

GBOU secondary school No. 626

Kazutina O.P.

Moscow 2012

"Arrangement of coefficients in chemical equations"

The teacher, being the main character in the organization of the cognitive activity of students, is constantly in search of ways to improve the effectiveness of learning. The organization of effective learning is possible only with the knowledge and skillful use of various forms of the pedagogical process.

1. A modern person must have not only the sum of knowledge and skills, but also the ability to perceive the world as a single, complex, constantly evolving whole.

Algorithm of work on preparation for the lesson

choice of topic, definition of goal setting;

content selection;

determining the means and ways of developing a positive motivational attitude for students to work in the lesson;

specifying the equipment of the lesson with the necessary visual and didactic material;

lesson plan development

An example of a chemistry lesson "Arrangement of coefficients in a chemical equation" for teachers

Target: answer the question: "why do we need to arrange the coefficients in the chemical equation"

Tasks:

The problem of the need to place the coefficients

Algorithm for placing coefficients

Proof of the meaning of the arrangement of coefficients

During the classes:

A modern student, if he studies, treats the knowledge he receives and processes with pragmatism. Therefore, the material provided should fit in the head logically and concisely.

To achieve this, the teacher should always pay attention to For what you need to learn one or the other action in the lesson. That is, the teacher must explain. And then, in a good way, wait for the right questions on a new topic.

The law of conservation of mass of substances

The famous English chemist R. Boyle, calcining various metals in an open retort and weighing them before and after heating, found that the mass of metals becomes larger. Based on these experiments, he did not take into account the role of air and made the wrong conclusion that the mass of substances changes as a result of chemical reactions. R. Boyle argued that there is some kind of "fiery matter", which, in the case of heating the metal, combines with the metal, increasing the mass.

Mg + O 2  MgO

24 g 40 g
M. V. Lomonosov, unlike R. Boyle, calcined metals not in the open air, but in sealed retorts and weighed them before and after calcination. He proved that the mass of substances before and after the reaction remains unchanged and that when calcined, some part of the air is added to the metal. (Oxygen had not yet been discovered at that time.) He formulated the results of these experiments in the form of a law: "All changes that occur in nature are such a state of being that how much of what is taken away from one body will be added to another." This law is currently formulated as follows:
The mass of substances that entered into chemical reaction, is equal to the mass of formed substances

Mg + O 2  MgO

24 g 32 g 40 g

Question: the law is not fulfilled (since the masses of the initial and final substances are not equal).

The solution to this problem is the arrangement of coefficients (integer numbers indicating the number of molecules):

2Mg + O 2  2MgO

48 g 32 g 80 g - the masses before and after are equal due to the fact that the number of atoms of the elements is also equal before and after the reaction.

Thus, having proved to students the need to equalize the mass coefficients, you can even do without some of the previous topics: formulating substances by valence, calculating mass, amount of substance ... Also the story that the law of conservation of mass of matter 20 years later "rediscovered" A. Lavoisier, clarifying it on the one hand, but completely ignoring M.V. Lomonosov with ethical, you can leave on independent study in the form of a report, for example.

So, for the successful completion of tasks of this kind, it is necessary to learn the condition: the number of atoms before the reaction db is equal to the number of atoms after the reaction: we will decide together:

H 2 S + 3O 2  SO 2 + 2H 2 O (double the oxygens on the right. Count them on the left)

CH 4 + 2O 2  CO 2 + 2H 2 O

We have placed the coefficients in the equations of combustion of two gases