How to put coefficients in chemical equations. Algorithm for arranging coefficients in the equations of ovr
There are several methods for determining coefficients in the equations of redox reactions. We use the electronic balance method, in which the compilation of the complete OVR equation is carried out in the following sequence:
1. Draw up a reaction diagram, indicating the substances that reacted and the substances resulting from the reaction, for example:
2. Determine the oxidation state of atoms and write its sign and value above the symbols of the elements, noting the elements whose oxidation state has changed:
3. Write down the electronic equations for the oxidation and reduction reactions, determine the number of electrons donated by the reducing agent and accepted by the oxidizing agent, and then equalize them by multiplying by the appropriate coefficients:
4. The obtained coefficients corresponding to the electronic balance are transferred to the main equation:
5. Equalize the number of atoms and ions that do not change the oxidation state (in sequence: metals, nonmetals, hydrogen):
6. Check the correctness of the selection of coefficients based on the number of oxygen atoms on the left and right sides of the reaction equation - they should be equal (in this equation 24 = 18 + 2 + 4, 24 = 24).
Let's look at a more complex example:
Let us determine the oxidation states of atoms in molecules:
Let's create electronic equations for oxidation and reduction reactions and equalize the number of given and accepted electrons:
Let's transfer the coefficients to the main equation:
Let's equalize the number of atoms that do not change the oxidation state:
By counting the number of oxygen atoms on the right and left sides of the equation, we will make sure that the coefficients are chosen correctly.
The most important oxidizing and reducing agents
The redox properties of elements depend on the structure of the electronic shell of the atoms and are determined by their position in the periodic table of Mendeleev.
Metals, having 1-3 electrons at the outer energy level, easily give them up and exhibit only reducing properties. Nonmetals (elements of groups IV-VII) can both donate and accept electrons, so they can exhibit both reducing and oxidizing properties. In periods with an increase in the atomic number of an element, the reducing properties of simple substances weaken, and the oxidizing properties increase. In groups with increasing serial number, the reducing properties increase, and the oxidizing properties weaken. Thus, of simple substances, the best reducing agents are alkali metals, aluminum, hydrogen, carbon; the best oxidizing agents are halogens and oxygen.
The redox properties of complex substances depend on the degree of oxidation of the atoms that make up them. Substances containing atoms with the lowest oxidation state exhibit reducing properties. The most important reducing agents are carbon monoxide 
, hydrogen sulfide 
, iron(II) sulfate 
.Substances containing atoms with the highest oxidation state exhibit oxidizing properties. The most important oxidizing agents are potassium permanganate 
, potassium dichromate 
, hydrogen peroxide 
, Nitric acid 
, concentrated sulfuric acid 
.
Substances containing atoms with intermediate oxidation states can behave as oxidizing or reducing agents depending on the properties of the substances with which they interact and the reaction conditions. So in reaction with 
sulfurous acid exhibits reducing properties:
and when interacting with hydrogen sulfide it is an oxidizing agent:
In addition, for such substances, self-oxidation-self-reduction reactions are possible, occurring with a simultaneous increase and decrease in the oxidation state of atoms of the same element, for example:
The strength of many oxidizing and reducing agents depends on the pH of the medium. For example, 
in an alkaline environment it is reduced to 
, in neutral to 
, in the presence of sulfuric acid - up to 
.
Algorithm
Arrangement of coefficients in chemical reaction equations
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 in the left and right parts schemes, 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 → →2A1 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?
1. Let's draw up a reaction diagram:
Lesson objectives.Educational. Introduce students to a new classification of chemical reactions based on changes in the oxidation states of elements - oxidation-reduction reactions (ORR); teach students to arrange coefficients using the electronic balance method.
Developmental. Continue the development of logical thinking, the ability to analyze and compare, and the formation of interest in the subject.
Educational. To form the scientific worldview of students; improve work skills.
Methods and methodological techniques. Story, conversation, demonstration of visual aids, independent work students.
Equipment and reagents. Reproduction with the image of the Colossus of Rhodes, algorithm for arranging coefficients using the electronic balance method, table of typical oxidizing and reducing agents, crossword puzzle; Fe (nail), solutions of NaOH, CuSO4.
DURING THE CLASSES
Introductory part
(motivation and goal setting)
Teacher. In the 3rd century. BC. On the island of Rhodes, a monument was built in the form of a huge statue of Helios (the Greek god of the Sun). The grandiose design and perfect 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 we know 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 an 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. About 50 years after its completion, the Colossus collapsed. During the earthquake it broke at the level of the knees.
