Chemistry how to arrange coefficients in chemical equations. How to arrange coefficients in chemical equations? Chemical Equations

In order to figure out how to equalize a chemical equation, you first need to know the purpose of this science.

Definition

Chemistry studies substances, their properties, and transformations. If there is no change in color, precipitation, release of a gaseous substance, then no chemical interaction occurs.

For example, when filing an iron nail with a file, the metal simply turns into powder. In this case, no chemical reaction occurs.

The calcination of potassium permanganate is accompanied by the formation of manganese oxide (4), the release of oxygen, that is, an interaction is observed. In this case, a completely natural question arises about how to properly equalize chemical equations. We will analyze all the nuances associated with such a procedure.

Specificity of chemical transformations

Any phenomena that are accompanied by a change in the qualitative and quantitative composition of substances are chemical transformations. In molecular form, the process of iron combustion in the atmosphere can be expressed using signs and symbols.

The method of placing the coefficients

How to equalize coefficients in chemical equations? In the course of chemistry high school understands the method of electronic balance. Let's consider the process in more detail. To begin with, in the initial reaction, it is necessary to arrange the oxidation states of each chemical element.

Exist certain rules, by which they can be determined for each element. In simple substances, the oxidation states will be zero. In binary compounds, it is positive for the first element, corresponds to highest valence. For the latter, this parameter is determined by subtracting the group number from eight and has a minus sign. Formulas consisting of three elements have their own nuances for calculating oxidation states.

For the first and last element, the order is similar to the definition in binary compounds, and an equation is made to calculate the central element. The sum of all indicators must be equal to zero, based on this, the indicator for the middle element of the formula is calculated.

Let's continue the conversation about how to equalize chemical equations using the electron balance method. After the oxidation states are set, it is possible to determine those ions or substances that have changed their value during the chemical interaction.

The plus and minus signs indicate the number of electrons that were accepted (given away) in the process of chemical interaction. Between the numbers obtained, find the least common multiple.

When dividing it into received and given electrons, coefficients are obtained. How to balance a chemical equation? The figures obtained in the balance sheet must be placed in front of the corresponding formulas. A prerequisite is to check the number of each element in the left and right parts. If the coefficients are placed correctly, their number should be the same.

The law of conservation of mass of substances

Arguing over how to equalize a chemical equation, it is necessary to use this law. Given that the mass of those substances that entered into a chemical reaction is equal to the mass of the resulting products, it becomes possible to set coefficients in front of the formulas. For example, how to equalize a chemical equation if simple substances calcium and oxygen interact, and after the process is completed, an oxide is obtained?

To cope with the task, it must be taken into account that oxygen is a diatomic molecule with a covalent non-polar bond, so its formula is written in the following form - O2. On the right side, when compiling calcium oxide (CaO), the valencies of each element are taken into account.

First you need to check the amount of oxygen in each part of the equation, as it is different. According to the law of conservation of mass of substances, a factor of 2 must be put in front of the product formula. Next, calcium is checked. In order for it to be equalized, we put a factor of 2 in front of the original substance. As a result, we get the record:

• 2Ca+O2=2CaO.

Analysis of the reaction by the electronic balance method

How to equalize chemical equations? Examples of RIAs will help answer this question. Suppose that it is necessary to place the coefficients in the proposed scheme using the electronic balance method:

• CuO + H2=Cu + H2O.

To begin with, for each of the elements in the initial substances and interaction products, we will place the values ​​of the oxidation states. We get the following form of the equation:

• Cu(+2)O(-2)+H2(0)=Cu(0)+H2(+)O(-2).

The indicators have changed for copper and hydrogen. It is on their basis that we will draw up an electronic balance:

• Cu(+2)+2e=Cu(0) 1 reducing agent, oxidation;
• H2(0)-2e=2H(+) 1 oxidizing agent, reduction.

Based on the coefficients obtained in the electronic balance, we obtain the following record of the proposed chemical equation:

• CuO+H2=Cu+H2O.

Let's take another example that involves setting coefficients:

• H2+O2=H2O.

In order to equalize this scheme on the basis of the law of conservation of substances, it is necessary to start with oxygen. Considering that a diatomic molecule entered into the reaction, it is necessary to put a factor of 2 before the formula of the interaction product.

