Compilation of the equations of chemical reactions. How to place coefficients in chemical equations

Class: 8

Presentation to the lesson

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The purpose of the lesson: Help the learning to form knowledge of the chemical equation as a conditional recording of a chemical reaction with the help of chemical formulas.

Tasks:

Educational:

systematize the previously studied material;
training the ability to draw up the equations of chemical reactions.

Educational:

rise communication skills (work in a pair, the ability to listen and hear).

Developing:

develop educational and organizational skills aimed at performing the task;
develop analytical thinking skills.

Type of lesson: combined.

Equipment: Computer, Multimedia Projector, Screen, Estimated sheets, Reflection card, "Chemical Signs Set", Notebook with Printed Foundation, Reagents: Sodium Hydroxide, Iron Chloride (III), Alcohol, Holder, Match, Watman Sheet, Multicolored Chemical Signs.

Presentation of the lesson (Appendix 3)

The structure of the lesson.

І. Organizing time.
ІІ. Actualization of knowledge and skills.
ІІІ. Motivation and goal.
ІV. Studying a new material:
4.1 Aluminum combustion reaction in oxygen;
4.2 Reaction of decomposition of iron hydroxide (III);
4.3 Algorithm for the placement of coefficients;
4.4 minute relaxation;
4.5 Practice coefficients;
V. Fastening the knowledge gained.
Vі. Summing up the lesson and estimates.
Vі. Homework.
Vііі. The final word of the teacher.

During the classes

Chemical People
It is determined by kind of elementary
components
The number of them I.
chemical structure.
D.I. Inendeev

Teacher. Hello guys. Sit down.
Please note: you have a notebook on your table with a printed basis (Appendix 2), In which you will work today, and the estimated sheet, in it you will fix your achievements, sign it.

Actualization of knowledge and skills.

Teacher. We met with physical and chemical phenomena, chemical reactions and signs of their flow. He studied the law of preserving the mass of substances.
Let's check your knowledge. I suggest you open a notebook with a printed basis and execute the task 1. You are given for 5 minutes to execute the task.

Test on "Physical and chemical phenomena. The law of preserving the mass of substances. "

1. Why do chemical reactions differ from physical phenomena?

Change shape, aggregate state of matter.
The formation of new substances.
Change location.

2. What are the signs of a chemical reaction?

The formation of precipitate, color change, gas isolation.

Magnetization, evaporation, oscillation.

Growth and development, movement, reproduction.

3. In accordance with what law are the chemical reaction equations?

The law of constancy of the composition of the substance.
The law of preserving the mass of substance.
Periodic law.
The law of speakers.
The law of global gravity.

4. The law of preserving the mass of the substance opened:

DI. Mendeleev.
C. Darwin.
M.V. Lomonosov.
I. Newton.
A.I. Butlers.

5. The chemical equation is called:

Conditional recording of a chemical reaction.

Conditional record of the composition of the substance.

Record the condition of the chemical problem.

Teacher. You performed work. I suggest you to check it. Change notebooks and carry out mutual test. Attention on the screen. For each correct answer - 1 point. The total number of points will enlist in the estimated sheets.

Motivation and goal.

Teacher.Using this knowledge, we will compose the equations of chemical reactions today, revealing the problem "Is the law of preserving the mass of substances the basis for compiling the equations of chemical reactions"

Studying a new material.

Teacher. We used to assume that the equation is a mathematical example, where there is an unknown, and this unknown must be calculated. But in the chemical equations, nothing unknown does not happen: they are simply written by all formulas: what substances react and which are obtained during this reaction. Let's see experience.

(Reaction of the compound of sulfur and iron.) Appendix 3

Teacher. From the point of view of mass substances, the equation of the reaction of the compound of iron and sulfur is understood as follows

Iron + sulfur → Iron sulfide (II) (Task 2 TPO)

But in the chemistry, words are reflected in chemical signs. Write this equation with chemical symbols.

Fe + S → Fes

(One student writes on the board, the rest in TPO.)

Teacher. Read now.
Students. Iron molecule interacts with sulfur molecule, one molecule of iron sulfide (II) is obtained.
Teacher. In this reaction, we see that the amount of starting materials is equal to the amount of substances in the reaction product.
It should always be remembered that in the preparation of the equations of reactions, no atom should be lost or suddenly appear. Therefore, sometimes by writing all formulas in the reaction equation, it is necessary to equalize the number of atoms in each part of the equation - to expose the coefficients. Let's see another experience

(Aluminum combustion in oxygen.) Appendix 4

Teacher.We write the equation of the chemical reaction (task 3 in TPO)

Al + O 2 → Al +3 O -2

To record the right formula of oxide, remember that

Students. Oxygen in oxides has the degree of oxidation -2, aluminum - the chemical element with a constant degree of oxidation +3. Nok \u003d 6.

Al + O 2 → Al 2 O 3

Teacher.We see that 1 aluminum atom is entering the reaction, two aluminum atoms are formed. Two oxygen atoms comes, three oxygen atoms are formed.
Simple and beautiful, but disrespectful to the law of preserving the mass of substances - it is different before and after the reaction.
Therefore, we need to place the coefficients in this equation of chemical reaction. For this we find the NOC for oxygen.

