Algorithm
Arrangement of coefficients in equations chemical reactions
Chemistry teacher MBOU secondary school No. 2
Volodchenko Svetlana Nikolaevna
Ussuriysk
ARRANGEMENT OF COEFFICIENTS IN EQUATIONS OF CHEMICAL REACTIONS
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,NZRO4, 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) 2СО + 02 2) С02 + 2Н.О. 3)4NO2 + 2H2O + O2
Algorithm for arranging coefficients in chemical reaction equations
А1 + О2→ А12О3A1-1 atom A1-2
O-2 atom O-3
2. Among the elements with different numbers atoms on the left and right sides of the diagram, 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, obtain the coefficient for the left side of the equation
6:2 = 3
Al + ZO 2 →Al 2 ABOUT 3
5. Divide the LCM by the number of atoms of this element on the right side of the equation, obtain the coefficient for the right side of the equation
6:3 = 2
A1+ O 2 →2A1 2 O3
6. If the set coefficient has changed the number of atoms of another element, then repeat steps 3, 4, 5 again.
A1 + ZO 2 → →2А1 2 ABOUT 3
A1 -1 atom A1 - 4
LCM = 4
4:1=4 4:4=1
4A1 + ZO 2 → →2A1 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. The number of atoms must be equalized using coefficients.
1)2Mg+O2 →2MgO
2) CaCO3 + 2HCl→CaCl2 + N2 O + CO2
Task 2 Place 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 → SaS 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; N 2 + WITH l 2 → NS l;
9.N 2 + O 2 → NO; CO 2 + WITH → CO ;
10. HI → N 2 → + 1 2 ; Mg+ NS l → MgCl 2 + N 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. Mg 2 Si + HCl → MgCl 2 + SiH 4
1.What is the equation of a chemical reaction?
2.What is written on the right side of the equation? And on the left?
3.What does the “+” sign mean in an equation?
4. Why are coefficients placed in chemical equations?
The equation of a reaction in chemistry is called the notation chemical process using chemical formulas and mathematical symbols.
This notation is a diagram of a chemical reaction. When the "=" sign appears, it is called an "equation". Let's try to solve it.
Example of analysis of simple reactions
There is one atom in calcium, since the coefficient is not worth it. The index is also not written here, which means one. On the right side of the equation, Ca is also one. We don't need to work on calcium.
Let's 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 there are 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.
After we set the coefficient, we got 2 calcium atoms. There is only one on the left side. This means that now we must put 2 in front of calcium.
Now let's check the result. If the number of atoms of an element is equal on both sides, then we can put the “equal” sign.
Another clear example: two hydrogens on the left, and after the arrow we also have two hydrogens.
- There are two oxygens before the arrow, but there are no indices after the arrow, which means there is one.
- There is more on the left and less on the right.
- We put coefficient 2 in front of water.
We multiplied the entire formula by 2, and now the amount of hydrogen has changed. We multiply the index by the coefficient, and we get 4. And on the left side there are two hydrogen atoms left. And to get 4, we have to multiply hydrogen by two.
This is the case when the element in one and the other formula is on the same side, up to the arrow.
One sulfur ion on the left, and one ion on the right. Two oxygen particles, plus two more oxygen particles. This means that there are 4 oxygens on the left side. On the right there are 3 oxygens. That is, on one side there is an even number of atoms, and on the other, an odd number. If we multiply the odd number by two times, we get an even number. First we bring it to an even value. To do this, multiply the entire formula after the arrow by two. After multiplication, we get six oxygen ions, and also 2 sulfur atoms. On the left we have one microparticle of sulfur. Now let's equalize it. We put the equations on the left before gray 2.
Called.
Complex reactions
This example is more complex because there are more elements of matter.
This is called a neutralization reaction. What needs to be equalized here first:
- On the left side is one sodium atom.
- On the right side, the index says that there are 2 sodium.
The conclusion suggests itself is that you need to multiply the entire formula by two.
Now let's see how much sulfur there is. One on the left and right sides. Let's 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 number. To do this, we multiply the formula of water by 2, that is, from one oxygen atom we make 2.
Now there are already 6 oxygen atoms on the right side. There are also 6 atoms on the left side. Let's check the hydrogen. Two hydrogen atoms and 2 more hydrogen atoms. So there will be four hydrogen atoms on the left side. And on the other side there are also four hydrogen atoms. All elements are equal. We put the equal sign.