• 2H2+O2=2H2O.

Conclusion

Based on the electronic balance, you can place the coefficients in any chemical equations. Graduates of the ninth and eleventh grades educational institutions, choosing an exam in chemistry, in one of the tasks of the final tests they offer similar tasks.

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

1. Let's make a reaction scheme:

Lesson objectives.Educational. To acquaint students with a new classification of chemical reactions on the basis of changes in the oxidation states of elements - with redox reactions (ORD); teach students to arrange coefficients using the electronic balance method.

Developing. Continue development logical thinking, the ability to analyze and compare, the formation of interest in the subject.

Educational. To form a scientific worldview of students; improve work skills.

Methods and methodological techniques. Story, conversation, demonstration of visual aids, independent work students.

Equipment and reagents. Reproduction depicting the Colossus of Rhodes, the algorithm for placing coefficients according to the electronic balance method, a table of typical oxidizing and reducing agents, a crossword puzzle; Fe (nail), solutions of NaOH, CuSO4.

DURING THE CLASSES

Introduction

(motivation and goal setting)

Teacher. In the III century. BC. on the island of Rhodes, a monument was built in the form of a huge statue of Helios (among the Greeks - the god of the Sun). The grandiose idea and perfection of execution of the Colossus of Rhodes - one of the wonders of the world - amazed everyone who saw it.

We do not know exactly what the statue looked like, but it is known that it was made of bronze and reached a height of about 33 m. The statue was created by the sculptor Haret and took 12 years to build.

The bronze shell was attached to the iron frame. The hollow statue began to be built from the bottom and, as it grew, it was filled with stones to make it more stable. Approximately 50 years after the completion of construction, the Colossus collapsed. During the earthquake, he broke at the level of his knees.

Scientists believe that the true reason for the fragility of this miracle was the corrosion of the metal. And at the heart of the corrosion process are redox reactions.

Today in the lesson you will get acquainted with redox reactions; learn about the concepts of "reducing agent" and "oxidizing agent", about the processes of reduction and oxidation; learn how to arrange the coefficients in the equations of redox reactions. Write in your workbooks the number, the topic of the lesson.

Learning new material

The teacher makes two demonstration experiments: the interaction of copper (II) sulfate with alkali and the interaction of the same salt with iron.

Teacher. Write down the molecular equations of the reactions performed. In each equation, arrange the oxidation states of the elements in the formulas of the starting materials and reaction products.

The student writes the reaction equations on the board and arranges the oxidation states:

Teacher. Did the oxidation states of the elements change in these reactions?

Student. In the first equation, the oxidation states of the elements did not change, but in the second they changed - in copper and iron.

Teacher. The second reaction is redox. Try to define redox reactions.

Student. Reactions, as a result of which the oxidation states of the elements that make up the reactants and reaction products change, are called redox reactions.

Students write down in a notebook under the dictation of the teacher the definition of redox reactions.

Teacher. What happened as a result of the redox reaction? Before the reaction, iron had an oxidation state of 0, after the reaction it became +2. As you can see, the oxidation state has increased, therefore, iron gives up 2 electrons.

Copper has an oxidation state of +2 before the reaction, and 0 after the reaction. As you can see, the oxidation state has decreased. Therefore, copper accepts 2 electrons.

Iron donates electrons, it is a reducing agent, and the process of electron transfer is called oxidation.

Copper accepts electrons, it is an oxidizing agent, and the process of adding electrons is called reduction.

We write the schemes of these processes:

So, give the definition of the concepts of "reducing agent" and "oxidizing agent".

Student. Atoms, molecules or ions that donate electrons are called reducing agents.

Atoms, molecules, or ions that accept electrons are called oxidizing agents.

Teacher. What is the definition of reduction and oxidation processes?

Student. Recovery is the process of adding electrons to an atom, molecule or ion.

Oxidation is the process by which electrons are transferred by an atom, molecule, or ion.

Students write the definitions in a notebook under dictation and complete the drawing.

Remember!

Donate electrons - oxidize.

Take electrons - recover.

Teacher. Oxidation is always accompanied by reduction, and vice versa, reduction is always associated with oxidation. The number of electrons donated by the reducing agent is equal to the number of electrons attached by the oxidizing agent.