Students.Nok \u003d 6.

Teacher.In front of oxygen formulas and aluminum oxide, we put the coefficients so that the number of oxygen atoms on the left and right was equal to 6.

Al + 3 O 2 → 2 Al 2 O 3

Teacher.Now we obtain that, as a result of the reaction, four aluminum atoms are formed. Therefore, in front of the aluminum atom in the left part, we set the coefficient 4

Al + 3o 2 → 2Al 2 O 3

Once again, we recount all atoms to the reaction and after it. We put equally.

4Al + 3O 2 _ \u003d 2 al 2 o 3

Teacher.Consider another example

(A teacher demonstrates experience in the decomposition of iron hydroxide (III).)

Fe (OH) 3 → Fe 2 O 3 + H 2 O

Teacher.We put the coefficients. The reaction comes 1 atom of iron, two iron atoms are formed. Consequently, in front of the iron hydroxide formula (3) we put the coefficient 2.

Fe (OH) 3 → Fe 2 O 3 + H 2 O

Teacher.We obtain that 6 hydrogen atoms (2x3) takes into the reaction, 2 hydrogen atom is formed.

Students. Nok \u003d 6. 6/2 \u003d 3. Consequently, in the water formula, we put the coefficient 3

2Fe (OH) 3 → Fe 2 O 3 + 3 H 2 O

Teacher. We consider oxygen.

Students.Left - 2x3 \u003d 6; Right - 3 + 3 \u003d 6

Students.The number of oxygen atoms entered into the reaction is equal to the amount of oxygen atoms formed during the reaction. You can put equals.

2Fe (OH) 3 \u003d Fe 2 O 3 +3 H 2 O

Teacher.Now let's summarize everything previously and get acquainted with the algorithm of the coefficients of coefficients in the chemical reaction equations.

Calculate the number of atoms of each element in the right and left part of the chemical reaction equation.

Determine which element the number of atoms changes, find the NOC.

Divide the NOC on the indexes - get the coefficients. Put them in front of formulas.

Recalcute the number of atoms, if necessary, repeat.

Last to check the number of oxygen atoms.

Teacher. You worked well and probably tired. I suggest you relax, close your eyes and remember any pleasant moments of life. Each of you are different. Now open your eyes and make circular movements them first clockwise, then opp. Now intensively move the eyes horizontally: to the right - left, and vertical: up - down.
And now we activate mental activity and massage the ush.

Teacher.We continue work.
In notebooks with the printed basis, you will perform the task 5. Work you will be in pairs. You need to place the coefficients in the equations of chemical reactions. 10 minutes is given to the task.

P + CL 2 → PCL 5
Na + S → Na 2 s
HCl + Mg → MgCl 2 + H 2
N 2 + H 2 → NH 3
H 2 O → H 2 + O 2

Teacher.Check the task execution ( the teacher polls and displays the correct answers to the slide). For each correctly supplied coefficient - 1 point.
With the task you coped. Well done!

Teacher.Now let's go back to our problem.
Guys, how do you think, is the law of preserving the mass of substances the basis for compiling the equations of chemical reactions.

Students. Yes, during the lesson, we proved that the law of preserving the mass of substances is the basis for the preparation of the equations of chemical reactions.

Consolidation of knowledge.

Teacher.We studied all basic questions. Now we will perform a small test that will allow you to see how you have mastered the topic. You must answer it only "yes" or "no". 3 minutes is given to work.

Approval.

In the CA + CL 2 reaction 2 → CaCl 2 coefficients are not needed.(Yes)
In the reaction Zn + HCl → ZnCl 2 + H 2 coefficient of zinc 2. (Not)
In the reaction Ca + O 2 → Cao, the coefficient of calcium oxide 2.(Yes)
In the reaction CH 4 → C + H 2 coefficients are not needed.(Not)
In the reaction Cuo + H 2 → Cu + H 2 O the coefficient of copper 2. (Not)
In the C + O 2 → CO reaction, the coefficient 2 should be put in carbon oxide (II), and carbon. (Yes)
In the CUCl 2 + Fe → Cu + FECL 2 reaction, the coefficients are not needed.(Yes)

Teacher. Check the performance of work. For each correct answer - 1 point.

The outcome of the lesson.

Teacher.You coped well with the task. Now calculate the total number of scored points for the lesson and make an estimate according to the rating that you see on the screen. Take me the estimated sheets to exhibit your marks.

Homework.

Teacher.Our lesson approached the end, during which we were able to prove that the law of preserving the mass of substances is the basis for the preparation of the equations of reactions, and learned how to make the equations of chemical reactions. And, like a final point, write down your homework

§ 27, UPR. 1 - for those who grate "3"
uPR. 2- For those who received the rating "4"
uPR. 3 - for those who have received an estimate“5”

The final word of the teacher.

Teacher. I thank you for the lesson. But before you leave the Cabinet, pay attention to the table (The teacher shows on the Watman Sheet with the image of the table and multicolored chemical signs).You see chemical signs of different colors. Each color symbolizes your mood .. I suggest you to make your table of chemical elements (it will differ from PSHE D.I. Indeleeva) - lesson mood table. To do this, you must approach the noth leaf, take one chemical element, according to the characteristic you see on the screen, and attach to the table of the table. I will make it first, showing you my comfort from working with you.

