Equalize the reaction scheme. How to balance chemical equations

Write down the chemical equation. As an example, consider the following reaction:

C 3 H 8 + O 2 –> H 2 O + CO 2
This reaction describes the combustion of propane (C 3 H 8) in the presence of oxygen to form water and carbon dioxide (carbon dioxide).

Write down the number of atoms of each element. Do this for both sides of the equation. Notice the subscripts next to each element to determine the total number of atoms. Write down the symbol for each element in the equation and note the corresponding number of atoms.

For example, on the right side of the equation under consideration, as a result of addition, we get 3 oxygen atoms.
On the left side we have 3 carbon atoms (C 3), 8 hydrogen atoms (H 8) and 2 oxygen atoms (O 2).
On the right side we have 1 carbon atom (C), 2 hydrogen atoms (H 2) and 3 oxygen atoms (O + O 2).

Leave hydrogen and oxygen for later, as they are part of several compounds on the left and right side. Hydrogen and oxygen are part of several molecules, so it's best to balance them last.

Before balancing hydrogen and oxygen, you will have to count the atoms again, as additional factors may be needed to balance other elements.

Start with the least frequently occurring element. If you need to balance several elements, choose one that is part of one molecule of reactants and one molecule of reaction products. So the first thing to do is to balance the carbon.

For balance, add a factor before the single carbon atom. Place a factor in front of the single carbon on the right side of the equation to balance it with the 3 carbons on the left side.

C 3 H 8 + O 2 –> H 2 O + 3 CO 2
The factor 3 in front of the carbon on the right side of the equation indicates that there are three carbon atoms, which correspond to the three carbon atoms included in the propane molecule on the left side.
In a chemical equation, you can change the coefficients in front of atoms and molecules, but the subscripts must remain unchanged.

Then balance the hydrogen atoms. After you equalized the number of carbon atoms on the left and right side, hydrogen and oxygen remained unbalanced. The left side of the equation contains 8 hydrogen atoms, the same number should be on the right side. Achieve this with a ratio.

C 3 H 8 + O 2 –> 4 H 2 O + 3CO 2
We've added a factor of 4 on the right side because the subscript shows we already have two hydrogens.
If you multiply the factor 4 by the subscript 2, you get 8.
As a result, 10 oxygen atoms are obtained on the right side: 3x2=6 atoms in three 3CO 2 molecules and four more atoms in four water molecules.

Reactions between various kinds of chemicals and elements are one of the main subjects of study in chemistry. To understand how to draw up a reaction equation and use them for your own purposes, you need a fairly deep understanding of all the patterns in the interaction of substances, as well as processes with chemical reactions.

Writing Equations

One way to express a chemical reaction is a chemical equation. It contains the formula of the starting substance and product, the coefficients that show how many molecules each substance has. All known chemical reactions are divided into four types: substitution, combination, exchange and decomposition. Among them are: redox, exogenous, ionic, reversible, irreversible, etc.

Learn more about how to write equations for chemical reactions:

It is necessary to determine the name of the substances interacting with each other in the reaction. We write them on the left side of our equation. As an example, consider the chemical reaction that took place between sulfuric acid and aluminum. We have the reagents on the left: H2SO4 + Al. Next, write the equal sign. In chemistry, you can see an arrow sign that points to the right, or two opposite arrows that mean "reversibility." The result of the interaction of metal and acid is salt and hydrogen. Write the products obtained after the reaction after the “equal” sign, that is, on the right. H2SO4+Al= H2+Al2(SO4)3. So, we can see the reaction scheme.
To compile a chemical equation, it is imperative to find the coefficients. Let's go back to the previous diagram. Let's look at the left side of it. Sulfuric acid contains hydrogen, oxygen and sulfur atoms in an approximate ratio of 2:4:1. On the right side there are 3 sulfur atoms and 12 oxygen atoms in the salt. There are two hydrogen atoms in a gas molecule. On the left side, the ratio of these elements is 2:3:12
To equalize the number of oxygen and sulfur atoms that are in the composition of aluminum (III) sulfate, it is necessary to put a factor of 3 in front of the acid on the left side of the equation. Now we have 6 hydrogen atoms on the left side. In order to equalize the number of elements of hydrogen, you need to put 3 in front of hydrogen on the right side of the equation.
Now it remains only to equalize the amount of aluminum. Since the composition of the salt includes two metal atoms, then on the left side in front of aluminum we set the coefficient 2. As a result, we will get the reaction equation of this scheme: 2Al + 3H2SO4 = Al2 (SO4) 3 + 3H2

Having understood the basic principles of how to write an equation for the reaction of chemicals, in the future it will not be difficult to write down any, even the most exotic, from the point of view of chemistry, reaction.