To select the coefficients in the equations of redox reactions, two methods are used - electron balance and electron-ion balance (half-reaction method).

We will consider only the electronic balance method. To do this, we use the algorithm for arranging the coefficients using the electronic balance method (drawn up on a piece of drawing paper).

EXAMPLE Arrange the coefficients in this reaction scheme using the electron balance method, determine the oxidizing agent and reducing agent, indicate the processes of oxidation and reduction:

Fe2O3 + CO Fe + CO2.

We will use the algorithm for placing the coefficients using the electronic balance method.

3. Let's write out the elements that change the degree of oxidation:

4. Compose electronic equations, determining the number of given and received electrons:

5. The number of given and received electrons must be the same, because not charged either starting materials, nor reaction products. We equalize the number of given and received electrons by choosing the least common multiple (LCM) and additional factors:

6. The resulting multipliers are coefficients. We transfer the coefficients to the reaction scheme:

Fe2O3 + 3CO = 2Fe + 3CO2.

Substances that are oxidizing or reducing agents in many reactions are called typical.

A table made on a Whatman sheet is posted.

Teacher. Redox reactions are very common. They are associated not only with corrosion processes, but also with fermentation, decay, photosynthesis, and metabolic processes occurring in a living organism. They can be observed during the combustion of fuel.

How to equalize a chemical equation: rules and algorithm

Redox processes accompany the cycles of substances in nature.

Did you know that about 2 million tons of nitric acid are formed in the atmosphere every day, or
700 million tons per year, and in the form of a weak solution fall to the ground with rain (man produces only 30 million tons of nitric acid per year).

What happens in the atmosphere?

Air contains 78% nitrogen by volume, 21% oxygen and 1% other gases. Under the action of lightning discharges, and an average of 100 lightning flashes on Earth every second, nitrogen molecules interact with oxygen molecules to form nitric oxide (II):

Nitric oxide (II) is easily oxidized by atmospheric oxygen to nitric oxide (IV):

The resulting nitric oxide (IV) interacts with atmospheric moisture in the presence of oxygen, turning into nitric acid:

NO2 + H2O + O2 HNO3.

All these reactions are redox reactions.

Exercise . Arrange the coefficients in the above reaction schemes using the electronic balance method, indicate the oxidizing agent, reducing agent, oxidation and reduction processes.

Solution

1. Let's determine the oxidation states of the elements:

2. We underline the symbols of elements whose oxidation states change:

3. Let's write out the elements that have changed their oxidation states:

4. Compose electronic equations (determine the number of given and received electrons):

5. The number of given and received electrons is the same.

6. Let's transfer the coefficients from electronic circuits to the reaction scheme:

Next, students are invited to independently arrange the coefficients using the electronic balance method, determine the oxidizing agent, reducing agent, indicate the processes of oxidation and reduction in other processes occurring in nature.

The other two reaction equations (with coefficients) are:

Checking the correctness of the tasks is carried out using a codoscope.

Final part

The teacher asks students to solve a crossword puzzle based on the material studied. The result of the work is submitted for verification.

Having guessed crossword, you will find out that the substances KMnO4, K2Cr2O7, O3 are strong ... (along the vertical (2)).

Horizontally:

1. What process does the scheme reflect:

3. Reaction

N2 (g.) + 3H2 (g.) 2NH3 (g.) + Q

is redox, reversible, homogeneous, … .

4. ... carbon(II) is a typical reducing agent.

5. What process does the scheme reflect:

6. For the selection of coefficients in the equations of redox reactions, the method of electronic ... is used.

7. According to the diagram, aluminum gave ... an electron.

8. In reaction:

H2 + Cl2 = 2HCl

hydrogen H2 - ... .

9. What type of reactions are always only redox reactions?

10. The oxidation state of simple substances is ....

11. In reaction:

reducer...

Homework assignment.

According to O.S. Gabrielyan's textbook "Chemistry-8" § 43, p. 178–179, ex. 1, 7 in writing. A task (at home). Constructors of the first spaceships and submarines faced a problem: how to maintain a constant composition of the air on the ship and space stations? Get rid of excess carbon dioxide and replenish oxygen? The solution has been found.