F I was comfortable at the lesson, I received an answer to all questions that interest me.

F In the lesson, I reached the goal half.
F I was bored in a lesson, I did not recognize anything new.

1) In order to place the coefficients in the chemical reaction equation online insert the equation and click "Equal"

2) Chemical elements symbols should be recorded strictly in the form in which they are featured in the Mendeleev table. Those. The first letter in the symbol designation of any chemical element must be the title, and the second line. For example, the symbol of the chemical element of manganese should be written as Mn, but not in any case as Mn and Mn;

3) occasionally arise situations where the formulas of the reagents and products are written absolutely correctly, but the coefficients are still not settled. This may occur in cases where the coefficients in the equation can be placed in two or more methods. The most likely the emergence of such a problem with the oxidation reactions of organic substances under which the carbon skeleton is torn. In this case, try replacing unchangeable fragments of organic molecules to some arbitrary symbol, for example, the phenyl radical C 6 H 5 can be designated as pH or X. For example, the following equation:

C 6 H 5 CH 2 CH 3 + KMNO 4 + H 2 SO 4 → C 6 H 5 COOH + CO 2 + K 2 SO 4 + MNSO 4 + H 2 O

it will not be balanced, as it is possible to various coefficients alignment. However, by entering the designation C 6 H 5 \u003d pH, the placement of the coefficients occurs correctly:

5Phch 2 CH 3 + 12KMNO 4 + 18H 2 SO 4 → 5PhCooh + 5CO 2 + 6K 2 SO 4 + 12mNSO 4 + 28H 2 O

Note

The equation is allowed to separate the formulas of the reagents on the product formulas to use as a sign of equality (\u003d) and the arrow (→), as well as the random recording of individual letters of symbols of chemical elements not with Latin, and Cyrillic in the case of their identical writing, such as symbols C, H, O, P.

Carefully examine the algorithms and write down in the notebook, decide independently proposed tasks

I. Using the algorithm, decide the following tasks yourself:

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

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

Algorithm №1

Calculation of the amount of substance according to a known number of substance participating in the reaction.

Example. Calculate the amount of substance of oxygen, highlighted as a result of water decomposition by the amount of substance 6 mol.

	Registration of the task
1. Write the task condition	*Dano* : ν (H 2 O) \u003d 6mol _____________ *To find* : ν (O 2) \u003d?
	*Decision* : M (o 2) \u003d 32g / mol
and put the coefficients	2N 2 O \u003d 2N 2 + O 2
, and under the formulas -
5. To calculate the desired amount of substance, let us make a relationship
6. Record the answer	*Answer: ν (O 2) \u003d 3mol*

II. Using the algorithm, the following tasks decreased:

1. Calculate the mass of the sulfur needed to obtain sulfur oxide ( S +. O 2 \u003d. SO 2).

2. Calculate the lithium mass required to obtain lithium chloride by the amount of substance 0.6 mol (2 Li +. Cl 2 \u003d 2 LICL).

Algorithm №2.

Calculation of the mass of matter according to the number of other substances involved in the reaction.

Example: Calculate the mass of aluminum necessary to obtain aluminum oxide by the amount of substance 8 mol.

Sequence of action	Registration of problem solving
1. Write the task condition	*Given:* ν( Al 2 O. 3 ) \u003d 8mol ___________ *To find:* m.( Al)=?
2. Calculate the molar masses of substances, about which, there is a question in the task	M.( Al 2 O. 3 ) \u003d 102g / mol
3. Write the reaction equation and put the coefficients	4 AL + 3O 2 \u003d 2AL 2 O 3
4. Above the formulas of substances *the number of substances from the condition of the problem* , and under the formulas - stoichiometric coefficients , displayed by the reaction equation
5. Calculate the amount of substance whose mass it is required to find. To do this, make a ratio.
6. Calculate the mass of the substance you want to find	m.= ν ∙ M., m.(Al)= ν (Al)∙ M.(Al) \u003d 16mol ∙ 27g / mol \u003d 432g
7. Record the answer	*Answer:* m. *(AL) \u003d* *432 g*

III. Using the algorithm, the following tasks decreased:

1. Calculate the amount of substance of sodium sulphide, if in a reaction with a sodium-fed by a gramass 12.8 g (2 Na +. S \u003d. Na 2. S).

2. Calculate the amount of substance-forming copper, if copper oxide is entered into the reaction with hydrogen ( Ii) weighing 64 g ( Cuo +. H 2 \u003d. Cu +. H 2. O).

Carefully examine the algorithm and write down in the notebook

Algorithm number 3.

Calculation of the amount of substance at a well-known mass of another substance involved in the reaction.

Example.Calculate the amount of copper oxide substance (I. ) If copper takes a mass of 19.2g to the oxygen reaction.