A reaction equation in chemistry is a record of a chemical process using chemical formulas and mathematical signs.

Such a record is a scheme of a chemical reaction. When the “=” sign appears, it is called an “equation”. Let's try to solve it.

In contact with

An example of parsing simple reactions

Calcium has one atom, since the coefficient is not worth it. The index is also not written here, which means it is one. On the right side of the equation, Ca is also one. We don't need to work on calcium.

Video: Coefficients in the equations of chemical reactions.

We look at the next element - oxygen. Index 2 indicates that there are 2 oxygen ions. There are no indices on the right side, that is, one particle of oxygen, and on the left - 2 particles. What are we doing? No additional indices or corrections can be made to the chemical formula, since it is written correctly.

The coefficients are what is written before the smallest part. They have the right to change. For convenience, we do not rewrite the formula itself. On the right side, we multiply one by 2 to get 2 oxygen ions there as well.

After we set the coefficient, we got 2 calcium atoms. There is only one on the left side. So now we have to put 2 in front of calcium.

Now let's check the result. If the number of element atoms is equal on both sides, then we can put an "equal" sign.

Another good example: two hydrogens on the left, and after the arrow we also have two hydrogens.

Two oxygens before the arrow, and after the arrow there are no indices, which means one.
More on the left, less on the right.
We put a factor of 2 in front of the water.

We multiplied the whole formula by 2, and now we have changed the amount of hydrogen. We multiply the index by the coefficient, and it turns out 4. And on the left side there are two hydrogen atoms. And to get 4, we have to multiply hydrogen by two.

Video: Arrangement of coefficients in a chemical equation

Here is the case when the element in one and the other formula is on the one hand, up to the arrow.

One sulfur ion on the left and one sulfur ion on the right. Two particles of oxygen, plus two more particles of oxygen. So there are 4 oxygens on the left side. On the right is 3 oxygen. That is, on the one hand, an even number of atoms is obtained, and on the other, an odd number. If we multiply an odd number by 2, we get an even number. We bring it to an even value first. To do this, multiply by two the entire formula after the arrow. After multiplication, we get six oxygen ions, and even 2 sulfur atoms. On the left, we have one microparticle of sulfur. Now let's equalize it. We put equations on the left in front of gray 2.

Called.

Complex reactions

This example is more complex, as there are more elements of matter.

This is called a neutralization reaction. What needs to be equalized here first of all:

On the left side is one sodium atom.
On the right side, the index says that there are 2 sodium.

The conclusion suggests itself that it is necessary to multiply the entire formula by two.

Video: Compilation of equations of chemical reactions

Now let's see how much sulfur. One on the left and right side. Pay attention to oxygen. On the left side we have 6 oxygen atoms. On the other hand - 5. Less on the right, more on the left. An odd number must be brought to an even value. To do this, we multiply the water formula by 2, that is, we make 2 from one oxygen atom.

Now on the right side there are already 6 oxygen atoms. There are also 6 atoms on the left side. Checking hydrogen. Two hydrogen atoms and 2 more hydrogen atoms. That is, there will be four hydrogen atoms on the left side. And on the other side also four hydrogen atoms. All elements are balanced. We put an "equal" sign.

Video: Chemical equations. How to write chemical equations.

Next example.

Here the example is interesting in that parentheses have appeared. 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. On the left, there is one nitrogen, and on the right, taking into account the brackets, there are two.

There are two hydrogen atoms on the right, but four are needed. We get out of the situation by simply multiplying the water by two, resulting in four hydrogens. Great, hydrogen equalized. There is oxygen left. Before the reaction, there are 8 atoms, after - also 8.