Potassium superoxide KO2 forms oxygen as a result of interaction with carbon dioxide:

As you can see, this is a redox reaction. Oxygen is both an oxidizing agent and a reducing agent in this reaction.

In a space expedition, every gram of cargo counts. Calculate the supply of potassium superoxide that must be taken on a space flight if the flight is designed for 10 days and if the crew consists of two people. It is known that a person exhales 1 kg of carbon dioxide per day.

(Answer. 64.5 kg KO2. )

Exercise ( elevated level difficulties). Write down the equations for the redox reactions that could have led to the destruction of the Colossus of Rhodes. Keep in mind that this giant statue stood in a port city on an island in the Aegean off the coast of modern Turkey, where the humid Mediterranean air is saturated with salts. It was made of bronze (an alloy of copper and tin) and mounted on an iron frame.

Literature

Gabrielyan O.S.. Chemistry-8. Moscow: Bustard, 2002;
Gabrielyan O.S., Voskoboynikova N.P., Yashukova A.V. Handbook of the teacher. 8th grade. Moscow: Bustard, 2002;
Cox R., Morris N. Seven wonders of the world. Ancient world, the Middle Ages, our time. M.: BMM AO, 1997;
Small children's encyclopedia. Chemistry. M.: Russian encyclopedic partnership, 2001; Encyclopedia for children "Avanta +". Chemistry. T. 17. M.: Avanta+, 2001;
Khomchenko G.P., Sevastyanova K.I. Redox reactions. Moscow: Education, 1989.

S.P. Lebesheva,
high school chemistry teacher no. 8
(Baltiysk, Kaliningrad region)

Rules for selection of coefficients:

- if the number of atoms of an element in one part of the reaction scheme is even, and odd in the other, then the coefficient 2 must be put in front of the formula with an odd number of atoms, and then the number of all atoms must be equalized.

- the placement of coefficients should begin with the most complex substance in composition and do this in the following sequence:

first you need to equalize the number of metal atoms, then acidic residues (non-metal atoms), then hydrogen atoms, and lastly, oxygen atoms.

- if the number of oxygen atoms in the left and right parts of the equation is the same, then the coefficients are determined correctly.

- after that, the arrow between the parts of the equation can be replaced with an equal sign.

— coefficients in the equation chemical reaction must not have common divisors.

Example. Let's make an equation for the chemical reaction between iron (III) hydroxide and sulfuric acid with the formation of iron (III) sulfate.

1. Let's make a reaction scheme:

Fe(OH)3 + H2SO4 → Fe2(SO4)3 + H2O

2. We select the coefficients for the formulas of substances. We know that we must start with the most complex substance and consistently equate in the whole scheme, first the metal atoms, then the acid residues, then the hydrogen, and finally the oxygen. In our scheme, the most complex substance is Fe2(SO4)3. It contains two iron atoms, and Fe(OH)3 contains one iron atom. So, before the formula Fe (OH) 3 it is necessary to put the coefficient 2:

2Fe(OH)3 + H2SO4 → Fe2(SO4)3 + H2O

Now we equalize the number of acid residues SO4. The Fe2(SO4)3 salt contains three SO4 acid residues. So, on the left side, before the H2SO4 formula, we put the coefficient 3:

2Fe(OH)3 + 3H2SO4 → Fe2(SO4)3 + H2O.

Now we equalize the number of hydrogen atoms. On the left side of the diagram in iron hydroxide 2Fe (OH) 3 - 6 hydrogen atoms (2

3), in sulfuric acid 3H2SO4 - also 6 hydrogen atoms.

How to arrange coefficients in chemical equations

In total, there are 12 hydrogen atoms on the left side. So, on the right side, we put a factor of 6 in front of the H2O water formula - and now there are also 12 hydrogen atoms on the right side:

2Fe(OH)3 + 3H2SO4 → Fe2(SO4)3 + 6H2O.

It remains to equalize the number of oxygen atoms. But this is no longer necessary, because the left and right parts of the diagram already have the same number of oxygen atoms - 18 in each part. This means that the circuit is written in full, and we can replace the arrow with an equal sign:

2Fe(OH)3 + 3H2SO4 = Fe2(SO4)3 + 6H2O.