Sequence of action	Registration of the task
1. Write the task condition	*Given:* m.( Cu.) \u003d 19.2g ___________ *To find:* ν( Cu. 2 O.)=?
2. Calculate the molar masses of substances, about which, there is a question in the task	M (Cu.) \u003d 64g / mol
3. Find the amount of substance whose mass given in the condition of the problem
and put the coefficients	4 Cu.+ O. 2 =2 Cu. 2 O.
*the number 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, let us make a relationship
7. We write the answer	*Answer: ν (* *Cu.* 2 O. *) \u003d 0.15 mol*

Carefully examine the algorithm and write down in the notebook

IV. Using the algorithm, the following tasks decreased:

1. Calculate the mass of oxygen necessary for reaction with iron weighing 112 g

(3 Fe + 4. O 2 \u003d. Fe 3. O 4).

Algorithm number 4.

Calculation of the mass of matter at a well-known mass of another substance participating in the reaction

Example.Calculate the mass of oxygen necessary for the combustion of phosphorus, weighing 0.31g.

Sequence of action

Designation

1. Write the task condition

Given:

m.( P.) \u003d 0.31

_________

To find:

m.( O. 2 )=?

2. Calculate the molar masses of substances,

about which, there is a question in the task

M (P.) \u003d 31g / mol

M.( O. 2 ) \u003d 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. Above formulas substant

the number 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 \u003d 0.4g

8. We write the answer

Answer: m. ( O. 2 ) \u003d 0.4g

Tasks for self solutions

3. Calculate the mass of the sulfur needed to obtain sulfur oxide ( Iv) the amount of substance is 4 mol ( S +. O 2 \u003d. SO 2).

4. Calculate the lithium mass required to produce lithium chloride with the amount of substance 0.6 mol (2 Li +. Cl 2 \u003d 2 LICL).

5. Calculate the amount of substance sodium sulfide, if a sulfur of 12.8 g (2) enters the reaction with sodium (2 Na +. S \u003d. Na 2. S).

6. Calculate the amount of substance formed copper if copper oxide is entered into the reaction with hydrogen ( Ii) weighing 64 g ( Cuo +. H 2 \u003d.

Methods of solving tasks in chemistry

When solving tasks, you must be guided by several simple rules:

Carefully read the condition of the problem;
Write down that is given;
Translate if it is necessary, units of physical quantities in the SI system (some generate units are allowed, for example liters);
Write, if necessary, the reaction equation and place the coefficients;
Solve the problem using the concept of the amount of substance, and not the method of drawing up proportions;
Record the answer.

In order to successfully prepare for chemistry, it is possible to carefully consider solutions to the tasks in the text, as well as to solve their independently enough. It is in the process of solving problems that the main theoretical provisions of the Chemistry Course will be enshrined. Relieving tasks are necessary throughout the time of study of chemistry and preparation for the exam.

You can use tasks on this page, or you can download a good collection of tasks and exercises with a solution of typical and complicated tasks (M. I. Lebedeva, I. A. Ankudimova): Download.

Mole, molar mass

The molar mass is the ratio of the mass of the substance to the amount of substance, i.e.

M (x) \u003d m (x) / ν (x), (1)

where M (x) is the molar mass of the substance X, M (x) - the mass of the substance X, ν (x) is the amount of the substance X. The unit of the molar mass - kg / mol, however, it is usually used a unit of g / mol. Unit of mass - g, kg. The unit of the amount of substance is mole.

Any the task in chemistry is solved Through the amount of substance. It is necessary to remember the main formula:

ν (x) \u003d m (x) / m (x) \u003d v (x) / v m \u003d n / n a, (2)

where V (x) is the volume of the substance X (L), V M is the molar volume of the gas (l / mol), N is the number of particles, N A is constant avogadro.

1. Determine the mass Sodium iodide NAI with a substance of 0.6 mol.

Dano: ν (NAI) \u003d 0.6 mol.

To find: M (NAI) \u003d?

Decision. The molar mass of sodium iodide is:

M (NAI) \u003d M (Na) + m (i) \u003d 23 + 127 \u003d 150 g / mol

We determine the mass of NAI:

m (NAI) \u003d ν (NAI) M (NAI) \u003d 0.6 150 \u003d 90

2. Determine the amount of substance Atomic boron contained in sodium tetraborate Na 2 B 4 O 7 weighing 40.4 g

Dano: M (Na 2 B 4 O 7) \u003d 40.4

To find: ν (b) \u003d?

Decision. The molar mass of the sodium tetraborate is 202 g / mol. Determine the amount of substance Na 2 B 4 O 7:

ν (Na 2 B 4 O 7) \u003d M (Na 2 B 4 O 7) / M (Na 2 B 4 O 7) \u003d 40.4 / 202 \u003d 0.2 mol.

Recall that 1 mol of the sodium tetragano molecule contains 2 mols of sodium atoms, 4 mole of boron atoms and 7 mol of oxygen atoms (see sodium tetraborate formula). Then the amount of the atomic boron substance is: ν (b) \u003d 4 ν (Na 2 B 4 O 7) \u003d 4 0.2 \u003d 0.8 mol.

Calculations for chemical formulas. Mass fraction.