Next example.
Here the example is interesting because parentheses appear. They say that if a factor is behind a parenthesis, then each element in the parentheses is multiplied by it. You need to start with nitrogen, since there is less of it 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 this by simply multiplying 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 set “equal”.
Next up is barium. It is equalized, you don’t need to touch it. Before the reaction there are two chlorines, after it there is only one. What needs to be done? Place 2 in front of the chlorine after the reaction.
Now, due to the coefficient that was just set, after the reaction we got two sodiums, and before the reaction we also got two. Great, everything else is equalized.
You can also equalize reactions 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 oxidation occurred and where reduction occurred.
In order to figure out how to balance 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, or release of a gaseous substance, then no chemical interaction occurs.
For example, when filing an iron nail, the metal simply turns into powder. In this case, no chemical reaction occurs.
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 correctly equalize chemical equations. Let's look at all the nuances associated with such a procedure.
Specifics of chemical transformations
Any phenomena that are accompanied by a change in the qualitative and quantitative composition of substances are classified as chemical transformations. In molecular form, the process of burning iron in the atmosphere can be expressed using signs and symbols.
Methodology for setting coefficients
How to match odds in chemical equations? Up to date with chemistry high school The electronic balance method is discussed. Let's look at the process in more detail. To begin with, in the initial reaction it is necessary to arrange the oxidation states of each chemical element.
There are certain rules by which they can be determined for each element. In simple substances, the oxidation states will be zero. In binary compounds, the first element has a positive value, corresponding to the 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 in calculating oxidation states.
For the first and last element, the order is similar to the definition in binary compounds, and an equation is drawn up 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 electronic balance method. After the oxidation states have been established, it is possible to determine those ions or substances that changed their value during chemical interaction.
The plus and minus signs must indicate the number of electrons that were accepted (donated) during the chemical interaction. The least common multiple is found between the resulting numbers.
When dividing it into received and donated electrons, the coefficients are obtained. How to balance a chemical equation? The figures obtained in the balance sheet must be placed before the corresponding formulas. A prerequisite is to check the quantity of each element on the left and right sides. If the coefficients are placed correctly, their number should be the same.
Law of conservation of mass of substances
When discussing how to balance a chemical equation, it is this law that must be used. Considering 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 balance a chemical equation if the simple substances calcium and oxygen interact, and after the process is completed, an oxide is obtained?
To cope with the task, it is necessary to take into account that oxygen is a diatomic molecule with a covalent nonpolar bond, therefore its formula is written in the following form - O2. On the right side, when composing calcium oxide (CaO), the valence of each element is taken into account.
First you need to check the amount of oxygen in each side of the equation as it is different. According to the law of conservation of mass of substances, a coefficient of 2 must be placed in front of the product formula. Next, calcium is checked. In order for it to be equalized, we put a coefficient of 2 in front of the original substance. As a result, we get the entry:
- 2Ca+O2=2CaO.
Analysis of the reaction using the electronic balance method
How to balance chemical equations? Examples of OVR will help answer this question. Let us assume that it is necessary to arrange the coefficients in the proposed scheme using the electronic balance method:
- CuO + H2=Cu + H2O.
To begin with, we will assign the oxidation states for each of the elements in the starting substances and reaction products. We get next view equations:
- 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)+2е=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 entry for 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 based on the law of conservation of substances, it is necessary to start with oxygen. Considering that a diatomic molecule reacted, a coefficient of 2 must be placed in front of the formula of the reaction product.
- 2H2+O2=2H2O.
Conclusion
Based on the electronic balance, you can place coefficients in any chemical equations. Graduates of ninth and eleventh grades educational institutions Those choosing an exam in chemistry are offered similar tasks in one of the tasks of the final tests.
Instructions
Before starting the task itself, you need to understand that the number that is placed in front of a chemical element or the entire formula is a coefficient. And the figure is worth (and slightly) the index. Besides this, that:
The coefficient applies to all chemical symbols that appear after it in the formula
The coefficient is multiplied by the index (does not add up!)
The number of atoms of each element of the substances entering the reaction must coincide with the number of atoms of these elements included in the reaction products.
For example, writing the formula 2H2SO4 means 4 H (hydrogen) atoms, 2 S (sulfur) atoms and 8 O (oxygen) atoms.
1. Example No. 1. Consider the combustion of ethylene.
When organic matter burns, it produces carbon monoxide (IV) (carbon dioxide) and water. Let's try the coefficients sequentially.