Scientists believe that the real reason The fragility of this miracle was due to metal corrosion. And the corrosion process is based on redox reactions.
Today in the lesson you will learn about redox reactions; learn about the concepts of “reducing agent” and “oxidizing agent”, about the processes of reduction and oxidation; learn to place coefficients in equations of redox reactions. Write down the date and topic of the lesson in your workbooks.
Learning new material
The teacher performs 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 for the reactions performed. In each equation, arrange the oxidation states of the elements in the formulas of the starting substances and reaction products.
The student writes reaction equations on the board and assigns 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 - for copper and iron.
Teacher. The second reaction is a redox reaction. Try to define redox reactions.
Student. Reactions that result in changes in the oxidation states of the elements that make up the reactants and reaction products are called redox reactions.
Students write down in their notebooks, under the teacher’s dictation, 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 we 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 we can see, the oxidation state has decreased. Therefore, copper accepts 2 electrons.
Iron donates electrons, it is a reducing agent, and the process of transferring electrons is called oxidation.
Copper accepts electrons, it is an oxidizing agent, and the process of adding electrons is called reduction.
Let us write down the diagrams of these processes:
So, give a definition of the concepts “reducing agent” and “oxidizing agent”.
Student. Atoms, molecules or ions that donate electrons are called reducing agents.
Atoms, molecules or ions that gain electrons are called oxidizing agents.
Teacher. How can we define the processes of reduction and oxidation?
Student. Reduction is the process by which an atom, molecule, or ion gains electrons.
Oxidation is the process of transfer of electrons by an atom, molecule or ion.
Students write down definitions from dictation in a notebook and draw.
Remember!
Donate electrons and oxidize.
Take electrons - recover.
Teacher. Oxidation is always accompanied by reduction, and vice versa, reduction is always associated with oxidation. The number of electrons given up by the reducing agent is equal to the number of electrons gained by the oxidizing agent.
To select coefficients in the equations of redox reactions, two methods are used - electronic balance and electron-ion balance (half-reaction method).
We will consider only the electronic balance method. To do this, we use an algorithm for arranging coefficients using the electronic balance method (designed on a piece of Whatman paper).
EXAMPLE Arrange the coefficients in this reaction scheme using the electronic balance method, determine the oxidizing agent and the reducing agent, indicate the processes of oxidation and reduction:
Fe2O3 + CO Fe + CO2.
We will use the algorithm for arranging coefficients using the electronic balance method.
3. Let’s write down the elements that change oxidation states:
4. Let’s create electronic equations, determining the number of given and received electrons:
5. The number of given and received electrons must be the same, because Neither the starting materials nor the reaction products are charged. We equalize the number of given and received electrons by selecting the least common multiple (LCM) and additional factors:
6. The resulting multipliers are coefficients. Let's 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 piece of Whatman paper 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 fuel combustion.
How to balance a chemical equation: rules and algorithm
Redox processes accompany the cycles of substances in nature.
Did you know that approximately 2 million tons of nitric acid are formed in the atmosphere every day, or
700 million tons per year, and fall to the ground with rain in the form of a weak solution (humans produce only 30 million tons of nitric acid per year).
What is happening in the atmosphere?
Air contains 78% by volume nitrogen, 21% oxygen and 1% other gases. Under the influence of lightning discharges, and on Earth there are an average of 100 lightning flashes 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 nitrogen oxide (IV) reacts with atmospheric moisture in the presence of oxygen, turning into nitric acid:
NO2 + H2O + O2 HNO3.
All these reactions are redox.
Exercise . Arrange the coefficients in the given reaction schemes using the electronic balance method, indicate the oxidizing agent, the reducing agent, the processes of oxidation and reduction.
Solution
1. Let's determine the oxidation states of elements:
2. Let us emphasize the symbols of elements whose oxidation states change:
3. Let’s write down the elements that have changed their oxidation states:
4. Let’s create electronic equations (determine the number of given and received electrons):
5. The number of electrons given and received is the same.
6. Let's transfer the coefficients from electronic circuits into the reaction scheme:
Next, students are asked to independently arrange the coefficients using the electronic balance method, determine the oxidizing agent, the reducing agent, and indicate the processes of oxidation and reduction in other processes occurring in nature.
The other two reaction equations (with coefficients) have the form:
The correctness of tasks is checked using an overhead projector.
Final part
The teacher asks students to solve a crossword puzzle based on the material they have studied. The result of the work is submitted for verification.