The mass fraction of the substance is the ratio of the mass of this substance in the system by the mass of the entire system, i.e. ω (x) \u003d m (x) / m, where ω (x) is the mass fraction of the substance x, m (x) - the mass of the substance X, M is the mass of the entire system. Mass fraction is a dimensionless value. It is expressed in fractions from one or percentage. For example, the mass fraction of atomic oxygen is 0.42, or 42%, i.e. Ω (O) \u003d 0.42. The mass fraction of atomic chlorine in sodium chloride is 0.607, or 60.7%, i.e. Ω (CL) \u003d 0.607.

3. Determine the mass share Crystallization Water in Barium Chloride Dihydrate BACL 2 2H 2 O.

Decision: The Molar Mass BACL 2 2H 2 O is:

M (BACL 2 2H 2 O) \u003d 137+ 2 35.5 + 2 18 \u003d 244 g / mol

From the BACL 2 2H 2H formula, it follows that 1 mole of the barium chloride dihydrate contains 2 mol H 2 O. From here, it is possible to determine the mass of water contained in BACL 2 2H 2 O:

m (H 2 O) \u003d 2 18 \u003d 36

We find the mass fraction of crystallization water in the dihydrate of barium chloride BACL 2 2H 2 O.

ω (H 2 O) \u003d M (H 2 O) / M (BACL 2 2H 2 O) \u003d 36/244 \u003d 0.1475 \u003d 14.75%.

4. From a rock sample with a mass of 25 g containing a mineral ARGEntitis AG 2 S, a silver mass of 5.4 g. Determine the mass share Argentita in the sample.

Dano: m (Ag) \u003d 5.4 g; m \u003d 25

To find: Ω (AG 2 S) \u003d?

Decision: Determine the amount of silver substance located in Argentita: ν (AG) \u003d M (Ag) / M (Ag) \u003d 5.4 / 108 \u003d 0.05 mol.

It follows from the AG 2 S formula that the amount of a substance of argentitis is two times less than the amount of silver substance. Determine the amount of the substance of the Argentita:

ν (Ag 2 S) \u003d 0.5 ν (Ag) \u003d 0.5 0.05 \u003d 0.025 mol

Calculate the mass of argentitis:

m (Ag 2 S) \u003d ν (Ag 2 S) M (Ag 2 S) \u003d 0.025 248 \u003d 6.2 g

Now we determine the mass fraction of argentitis in the sample of rock, weighing 25 g.

Ω (Ag 2 S) \u003d M (Ag 2 S) / m \u003d 6.2 / 25 \u003d 0.248 \u003d 24.8%.

Output of the formulas of compounds

5. Determine the simplest compound formula Potassium with manganese and oxygen, if the mass fractions of the elements in this substance are 24.7, 34.8 and 40.5%, respectively.

Dano: ω (k) \u003d 24.7%; Ω (Mn) \u003d 34.8%; Ω (O) \u003d 40.5%.

To find: Connection formula.

Decision: For calculations, we choose a mass of compound equal to 100 g, i.e. M \u003d 100 g. Mass of potassium, manganese and oxygen will be:

m (k) \u003d m Ω (k); m (k) \u003d 100 0.247 \u003d 24.7 g;

m (Mn) \u003d m Ω (Mn); M (Mn) \u003d 100 0.348 \u003d 34.8 g;

m (O) \u003d M Ω (O); M (O) \u003d 100 0.405 \u003d 40.5 g

We determine the amounts of substances of atomic potassium, manganese and oxygen:

ν (k) \u003d m (k) / m (k) \u003d 24.7 / 39 \u003d 0.63 mol

ν (Mn) \u003d M (Mn) / M (Mn) \u003d 34.8 / 55 \u003d 0.63 mol

ν (O) \u003d M (O) / M (O) \u003d 40.5 / 16 \u003d 2.5 mol

We find the ratio of the amounts of substances:

ν (k): ν (Mn): ν (O) \u003d 0.63: 0.63: 2.5.

Dividing the right-hand side of equality for a smaller number (0.63) we obtain:

ν (k): ν (Mn): ν (O) \u003d 1: 1: 4.

Consequently, the simplest formula of the KMNO 4 compound.

6. With a combustion of 1.3 g of substance, 4.4 g of carbon oxide (IV) and 0.9 g of water were formed. Find a molecular formula Substances, if its hydrogen density is 39.

Dano: m (V-Ba) \u003d 1.3 g; m (CO 2) \u003d 4.4 g; m (H 2 O) \u003d 0.9 g; D H2 \u003d 39.

To find: The formula of substance.

Decision: Suppose that the desired substance contains carbon, hydrogen and oxygen, because With its combustion, CO 2 and H 2 O is formed. Then it is necessary to find the amounts of substances CO 2 and H 2 O to determine the amounts of substances of atomic carbon, hydrogen and oxygen.

ν (CO 2) \u003d M (CO 2) / M (CO 2) \u003d 4.4 / 44 \u003d 0.1 mol;

ν (H 2 O) \u003d M (H 2 O) / M (H 2 O) \u003d 0.9 / 18 \u003d 0.05 mol.