Education

How to arrange coefficients in chemical equations? Chemical Equations

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 at the first stage you understand how to write chemical equations, 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.

Related videos

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, we write the starting substances on the left side, and the resulting products on the right. After checking the number of atoms of each chemical element, we obtain 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 of the following types:

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.

KMnO4 + H2SO4 + KBr = MnSO4 + Br2 +…+…

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 + HNO3 = NO2 +…+…

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 H3PO4. OVR assumes the presence of a reducing agent, which will be nitrogen. It goes into nitric oxide (4), forming NO2

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

P0 gives 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 + 5HNO3 = 5NO2 + H2O + H3PO4

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.

H2S + HMnO4 = S + MnO2 +…

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 = H2SiF6 + 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 + 4HNO3 + 18HF = 3H2SiF6 + 4NO + 8H2O

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.).

CaC2 + 2H2O = Ca(OH)2 + C2H2

Calcium hydroxide interacts with hydrogen chloride, complete neutralization occurs:

Ca(OH)2 + 2HCl = CaCl2 + 2H2O

We calculate the mass of acid that is required for this process.

Coefficients and indices in chemical equations

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.

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OPTION 1

a) Na + O2 -> Na2O d) H2 + F2 -> HF
b) CaCO3-> CaO + CO2 e) H2O + K2O -> KOH
c) Zn + H2SO4 -> H2 + ZnSO4 f) Cu(OH)2 + HNO3 -> Cu(NO3)2 + H2O

Lesson 13

Write down the definitions:
a) compound reaction b) exothermic reaction c) irreversible reaction.

a) carbon interacts with oxygen and carbon monoxide (II) is formed;
b) magnesium oxide interacts with nitric acid and magnesium nitrate and water are formed;
c) iron (III) hydroxide decomposes into iron oxide (III) and water;
d) methane CH4 burns in oxygen and carbon monoxide (IV) and water are formed;
e) nitric oxide (V) when dissolved in water forms nitric acid.

4. Solve the problem according to the equation:
a) What volume of hydrogen fluoride is formed when hydrogen reacts with fluorine?
b) What mass of calcium oxide is formed during the decomposition of limestone containing 80% CaCO3?
c) What volume and mass of hydrogen will be released when interacting with zinc sulfuric acid containing 35% impurities?

OPTION 2

1. Arrange the coefficients, determine the type of chemical reaction, write down the names of the substances under the formulas:

a) P + O2 -> P2O5 d) H2 + N2 -> NH3
b) CaCO3 + HCl -> CaCl2 + H2O + CO2 e) H2O + Li2O -> LiOH
c) Mg + H2SO4 -> H2 + MgSO4 e) Ca(OH)2 + HNO3 -> Ca(NO3)2 + H2O

2. Write down the definitions:
a) decomposition reaction b) endothermic reaction c) catalytic reaction.

3. Write down the equations according to the description:
a) carbon interacts with oxygen and carbon monoxide (IV) is formed;
b) barium oxide reacts with nitric acid and barium nitrate and water are formed;
c) aluminum hydroxide decomposes into aluminum oxide and water;
d) ammonia NH3 burns in oxygen and nitrogen and water are formed;
e) phosphorus (V) oxide, when dissolved in water, forms phosphoric acid.

4. Solve the problem according to the equation:
a) What volume of ammonia is formed when hydrogen reacts with nitrogen?
b) What mass of calcium chloride is formed upon interaction with hydrochloric acid of marble containing 80% CaCO3?
c) What volume and mass of hydrogen will be released when interacting with magnesium sulfuric acid containing 30% impurities?

How to write chemical equations? First, it is important to arrange the oxidation state of each element in the compound. Let's say the task is given: "Arrange the coefficients in the reaction, fill in the gaps, determine the oxidizing agent and reducing agent." 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. All calculations for the problem are carried out taking into account stereochemical coefficients, which confirms their importance. Among the typical mistakes that schoolchildren make when placing coefficients in reactions of various types, there are many mathematical errors.

There are certain rules by which they can be determined for each element. Formulas consisting of three elements have their own nuances for calculating oxidation states. Let's continue the conversation about how to equalize chemical equations using the electron balance method. A prerequisite is to check the number of each element in the left and right parts. If the coefficients are placed correctly, their number should be the same.