Having solved crossword, you will learn that the substances KMnO4, K2Cr2O7, O3 are strong... (vertical (2)).
Horizontally:
1. What process does the diagram 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 diagram reflect:
6. To select coefficients in the equations of redox reactions, use the electronic... method.
7. According to the diagram, aluminum gave up ... an electron.
8. In reaction:
H2 + Cl2 = 2HCl
hydrogen H2 – ... .
9. What type of reactions are always only redox?
10. The oxidation state of simple substances is….
11. In reaction:
reducing agent –….
Homework assignment.
According to the textbook by O.S. Gabrielyan “Chemistry-8” § 43, p. 178–179, ex. 1, 7 in writing. Task (for home). Designers of the first spaceships and submarines were faced with a problem: how to maintain a constant air composition on the ship and space stations? Get rid of excess carbon dioxide and replenish oxygen? A solution has been found.
Potassium superoxide KO2 reacts with carbon dioxide to form oxygen:
As you can see, this is a redox reaction. Oxygen in this reaction is both an oxidizing agent and a reducing agent.
IN space expedition Every gram of cargo counts. Calculate the supply of potassium superoxide that must be taken on a space flight if the flight lasts 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 ( increased level difficulties). Write down the equations of redox reactions that could lead 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 Sea, off the coast of modern-day Turkey, where the humid Mediterranean air is laden 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. M.: Bustard, 2002;
Gabrielyan O.S., Voskoboynikova N.P., Yashukova A.V. Teacher's handbook. 8th grade. M.: Bustard, 2002;
Cox R., Morris N. Seven wonders of the world. Ancient world, 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. M.: Education, 1989.
S.P.Lebesheva,
chemistry teacher of secondary school No. 8
(Baltiysk, Kaliningrad region)
Rules for selecting odds:
- if the number of atoms of an element in one part of the reaction scheme is even and in the other odd, then a coefficient of 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 - acid residues (non-metal atoms), then hydrogen atoms, and lastly - oxygen atoms.
— if the number of oxygen atoms on the left and right sides of the equation is the same, then the coefficients are determined correctly.
- after this, the arrow between the parts of the equation can be replaced with an equal sign.
— the coefficients in the chemical reaction equation should not have common divisors.
Example. Let's create an equation for the chemical reaction between iron (III) hydroxide and sulfuric acid to form iron (III) sulfate.
1. Let's draw up a reaction diagram:
Fe(OH)3 + H2SO4 → Fe2(SO4)3 + H2O
2. Let's select coefficients for the formulas of substances. We know that we need to start with the most complex substance and sequentially equalize throughout the entire scheme, first the metal atoms, then the acid residues, then hydrogen and finally 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. This means that before the formula Fe(OH)3 you need to put a coefficient of 2:
2Fe(OH)3 + H2SO4 → Fe2(SO4)3 + H2O
Now let's equalize the number of acidic residues of SO4. The salt Fe2(SO4)3 contains three acidic SO4 residues. This means that on the left side before the formula H2SO4 we put a coefficient of 3:
2Fe(OH)3 + 3H2SO4 → Fe2(SO4)3 + H2O.
Now let's 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 there are also 6 hydrogen atoms.
How to place coefficients in chemical equations
There are a total of 12 hydrogen atoms on the left side. This means that on the right side we put the coefficient 6 in front of the water formula H2O - 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 diagram is written completely, and we can replace the arrow with an equal sign:
2Fe(OH)3 + 3H2SO4 = Fe2(SO4)3 + 6H2O.
Education
How to place coefficients in chemical equations? Chemical equations
Today we will talk about how to place coefficients in chemical equations. This question is of interest not only to high school students educational institutions, but also for guys who are just getting acquainted with the basic elements of complex and interesting science. If you understand how to write chemical equations at the first stage, you won’t have problems solving problems in the future. Let's figure it out from the very beginning.
What is the equation
It is usually understood as a conventional recording of a chemical reaction occurring between selected reagents. For such a process, indices, coefficients, and formulas are used.
Compilation algorithm
How to write chemical equations? Examples of any interactions can be written by summing up the original connections. The equal sign indicates that interaction occurs between the reacting substances. Next, the formula of the products is compiled according to valency (oxidation state).
Video on the topic
How to record a reaction
For example, if you need to write down chemical equations confirming the properties of methane, choose the following options:
- halogenation (radical interaction with element VIIA of D.I. Mendeleev’s periodic table);
- combustion in air 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 oxygen, an exothermic process occurs, resulting in the formation of carbon dioxide and water vapor.