We determine the amounts of substances of atomic carbon and hydrogen:

ν (c) \u003d ν (CO 2); ν (C) \u003d 0.1 mol;

ν (n) \u003d 2 ν (H 2 O); ν (n) \u003d 2 0.05 \u003d 0.1 mol.

Consequently, carbon and hydrogen mass will be equal:

m (c) \u003d ν (c) m (c) \u003d 0.1 12 \u003d 1.2 g;

m (n) \u003d ν (n) m (n) \u003d 0.1 1 \u003d 0.1 g

Determine the qualitative composition of the substance:

m (V-BA) \u003d M (C) + m (n) \u003d 1.2 + 0.1 \u003d 1.3 g

Consequently, the substance consists only of carbon and hydrogen (see the condition of the task). We will now define its molecular weight based on this tasks Density of substance on hydrogen.

M (V-BA) \u003d 2 D H2 \u003d 2 39 \u003d 78 g / mol.

ν (c): ν (n) \u003d 0.1: 0,1

Sharing the right-hand side of the equality by the number 0.1, we get:

ν (C): ν (H) \u003d 1: 1

We will take the number of carbon atoms (or hydrogen) for "x", then, multiplying "x" to atomic weights of carbon and hydrogen and equating this amount of the molecular weight of the substance, solving the equation:

12x + x \u003d 78. Hence x \u003d 6. Therefore, the formula of the substance from 6 H 6 is benzene.

Molar volume of gases. Laws of perfect gases. Volumetric share.

The molar volume of gas is equal to the ratio of the volume of gas to the amount of substance of this gas, i.e.

V m \u003d v (x) / ν (x),

where V m is the molar volume of the gas - a constant value for any gas under these conditions; V (x) - the volume of gas x; ν (x) - the amount of the substance of gas x. The molar volume of gases under normal conditions (normal pressure p \u003d 101 325 Pa ≈ 101.3 kPa and temperature TN \u003d 273.15 K ≈ 273 K) is v m \u003d 22.4 l / mole.

In the calculations associated with gases, often have to move from these conditions to normal or vice versa. At the same time, it is convenient to use the formula following from the combined gas law of Boyl-Mariott and Gay Loursak:

──── = ─── (3)

Where p is pressure; V - volume; T- temperature in the Kelvin scale; The "H" index indicates normal conditions.

The composition of gas mixtures is often expressed using a bulk fraction - the ratio of the volume of this component to the total volume of the system, i.e.

where φ (x) is the volume fraction of the component X; V (x) - the volume of the component X; V is the volume of the system. The volume fraction is a dimensionless value, it is expressed in fractions from one or in percent.

7. What volume It takes at a temperature of 20 o C and a pressure of 250 kPa ammonia weighing 51 g?

Dano: M (NH 3) \u003d 51 g; p \u003d 250 kPa; T \u003d 20 o C.

To find: V (NH 3) \u003d?

Decision: Determine the amount of ammonia substance:

ν (NH 3) \u003d M (NH 3) / M (NH 3) \u003d 51/17 \u003d 3 mole.

Ammonia volume under normal conditions is:

V (NH 3) \u003d V M ν (NH 3) \u003d 22.4 3 \u003d 67.2 liters.

Using formula (3), give the volume of ammonia to these conditions [Temperature T \u003d (273 +20) K \u003d 293 K]:

p H TV H (NH 3) 101.3 293 67.2

V (NH 3) \u003d ──────── \u003d ───────── \u003d 29.2 liters.

8. Determine volumewhich will take a gas mixture under normal conditions containing hydrogen, weighing 1.4 g and nitrogen, weighing 5.6 g.

Dano: m (n 2) \u003d 5.6 g; M (H 2) \u003d 1.4; Well.

To find: V (mixtures) \u003d?

Decision: We find the amount of substance of hydrogen and nitrogen:

ν (n 2) \u003d m (n 2) / m (n 2) \u003d 5.6 / 28 \u003d 0.2 mol

ν (H 2) \u003d M (H 2) / M (H 2) \u003d 1.4 / 2 \u003d 0.7 mol

Since under normal conditions, these gases do not interact with each other, the volume of the gas mixture will be equal to the amount of gases, i.e.

V (mixtures) \u003d V (N 2) + V (H 2) \u003d V M ν (n 2) + V M ν (H 2) \u003d 22.4 0.2 + 22.4 0.7 \u003d 20,16 l.

Calculations for chemical equations

Calculations for chemical equations (stoichiometric calculations) are based on the law of preserving the mass of substances. However, in real chemical processes, due to the incomplete flow of reaction and various losses of substances, the mass of products formed often is less than that which should be formed in accordance with the law of preserving the mass of substances. The yield of the reaction product (or the mass fraction of the output) is pronounced in percentage of the mass ratio of the actually obtained product to its mass, which should be formed in accordance with theoretical calculation, i.e.

η \u003d / m (x) (4)

Where η is the product output,%; M p (x) - the mass of the product X, obtained in the real process; M (x) - the calculated mass of the substance X.

In those tasks where the product yield is not specified, it is assumed that it is a quantitative (theoretical), i.e. η \u003d 100%.

9. What a mass of phosphorus should be burned for getting Phosphorus oxide (V) weighing 7.1 g?

Dano: m (p 2 O 5) \u003d 7.1 g

To find: m (p) \u003d?