Great, all the elements are equal, we can put "equal".

Last example.

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? Put 2 in front of chlorine after the reaction.

Video: Balancing chemical equations.

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.

Reactions can also be equalized using the electronic balance method. This method has a number of rules by which it can be implemented. 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.

Carefully study the algorithms and write down in a notebook, solve the proposed tasks on your own

I. Using the algorithm, solve the following problems yourself:

1. Calculate the amount of aluminum oxide substance formed as a result of the interaction of aluminum with a quantity of substance 0.27 mol with a sufficient amount of oxygen (4 Al+3 O 2 \u003d 2 Al 2 O 3).

2. Calculate the amount of sodium oxide substance formed as a result of the interaction of sodium with a quantity of substance 2.3 mol with a sufficient amount of oxygen (4 Na+ O 2 \u003d 2 Na 2 O).

Algorithm #1

Calculation of the amount of a substance from a known amount of a substance participating in a reaction.

Example.Calculate the amount of oxygen substance released as a result of the decomposition of water with a substance of 6 mol.

	Task design
1. Write down the condition of the problem	*Given* : ν (H 2 O) \u003d 6mol _____________ *To find* : ν(O 2) \u003d?
	*Decision* : M (O 2) \u003d 32g / mol
and put the coefficients	2H 2 O \u003d 2H 2 + O 2
, and under the formulas -
5. To calculate the desired amount of substance, make up the ratio
6. Write down the answer	*Answer: ν (O 2) \u003d 3mol*

II. Using the algorithm, solve the following problems on your own:

1. Calculate the mass of sulfur required to obtain sulfur oxide ( S+ O 2 = SO2).

2. Calculate the mass of lithium required to obtain lithium chloride with a substance of 0.6 mol (2 Li+ Cl 2 \u003d 2 LiCl).

Algorithm #2

Calculation of the mass of a substance from a known amount of another substance participating in the reaction.

Example:Calculate the mass of aluminum required to obtain aluminum oxide with a substance of 8 mol.

Sequence of actions	Formulation of a solution to the problem
1. Write down the condition of the problem	*Given:* ν( Al 2 O 3 )=8mol ___________ *To find:* m( Al)=?
2. Calculate the molar masses of substances, which are discussed in the problem	M( Al 2 O 3 )=102g/mol
3. Write the reaction equation and put the coefficients	4 Al + 3O 2 \u003d 2Al 2 O 3
4. We write over the formulas of substances *the amount of substances from the condition of the problem* , and under the formulas - stoichiometric coefficients , displayed by the reaction equation
5. Calculate the amount of a substance whose mass required to be found. To do this, we will make a ratio.
6. Calculate the mass of the substance to be found	m= ν ∙ M, m(Al)= ν (Al)∙ M(Al)=16mol∙27g/mol=432g
7. Write down the answer	*Answer:* m *(Al)=* *432 g*

III. Using the algorithm, solve the following problems on your own:

1. Calculate the amount of sodium sulfide substance if a sulfur mass of 12.8 g (2 Na+ S= Na 2 S).

2. Calculate the amount of substance formed copper if copper oxide reacts with hydrogen ( II ) weighing 64 g ( CuO+ H2= Cu+ H2 O).

Carefully study the algorithm and write it down in a notebook

Algorithm #3

Calculation of the amount of a substance given the known mass of another substance involved in the reaction.

Example.Calculate the amount of copper oxide substance ( I ), if copper weighing 19.2 g reacts with oxygen.

Sequence of actions	Task design
1. Write down the condition of the problem	*Given:* m( Cu)=19.2g ___________ *To find:* ν( Cu 2 O)=?
2. Calculate the molar masses of substances, which are discussed in the problem	M(Cu)=64g/mol
3. Find the amount of a substance whose mass given in the problem statement
and put the coefficients	4 Cu+ O 2 =2 Cu 2 O
*the amount of substances from the condition of the problem* , and under the formulas - stoichiometric coefficients , displayed by the reaction equation
6. To calculate the desired amount of substance, make up the ratio
7. Write down the answer	*Answer: v(* Cu 2 O *)=0.15 mol*

Carefully study the algorithm and write it down in a notebook

IV. Using the algorithm, solve the following problems on your own:

1. Calculate the mass of oxygen required to react with 112 g of iron

(3 Fe+4 O 2 = Fe3 O 4).