In order to correctly set the coefficients in chemical equations, the law of conservation of mass of substances is used. We begin the equalization 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 balanced using the electron balance or half-reaction method. We offer a sequence of actions designed to assign coefficients in the following types of reactions:
First, it is important to arrange the oxidation states of each element in the compound. When arranging them, it is necessary to take into account some rules:
- For a simple substance it is zero.
- In a binary compound their sum is 0.
- In a compound of three or more elements, the first one exhibits a positive value, the outermost ion - negative meaning degree of oxidation. The central element is calculated mathematically, taking into account that the total must be 0.
Next, select those atoms or ions whose oxidation state has changed. The plus and minus signs indicate the number of electrons (received, given). 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 place coefficients in chemical equations.
First example
Let's say the task is given: “Arrange the coefficients in the reaction, fill in the blanks, determine the oxidizing agent and the reducing agent.” Such examples are offered to school graduates who have chosen chemistry as their Unified State Examination.
KMnO4 + H2SO4 + KBr = MnSO4 + Br2 +…+…
Let's try to understand how to place coefficients in chemical equations offered to future engineers and doctors. 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 exhibits reducing properties.
We conclude that the missed 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 stage will be setting the coefficients in the equation.
Second example
Let's give another example to understand how to place coefficients in chemical equations of the redox type.
Suppose we are given the following diagram:
P + HNO3 = NO2 +…+…
Phosphorus, which by definition is a simple substance, exhibits reducing properties, increasing the oxidation state to +5. Therefore, one of the missing substances will be phosphoric acid H3PO4. ORR assumes the presence of a reducing agent, which will be nitrogen. It turns into nitric oxide (4), forming NO2
In order to put coefficients in this reaction, we will draw up an electronic balance.
P0 gives 5e = P+5
N+5 takes e = N+4
Considering that nitric acid and nitric oxide (4) must be preceded by a coefficient of 5, we obtain the finished reaction:
P + 5HNO3 =5NO2 + H2O + H3PO4
Stereochemical coefficients in chemistry make it possible to solve various calculation problems.
Third example
Considering that arranging coefficients causes difficulties for many high school students, it is necessary to practice the sequence of actions using specific examples. We offer another example of a task, the completion of which requires knowledge of the methodology for arranging coefficients in a 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 can we proceed to setting the coefficients.
After arranging the oxidation states of each element in the compounds, we can conclude that manganese exhibits oxidizing properties, reducing its valence. The reducing ability in the proposed reaction is demonstrated by sulfur, being reduced to a simple substance. After compiling the electronic balance, all we have to do is arrange the coefficients in the proposed process diagram. And it's 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 entered into the reaction must correspond to their number in the reaction products. Only in this case will it be possible to talk about the usefulness of the recorded chemical interaction, the possibility of its use for carrying out calculations and solving calculation problems different levels difficulties. Here is a variant of the task that involves placing the missing stereochemical coefficients in the reaction:
Si + …+ HF = H2SiF6 + NO +…
The difficulty of the task is that both the starting substances and the reaction products are missing. After setting the oxidation states of all elements, we see that the silicon atom in the proposed task exhibits reducing properties. Nitrogen (II) is present among the reaction products; one of the starting compounds is nitric acid. We logically determine that the missing product of the reaction is water. The final stage will be the placement of the resulting stereochemical coefficients into the reaction.
3Si + 4HNO3 + 18HF = 3H2SiF6 + 4NO + 8 H2O
Example of an equation problem
It is necessary to determine the volume of a 10% hydrogen chloride solution, the density of which is 1.05 g/ml, required to completely neutralize 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.s.). In order to cope with the task, you must first create an equation for the hydrolysis process of calcium carbide:
CaC2 + 2H2O = Ca (OH)2 + C2H2
Calcium hydroxide reacts with hydrogen chloride and complete neutralization occurs:
Ca(OH)2 + 2HCl = CaCl2 + 2H2O
We calculate the mass of acid that will be required for this process.
Coefficients and indices in chemical equations
Determine the volume of 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, drawing up an electronic balance, and determining an oxidizing agent and a reducing agent cause serious difficulties for modern graduates of secondary schools. Unfortunately, the degree of independence of modern graduates is almost minimal, so high school students do not practice the theoretical basis proposed by the teacher.