Algebraic method

Be sure to read about elemental analysis for a detailed look at empirical formulas and chemical analysis.

Chemistry studies substances, their properties, and transformations. In molecular form, the process of iron combustion in the atmosphere can be expressed using signs and symbols. According to the law of conservation of mass of substances, a factor of 2 must be put in front of the product formula. Next, calcium is checked. To begin with, for each of the elements in the initial substances and interaction products, we will place the values ​​of the oxidation states. The next step is to test the hydrogen.

Equalization of chemical reactions

Equalization of chemical reactions is necessary in order to obtain a complete one from a simple chemical equation. Let's start with carbon.

The law of conservation of mass excludes the appearance of new atoms and the destruction of old ones in the course of a chemical reaction. Pay attention to the index of each of the atoms, it is he who indicates their number. By adding indices in front of the molecules of substances on the right side of the equation, we also changed the number of oxygen atoms. Now the number of all carbon, hydrogen, and oxygen atoms is the same on both sides of the equation.

They say that if the factor is outside the bracket, then every element in the brackets is multiplied by it. You need to start with nitrogen, since it is less than oxygen and hydrogen. Great, hydrogen equalized. Next up is barium. It is leveled, it is not necessary to touch it. Before the reaction, there are two chlorine, after it - only one. What needs to be done? Now, due to the coefficient that has just been set, after the reaction, two sodium were obtained, and before the reaction, also two. Great, everything else is balanced. The next step is to arrange the oxidation states of all elements in each substance in order to understand where the oxidation occurred and where the reduction took place.

An example of parsing simple reactions

There are no indices on the right side, that is, one particle of oxygen, and on the left - 2 particles. No additional indexes or fixes in chemical formula cannot be entered, as it is written correctly. On the right side, we multiply one by 2 to get 2 oxygen ions there as well.

Before proceeding to the task itself, you need to learn that the number that is placed in front of a chemical element or the entire formula is called a coefficient. We start to analyze. Thus, it turned out the same number of atoms of each element before and after the equal sign. Be sure to keep in mind that the coefficient is multiplied by the index, and not added.

You are free to use any document for your own purposes, subject to the following conditions:

2) Symbols chemical elements should be written strictly in the form in which they appear in the periodic table.

Information card. "Algorithm for placement of coefficients in equations of chemical reactions."

3) Occasionally, situations arise when the formulas of the reactants and products are written absolutely correctly, but the coefficients are still not placed. The most likely occurrence of such a problem with oxidation reactions organic matter at which the carbon skeleton is torn.

The reaction equation must be able not only to write, but also to read. Therefore, sometimes, having written down 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. Count whether the atoms of each element are equal in the left and right sides of the equation.

For many schoolchildren, writing the equations of chemical reactions and correctly arranging the coefficients is not an easy task. But you just need to remember a few simple rules, and the task will cease to cause difficulties. The coefficient, that is, the number in front of the formula of the molecule chemical, applies to all characters, and is multiplied by each index of each character!

In lesson 13 "" from the course " Chemistry for dummies» consider what chemical equations are for; we will learn how to equalize chemical reactions by correctly placing the coefficients. This lesson will require you to know chemical bases from previous lessons. Be sure to read about elemental analysis for a detailed look at empirical formulas and chemical analysis.

As a result of the combustion reaction of methane CH 4 in oxygen O 2, carbon dioxide CO 2 and water H 2 O are formed. This reaction can be described chemical equation:

• CH 4 + O 2 → CO 2 + H 2 O (1)

Let's try to extract more information from the chemical equation than just an indication products and reagents reactions. The chemical equation (1) is NOT complete and therefore does not give any information about how many O 2 molecules are consumed per 1 CH 4 molecule and how many CO 2 and H2 O molecules are obtained as a result. But if we write numerical coefficients in front of the corresponding molecular formulas, which indicate how many molecules of each kind take part in the reaction, then we get full chemical equation reactions.

In order to complete the composition of the chemical equation (1), you need to remember one simple rule: the left and right sides of the equation must contain the same number of atoms of each kind, since no new atoms are created in the course of a chemical reaction and no existing ones are destroyed. This rule is based on the law of conservation of mass, which we discussed at the beginning of the chapter.