Algorithm No. 4

Calculation of the mass of a substance from the known mass of another substance participating in the reaction

Example.Calculate the mass of oxygen required for the combustion of phosphorus, with a mass of 0.31 g.

Sequence of actions

Making a task

1. Write down the condition of the problem

Given:

m( P)=0.31g

_________

To find:

m( O 2 )=?

2. Calculate the molar masses of substances,

which are discussed in the problem

M(P)=31g/mol

M( O 2 )=32g/mol

3. Find the amount of substance, the mass of which is given in the condition of the problem

4. Write the reaction equation

and put the coefficients

4 P+5 O 2 = 2 P 2 O 5

5. We write over the formulas of substances

the amount of substances from the condition of the problem ,

and under the formulas -

stoichiometric coefficients ,

displayed by the reaction equation

6. Calculate the amount of substance whose mass must be found

m( O 2 )= ν ( O 2 )∙ M( O 2 )=

0.0125mol∙32g/mol=0.4g

8. Write down the answer

Answer: m ( O 2 )=0.4g

TASKS FOR INDEPENDENT SOLUTION

3. Calculate the mass of sulfur required to obtain sulfur oxide ( IV ) the amount of substance 4 mol ( S+ O 2 = SO2).

4. Calculate the mass of lithium required to obtain lithium chloride with a substance of 0.6 mol (2 Li+ Cl 2 \u003d 2 LiCl).

5. Calculate the amount of sodium sulfide substance if sulfur reacts with sodium with a mass of 12.8 g (2 Na+ S= Na 2 S).

6. Calculate the amount of substance formed copper, if copper oxide reacts with hydrogen ( II ) weighing 64 g ( CuO+ H2=

Scheme of a chemical reaction.

There are several ways to write chemical reactions. You familiarized yourself with the “verbal” reaction scheme in § 13.

Here's another example:

sulfur + oxygen -> sulfur dioxide.

Lomonosov and Lavoisier discovered the law of conservation of mass of substances in a chemical reaction. It is formulated like this:

Let's explain why masses ash and calcined copper are different from the masses of paper and copper before it is heated.

In the process of burning paper, oxygen is involved, which is contained in the air (Fig. 48, a).

Therefore, two substances are involved in the reaction. In addition to ash, carbon dioxide and water (in the form of steam) are formed, which enter the air and dissipate.

Rice. 48. Reactions of paper (a) and copper (b) with oxygen

Antoine Laurent Lavoisier (1743-1794)

An outstanding French chemist, one of the founders of scientific chemistry. Academician of the Paris Academy of Sciences. Introduced quantitative (exact) research methods into chemistry. He experimentally determined the composition of air and proved that combustion is a reaction of a substance with oxygen, and water is a combination of Hydrogen with Oxygen (1774-1777).

Compiled the first table of simple substances (1789), actually proposing a classification of chemical elements. Independently of M. V. Lomonosov, he discovered the law of conservation of the mass of substances in chemical reactions.

Rice. 49. Experience confirming the law of Lomonosov - Lavoisier: a - the beginning of the experiment; b - the end of the experiment

Their mass exceeds the mass of oxygen. Therefore, the mass of ash is less than the mass of paper.

When copper is heated, air oxygen "combines" with it (Fig. 48, b). The metal turns into a black substance (its formula is CuO, and the name is cuprum (P) oxide). Obviously, the mass of the reaction product must exceed the mass of copper.

Comment on the experience shown in Figure 49 and draw a conclusion.

Law as a form of scientific knowledge.

The discovery of laws in chemistry, physics, and other sciences occurs after scientists conduct many experiments and analyze the results.

Law is a generalization of objective, human-independent connections between phenomena, properties, etc.

The law of conservation of mass of substances in a chemical reaction is the most important law of chemistry. It applies to all transformations of substances that occur both in the laboratory and in nature.