Among typical mistakes, which schoolchildren assume when arranging coefficients in reactions different types, a lot of mathematical errors. For example, not everyone knows how to find the least common multiple or correctly divide and multiply numbers. The reason for this phenomenon is a decrease in the number of hours allocated in educational schools to study this topic. In the basic chemistry curriculum, teachers do not have the opportunity to work with their students on issues related to the preparation of 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 e) Cu(OH)2 + HNO3 -> Cu(NO3)2 + H2O
Lesson 13. Writing chemical equations
Write down the definitions:
a) compound reaction b) exothermic reaction c) irreversible reaction.
a) carbon reacts with oxygen and carbon monoxide (II) is formed;
b) magnesium oxide reacts with nitric acid and magnesium nitrate and water are formed;
c) iron (III) hydroxide decomposes into iron (III) oxide and water;
d) methane CH4 burns in oxygen and produces carbon monoxide (IV) and water;
e) nitric oxide (V) when dissolved in water forms nitric acid.
4. Solve the problem using 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 zinc containing 35% impurities reacts with sulfuric acid?
OPTION 2
- 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 + Mg SO4 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 as described:
a) carbon reacts 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 using the equation:
a) What volume of ammonia is formed when hydrogen reacts with nitrogen?
b) What mass of calcium chloride is formed when interacting with hydrochloric acid marble containing 80% CaCO3?
c) What volume and mass of hydrogen will be released when magnesium containing 30% impurities reacts with sulfuric acid?
How to write chemical equations? First, it is important to arrange the oxidation states of each element in the compound. Let’s say the task is given: “Arrange the coefficients in the reaction, fill in the blanks, determine the oxidizing agent and the reducing agent.” One of the missing products is water, and the second will be potassium sulfate. After compiling the electronic balance, the final stage will be setting 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 arranging coefficients in reactions of various types, there are many mathematical errors.
Exist certain rules, by which they can be determined for each element. Formulas consisting of three elements have their own nuances in calculating oxidation states. Let's continue the conversation about how to equalize chemical equations using the electronic balance method. 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.
Algebraic method
Be sure to read about elemental analysis for an in-depth look at empirical formulas and chemical analysis.
Chemistry studies substances, their properties, and transformations. In molecular form, the process of burning iron in the atmosphere can be expressed using signs and symbols. 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. To begin with, we will assign oxidation states for each of the elements in the starting substances and reaction products. Next, the hydrogen is tested.
Equalizing chemical reactions
Equating 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 creation of new atoms and the destruction of old ones during a chemical reaction. Pay attention to the index of each atom; it indicates their number. By adding subscripts 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 a factor is behind the brackets, then each element in the brackets is multiplied by it. You need to start with nitrogen, since there is less of it than oxygen and hydrogen. Great, hydrogen equalized. 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? 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. Next action we must arrange the oxidation states of all elements in each substance in order to understand where oxidation occurred and where reduction occurred.
Example of analysis of simple reactions
WITH right side there are no indices, that is, one particle of oxygen, and on the left - 2 particles. No additional indexes or fixes in chemical formula You cannot enter it because it is written correctly. On the right side, we multiply one by 2 to get 2 oxygen ions there.
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 called a coefficient. Let's start analyzing. Thus, the result is 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 permitted to freely use any document for your own purposes, subject to the following conditions:
2) The symbols of chemical elements should be written strictly in the form in which they appear in the periodic table.
Information card. "Algorithm for arranging coefficients in chemical reaction equations."
3) Occasionally situations arise when the formulas of reactants and products are written absolutely correctly, but the coefficients are still not assigned. This problem is most likely to occur with oxidation reactions. organic matter in which the carbon skeleton is torn.
You need to be able to not only write the reaction equation, but also read it. Therefore, sometimes, having written all the formulas in the reaction equation, you have to equalize the number of atoms in each part of the equation and set the coefficients. Count whether there are equal numbers of atoms of each element on the left and right sides of the equation.
For many schoolchildren, writing equations of chemical reactions and correctly arranging coefficients is not an easy task. But you just need to remember a few simple rules, and the task will no longer cause difficulties. Coefficient, that is, the number in front of the formula of the molecule chemical substance, applies to all characters, and is multiplied by every index of every character!
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 → →2A1 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?
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 equalize coefficients in chemical equations? The high school chemistry course covers the electronic balance method. 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 valency. 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 they interact simple substances calcium and oxygen, 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 oxidation states for each of the elements in the starting substances and reaction products. 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)+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 the ninth and eleventh grades of educational institutions who choose an exam in chemistry are offered similar tasks in one of the tasks of the final tests.