C2H4 + O2 => CO2+ H2O
Let's start analyzing. 2 atoms of C (carbon) entered the reaction, but only 1 atom was obtained, which means we put 2 in front of CO2. Now their number is the same.
C2H4 + O2 => 2CO2+ H2O
Now let's look at H (hydrogen). 4 hydrogen atoms entered the reaction, but the result was only 2 atoms, therefore, we put 2 in front of H2O (water) - now we also get 4
C2H4 + O2 => 2CO2+ 2H2O
We count all O (oxygen) atoms formed as a result of the reaction (that is, after equality). 4 atoms in 2CO2 and 2 atoms in 2H2O - a total of 6 atoms. And before the reaction there are only 2 atoms, which means we put 3 in front of the oxygen molecule O2, which means there are also 6 of them.
C2H4 + 3O2 => 2CO2+ 2H2O
Thus, the result is the same number of atoms of each element before and after the equal sign.
C2H4 + 3O2 => 2CO2+ 2H2O
2. Example No. 2. Consider the reaction of aluminum with dilute sulfuric acid.
Al + H2SO4 => Al2 (SO4) 3 + H2
We look at the S atoms included in Al2 (SO4) 3 - there are 3 of them, but in H2SO4 (sulfuric acid) there is only 1, therefore, we also put 3 in front of sulfuric acid.
Al + 3H2SO4 => Al2 (SO4) 3 + H2
But now there are 6 H (hydrogen) atoms before the reaction, and only 2 after the reaction, which means we also put 3 in front of the H2 (hydrogen) molecule, so that in total we get 6.
Al + 3H2SO4 => Al2 (SO4) 3 + 3H2
Lastly, we look at. Since there are only 2 aluminum atoms in Al2 (SO4) 3 (aluminum sulfate), we put 2 in front of Al (aluminum) before the reaction.
2Al + 3H2SO4 => Al2 (SO4) 3 + 3H2
Now the number of all atoms before and after the reaction is the same. It turned out that arranging coefficients in chemical equations is not so difficult. Just practice and everything will work out.
Be sure to keep in mind that the coefficient is multiplied by the index and not added.
Sources:
- how elements react
- Test on the topic "Chemical Equations"
For many schoolchildren, write equations of chemical reactions and place them correctly odds no easy task. Moreover, for some reason the main difficulty for them is precisely the second part of it. It would seem that there is nothing complicated about it, but sometimes students give up, falling into complete confusion. But you just need to remember a few simple rules, and the task will no longer cause difficulties.
Instructions
Coefficient, that is, the number in front of the formula of a chemical molecule, to all symbols, and is multiplied by each symbol! It is multiplying, not adding! It may seem incredible, but some students add two numbers instead of multiplying them.
The number of atoms of each element of the starting substances (that is, located on the left side of the equation) must coincide with the number of atoms of each element of the reaction products (respectively, located on its right side).
RANGE OF COEFFICIENTS
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 participating 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) 2СО + 02 2) С02 + 2Н.О. 3)4NO2 + 2H2O + O2
Algorithm for arranging coefficients in chemical reaction equations
А1 + О2→ А12О3A1-1 atom A1-2
O-2 atom O-3
2. Among elements with different numbers of atoms in the left and right parts of the diagram, 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, obtain the coefficient for the left side of the equation
6:2 = 3
Al + ZO 2 →Al 2 ABOUT 3
5. Divide the LCM by the number of atoms of this element on the right side of the equation, obtain the coefficient for the right side of the equation
6:3 = 2
A1+ O 2 →2A1 2 O3
6. If the set coefficient has changed the number of atoms of another element, then repeat steps 3, 4, 5 again.
A1 + ZO 2 → →2А1 2 ABOUT 3
A1 -1 atom A1 - 4
LCM = 4
4:1=4 4:4=1
4A1 + ZO 2 → →2A1 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. The number of atoms must be equalized using coefficients.
1)2Mg+O2 →2MgO
2) CaCO3 + 2HCl→CaCl2 + N2 O + CO2
Task 2 Place 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 → SaS 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; N 2 + WITH l 2 → NS l;
9.N 2 + O 2 → NO; CO 2 + WITH → CO ;
10. HI → N 2 → + 1 2 ; Mg+ NS l → MgCl 2 + N 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 an equation?
4. Why are coefficients placed in chemical equations?