Chemical laws make it possible to predict the properties of substances and the course of chemical reactions, to regulate processes in chemical technology.

In order to explain the law, hypotheses are put forward, which are tested with the help of appropriate experiments. If one of the hypotheses is confirmed, a theory is created on its basis. In high school, you will become familiar with several theories that chemists have developed.

The total mass of substances during a chemical reaction does not change because the atoms of chemical elements do not appear and disappear during the reaction, but only their rearrangement occurs. In other words,
the number of atoms of each element before the reaction is equal to the number of its atoms after the reaction. This is indicated by the reaction schemes given at the beginning of the paragraph. Let's replace the arrows between the left and right sides with equal signs:

Such records are called chemical equations.

A chemical equation is a record of a chemical reaction using the formulas of reactants and products, which is consistent with the law of conservation of mass of substances.

There are many reaction schemes that do not correspond to the Lomonosov-Lavoisier law.

For example, the reaction scheme for the formation of water:

H 2 + O 2 -> H 2 O.

Both parts of the scheme contain the same number of hydrogen atoms, but a different number of oxygen atoms.

Let's turn this scheme into a chemical equation.

In order for there to be 2 oxygen atoms on the right side, we put a coefficient 2 in front of the water formula:

H 2 + O 2 -> H 2 O.

Now there are four Hydrogen atoms on the right. In order for the same number of Hydrogen atoms to be on the left side, we write the coefficient 2 in front of the hydrogen formula. We get the chemical equation:

2H 2 + O 2 \u003d 2H 2 0.

Thus, in order to turn a reaction scheme into a chemical equation, you need to choose the coefficients for each substance (if necessary), write them down in front of the chemical formulas, and replace the arrow with an equal sign.

Perhaps one of you will write this equation: 4H 2 + 20 2 \u003d 4H 2 0. In it, the left and right sides contain the same number of atoms of each element, but all coefficients can be reduced by dividing by 2. This should be done.

It is interesting

The chemical equation has much in common with the mathematical one.

Below are various ways of recording the considered reaction.

Turn the reaction scheme Cu + O 2 -> CuO into a chemical equation.

Let's perform a more difficult task: turn the reaction scheme into a chemical equation

On the left side of the scheme - I atom of Aluminum, and on the right - 2. Put a coefficient 2 in front of the metal formula:

There are three times more Sulfur atoms on the right than on the left. We write the coefficient 3 in front of the formula of the Sulfur compound on the left side:

Now, on the left side, the number of Hydrogen atoms is 3 2 = 6, and on the right - only 2. In order for them to be 6 on the right, we put the coefficient 3 in front of the hydrogen formula (6: 2 = 3):

Let us compare the number of oxygen atoms in both parts of the scheme. They are the same: 3 4 = 4 * 3. Let's replace the arrow with an equal sign:

findings

Chemical reactions are written using reaction schemes and chemical equations.

The reaction scheme contains the formulas of the reactants and products, and the chemical equation also contains the coefficients.

The chemical equation is consistent with the law of conservation of mass of Lomonosov-Lavoisier substances:

the mass of substances that entered into a chemical reaction is equal to the mass of substances formed as a result of the reaction.

Atoms of chemical elements do not appear or disappear during reactions, but only their rearrangement occurs.

?
105. What is the difference between a chemical equation and a reaction scheme?

106. Arrange the missing coefficients in the reaction records:

107. Turn the following reaction schemes into chemical equations:

108. Make the formulas of the reaction products and the corresponding chemical equations:

109. Instead of dots, write down the formulas of simple substances and make chemical equations:

Bear in mind that boron and carbon are made up of atoms; fluorine, chlorine, hydrogen and oxygen - from diatomic molecules, and phosphorus (white) - from four-atomic molecules.

110. Comment on the reaction schemes and turn them into chemical equations:

111. What mass of quicklime was formed during prolonged calcination of 25 g of chalk, if it is known that 11 g of carbon dioxide was released?

Popel P. P., Kriklya L. S., Chemistry: Pdruch. for 7 cells. zahalnosvit. navch. zakl. - K .: Exhibition Center "Academy", 2008. - 136 p.: il.

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