Decision: Record the combustion equation of phosphorus burning and set stoichiometric coefficients.

4p + 5O 2 \u003d 2p 2 O 5

Determine the amount of substance P 2 O 5, which has been in the reaction.

ν (P 2 O 5) \u003d M (P 2 O 5) / M (P 2 O 5) \u003d 7.1 / 142 \u003d 0.05 mol.

It follows from the reaction equation that ν (P 2 O 5) \u003d 2 ν (P), therefore, the amount of phosphorus substance required in the reaction is:

ν (P 2 O 5) \u003d 2 ν (p) \u003d 2 0.05 \u003d 0.1 mol.

From here we find the mass of phosphorus:

m (p) \u003d ν (p) m (p) \u003d 0.1 31 \u003d 3.1 g

10. In the excess of hydrochloric acid, magnesium mass was dissolved with a mass of 6 g and zinc weighing 6.5 g. What volume hydrogen measured under normal conditions stand out wherein?

Dano: m (Mg) \u003d 6 g; m (zn) \u003d 6.5 g; Well.

To find: V (H 2) \u003d?

Decision: Record the equations of the reaction of the interaction of magnesium and zinc with hydrochloric acid and arrange the stoichiometric coefficients.

Zn + 2 HCl \u003d ZnCl 2 + H 2

Mg + 2 HCl \u003d MgCl 2 + H 2

Determine the amounts of magnesium and zinc substances that have joined the reaction with hydrochloric acid.

ν (Mg) \u003d M (Mg) / M (Mg) \u003d 6/24 \u003d 0.25 mol

ν (zn) \u003d m (zn) / m (zn) \u003d 6.5 / 65 \u003d 0.1 mol.

From the equations of the reaction it follows that the amount of substance of metal and hydrogen is equal, i.e. ν (Mg) \u003d ν (H 2); ν (zn) \u003d ν (H 2), determine the amount of hydrogen, resulting from two reactions:

ν (H 2) \u003d ν (Mg) + ν (zn) \u003d 0.25 + 0.1 \u003d 0.35 mol.

We calculate the volume of hydrogen highlighted as a result of the reaction:

V (H 2) \u003d V M ν (H 2) \u003d 22.4 0.35 \u003d 7.84 liters.

11. When the sulfide of 2.8 l (normal conditions) is passed through an excess of copper sulfate solution (II), a sediment was formed by a mass of 11.4 g. Determine the output Reaction product.

Dano: V (H 2 S) \u003d 2.8 l; m (precipitate) \u003d 11.4 g; Well.

To find: η =?

Decision: Record the equation of reacting the reaction of hydrogen sulfide and copper sulfate (II).

H 2 S + CUSO 4 \u003d CUS ↓ + H 2 SO 4

Determine the amount of the substance of the hydrogen sulfide participating in the reaction.

ν (H 2 S) \u003d V (H 2 S) / V m \u003d 2.8 / 22.4 \u003d 0.125 mol.

It follows from the reaction equation that ν (H 2 S) \u003d ν (Cus) \u003d 0.125 mol. So you can find the theoretical mass of CUS.

m (CUS) \u003d ν (CUS) M (CUS) \u003d 0.125 96 \u003d 12

Now we determine the product output, using formula (4):

η \u003d / m (x) \u003d 11.4 100/12 \u003d 95%.

12. What weight Ammonium chloride is formed in the interaction of chloride produce weighing 7.3 g with ammonia weighing 5.1 g? What gas will remain in excess? Determine the mass of excess.

Dano: M (HCl) \u003d 7.3 g; M (NH 3) \u003d 5.1 g

To find: M (NH 4 CL) \u003d? M (excess) \u003d?

Decision: Record the reaction equation.

HCl + NH 3 \u003d NH 4 Cl

This task for "excess" and "disadvantage". We calculate the amounts of the substance of the chloride farmer and ammonia and determine which gas is in excess.

ν (HCl) \u003d M (HCl) / M (HCl) \u003d 7.3 / 36.5 \u003d 0.2 mol;

ν (NH 3) \u003d M (NH 3) / M (NH 3) \u003d 5.1 / 17 \u003d 0.3 mol.

Ammonia is in excess, so the calculation is carried out by deficiency, i.e. By chloride. It follows from the reaction equation that ν (HCl) \u003d ν (NH 4 Cl) \u003d 0.2 mol. We determine the mass of ammonium chloride.

m (NH 4 Cl) \u003d ν (NH 4 Cl) M (NH 4 Cl) \u003d 0.2 53.5 \u003d 10.7 g

We determined that ammonia is in excess (in the amount of substance an excess is 0.1 mol). We calculate the mass of excess ammonia.

m (NH 3) \u003d ν (NH 3) M (NH 3) \u003d 0.1 17 \u003d 1.7 g

13. The technical carbide of calcium weighing 20 g was treated with an excess of water, having received acetylene, with a passage of which, through an excess of bromine water, formed 1,1,2,2 -thetrabrometan weighing 86.5 g. Determine mass share CAC 2 in technical carbide.