It is necessary in order to obtain a complete one from a simple chemical equation. So, let's move on to the direct equation of reaction (1): look again at the chemical equation, exactly at the atoms and molecules on the right and left sides. It is easy to see that three types of atoms participate in the reaction: carbon C, hydrogen H and oxygen O. Let's count and compare the number of atoms of each type on the right and left sides of the chemical equation.

Let's start with carbon. On the left side, one C atom is part of the CH 4 molecule, and on the right side, one C atom is part of CO 2. Thus, the number of carbon atoms on the left and on the right side is the same, so we leave it alone. But for clarity, we put a coefficient of 1 in front of molecules with carbon, although this is not necessary:

• 1CH 4 + O 2 → 1CO 2 + H 2 O (2)

Then we proceed to counting hydrogen atoms H. On the left side there are 4 H atoms (in the quantitative sense H 4 = 4H) in the composition of the CH 4 molecule, and on the right side there are only 2 H atoms in the composition of the H 2 O molecule, which is two times less than on the left side of the chemical equation (2). Let's equalize! To do this, we put a factor of 2 in front of the H 2 O molecule. Now we will have 4 hydrogen molecules H in both the reagents and the products:

• 1CH 4 + O 2 → 1CO 2 + 2H 2 O (3)

Please note that the coefficient 2, which we wrote in front of the water molecule H 2 O to equalize the hydrogen H, doubles all the atoms that make up its composition, i.e. 2H 2 O means 4H and 2O. Okay, this seems to have been sorted out, it remains to calculate and compare the number of oxygen atoms O in the chemical equation (3). It immediately catches the eye that in the left side of the O atoms exactly 2 times less than in the right. Now you already know how to equalize chemical equations yourself, so I’ll immediately write down the final result:

• 1CH 4 + 2O 2 → 1CO 2 + 2H 2 O or CH 4 + 2O 2 → CO 2 + 2H 2 O (4)

As you can see, equalizing chemical reactions is not such a tricky thing, and it is not chemistry that is important here, but mathematics. Equation (4) is called complete equation chemical reaction, because the law of conservation of mass is observed in it, i.e. the number of atoms of each kind entering the reaction is exactly the same as the number of atoms of this kind at the end of the reaction. Each part of this complete chemical equation contains 1 carbon atom, 4 hydrogen atoms and 4 oxygen atoms. However, it is worth understanding a couple important points: a chemical reaction is a complex sequence of separate intermediate stages, and therefore it is impossible, for example, to interpret equation (4) in the sense that 1 methane molecule must simultaneously collide with 2 oxygen molecules. The processes occurring during the formation of reaction products are much more complicated. The second point: the complete reaction equation does not tell us anything about its molecular mechanism, i.e. about the sequence of events that occur on molecular level during its course.

Coefficients in the equations of chemical reactions

Another good example how to arrange correctly odds in the equations of chemical reactions: Trinitrotoluene (TNT) C 7 H 5 N 3 O 6 vigorously combines with oxygen, forming H 2 O, CO 2 and N 2. We write the reaction equation, which we will equalize:

• C 7 H 5 N 3 O 6 + O 2 → CO 2 + H 2 O + N 2 (5)

It is easier to write the complete equation based on two TNT molecules, since the left side contains an odd number of hydrogen and nitrogen atoms, and the right side contains an even number:

• 2C 7 H 5 N 3 O 6 + O 2 → CO 2 + H 2 O + N 2 (6)

Then it is clear that 14 carbon atoms, 10 hydrogen atoms and 6 nitrogen atoms must turn into 14 carbon dioxide molecules, 5 water molecules and 3 nitrogen molecules:

• 2C 7 H 5 N 3 O 6 + O 2 → 14CO 2 + 5H 2 O + 3N 2 (7)

Now both parts contain the same number of all atoms except oxygen. Of the 33 oxygen atoms present on the right hand side of the equation, 12 are supplied by the original two TNT molecules, and the remaining 21 must be supplied by the 10.5 O 2 molecules. Thus, the complete chemical equation will look like:

• 2C 7 H 5 N 3 O 6 + 10.5O 2 → 14CO 2 + 5H 2 O + 3N 2 (8)

You can multiply both sides by 2 and get rid of the non-integer factor 10.5:

• 4C 7 H 5 N 3 O 6 + 21O 2 → 28CO 2 + 10H 2 O + 6N 2 (9)

But this can not be done, since all the coefficients of the equation do not have to be integers. It is even more correct to make an equation based on one TNT molecule:

• C 7 H 5 N 3 O 6 + 5.25O 2 → 7CO 2 + 2.5H 2 O + 1.5N 2 (10)

The complete chemical equation (9) carries a lot of information. First of all, it indicates the starting substances - reagents, and products reactions. In addition, it shows that in the course of the reaction all atoms of each kind are individually preserved. If we multiply both sides of equation (9) by the Avogadro number N A =6.022 10 23 , we can state that 4 moles of TNT react with 21 moles of O 2 to form 28 moles of CO 2 , 10 moles of H 2 O and 6 moles of N 2 .

There is one more feature. Using the periodic table, we determine the molecular weights of all these substances:

• C 7 H 5 N 3 O 6 \u003d 227.13 g / mol
• O2 = 31.999 g/mol
• CO2 = 44.010 g/mol
• H2O = 18.015 g/mol
• N2 = 28.013 g/mol

Now equation 9 will also indicate that 4 227.13 g \u003d 908.52 g of TNT require 21 31.999 g \u003d 671.98 g of oxygen to complete the reaction and as a result 28 44.010 g \u003d 1232.3 g CO 2 are formed, 10 18.015 g = 180.15 g H 2 O and 6 28.013 g = 168.08 g N 2. Let's check whether the law of conservation of mass is fulfilled in this reaction:

	Reagents	Products
	908.52 g TNT	1232.3 g CO2
	671.98 g CO2	180.15 g H2O
		168.08 g N2
Total	1580.5 g	1580.5 g

But it is not necessary for individual molecules to participate in a chemical reaction. For example, the reaction of limestone CaCO3 and hydrochloric acid HCl, with the formation of an aqueous solution of calcium chloride CaCl2 and carbon dioxide CO2:

• CaCO 3 + 2HCl → CaCl 2 + CO 2 + H 2 O (11)

Chemical equation (11) describes the reaction of calcium carbonate CaCO 3 (limestone) and hydrochloric acid HCl to form an aqueous solution of calcium chloride CaCl 2 and carbon dioxide CO 2 . This equation is complete, since the number of atoms of each kind in its left and right sides is the same.

The meaning of this equation is macroscopic (molar) level is as follows: 1 mol or 100.09 g of CaCO 3 requires 2 mol or 72.92 g of HCl to complete the reaction, resulting in 1 mol of CaCl 2 (110.99 g / mol), CO 2 (44.01 g /mol) and H 2 O (18.02 g/mol). From these numerical data, it is easy to verify that the law of conservation of mass is satisfied in this reaction.

Interpretation of equation (11) on microscopic (molecular) level is not so obvious, since calcium carbonate is a salt, not a molecular compound, and therefore it is impossible to understand the chemical equation (11) in the sense that 1 molecule of calcium carbonate CaCO 3 reacts with 2 molecules of HCl. Moreover, the HCl molecule in solution generally dissociates (decomposes) into H + and Cl - ions. Thus, a more correct description of what happens in this reaction at the molecular level, gives the equation:

• CaCO 3 (solid) + 2H + (aq.) → Ca 2+ (aq.) + CO 2 (g.) + H 2 O (l.) (12)

Here, in parentheses, the physical state of each kind of particles is abbreviated ( tv.- hard, aq. is a hydrated ion in an aqueous solution, G.- gas, and.- liquid).

Equation (12) shows that solid CaCO 3 reacts with two hydrated H + ions to form a positive Ca 2+ ion, CO 2 and H 2 O. Equation (12), like other complete chemical equations, does not give an idea of ​​the molecular mechanism reactions and is less convenient for counting the amount of substances, however, it gives best description happening at the microscopic level.

Consolidate your knowledge of the formulation of chemical equations by independently analyzing the example with the solution:

I hope from lesson 13 " Compilation of chemical equations» you learned something new for yourself. If you have any questions, write them in the comments.