Dano: m \u003d 20 g; M (C 2 H 2 Br 4) \u003d 86.5

To find: Ω (SAC 2) \u003d?

Decision: Record the equations of interaction of calcium carbide with water and acetylene with bromine water and arrange the stoichiometric coefficients.

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

C 2 H 2 +2 BR 2 \u003d C 2 H 2 BR 4

We find the amount of substance of the tetrabromethane.

ν (C 2 H 2 BR 4) \u003d M (C 2 H 2 Br 4) / M (C 2 H 2 Br 4) \u003d 86.5 / 346 \u003d 0.25 mol.

From the equations of the reaction it follows that ν (C 2 H 2 br 4) \u003d ν (C 2 H 2) \u003d ν (SAC 2) \u003d 0.25 mol. From here we can find a mass of pure calcium carbide (without impurities).

m (SAC 2) \u003d ν (SAC 2) M (SAC 2) \u003d 0.25 64 \u003d 16

We determine the mass fraction of SC 2 in technical carbide.

Ω (SAC 2) \u003d M (SAC 2) / M \u003d 16/20 \u003d 0.8 \u003d 80%.

Solutions. Mass fraction of the component of the solution

14. In benzene, 170 ml was dissolved in a mass of 1.8 g. The density of benzene is 0.88 g / ml. Determine mass share sulfur in solution.

Dano: V (C 6 H 6) \u003d 170 ml; M (S) \u003d 1.8 g; ρ (C 6 C 6) \u003d 0.88 g / ml.

To find: Ω (s) \u003d?

Decision: To find the mass fraction of sulfur in the solution it is necessary to calculate the mass of the solution. We determine the mass of benzene.

m (C 6 C 6) \u003d ρ (C 6 C 6) V (C 6 H 6) \u003d 0.88 170 \u003d 149.6

We find a total mass of the solution.

m (p-ra) \u003d m (C 6 C 6) + m (S) \u003d 149,6 + 1.8 \u003d 151.4 g.

Calculate the mass fraction of sulfur.

ω (s) \u003d m (s) / m \u003d 1.8 / 151,4 \u003d 0.0119 \u003d 1.19%.

15. In water weighing 40 g dissolved iron cuneition FESO 4 7H 2 O weighing 3.5 g. Determine mass fraction of iron sulfate (II) In the resulting solution.

Dano: m (H 2 O) \u003d 40 g; M (FESO 4 7H 2 O) \u003d 3.5 g

To find: Ω (FESO 4) \u003d?

Decision: We will find a mass FESO 4 contained in FESO 4 7H 2 O. For this, we calculate the amount of substance FESO 4 7H 2 O.

ν (FESO 4 7H 2 O) \u003d M (FESO 4 7H 2 O) / M (FESO 4 7H 2 O) \u003d 3.5 / 278 \u003d 0,0125mol

It follows from the formula of the iron mood that ν (feso 4) \u003d ν (FESO 4 7H 2 O) \u003d 0.0125 mol. Let's calculate the mass of FESO 4:

m (FESO 4) \u003d ν (FESO 4) M (FESO 4) \u003d 0.0125 152 \u003d 1.91

Considering that the mass of the solution is consisted of the mass of iron vitriol (3.5 g) and the mass of water (40 g), we calculate the mass fraction of iron sulfate in solution.

Ω (FESO 4) \u003d M (FESO 4) / M \u003d 1.91 / 43,5 \u003d 0.044 \u003d 4.4%.

Tasks for self solutions

At 50 g of iodide methyl in hexane, the metallic sodium was carried out, while 1.12 liters of gas measured under normal conditions was separated. Determine the mass fraction of iodide methyl in solution. Answer: 28,4%.
Some alcohol was oxidized, while a monosular carboxylic acid was formed. When burning 13.2 g of this acid, carbon dioxide was obtained, for the complete neutralization of which 192 ml of a solution of KOV with a mass fraction of 28% was required. Kon's solution density is 1.25 g / ml. Determine the alcohol formula. Answer: Butanol.
The gas obtained by the interaction of 9.52 g of copper with 50 ml of 81% solution of nitric acid, 1.45 g / ml density was passed through 150 ml of 20% NaOH solution with a density of 1.22 g / ml. Determine the mass fractions of solutes. Answer: 12.5% \u200b\u200bNaOH; 6.48% Nano 3; 5.26% Nano 2.
Determine the volume of the surrounding gases in the explosion of 10 g of nitroglycerin. Answer: 7.15 liters.
A sample of organic matter weighing 4.3 g burned in oxygen. Reaction products are carbon monoxide (IV) with a volume of 6.72 liters (normal conditions) and water weighing 6.3 g. The density of the steam of the starting material according to hydrogen is 43. Determine the formula of substance. Answer: C 6 H 14.

Quite often, schoolchildren and students have to be compiled. ionic equations of reactions. In particular, this particular topic is devoted to the task 31 offered to the exam in chemistry. In this article, we will discuss in detail the algorithm for writing short and complete ion equations, we will analyze many examples of different levels of complexity.

Why do Ionic equations need

Let me remind you that when dissolving many substances in water (and not only in water!) The dissociation process is occurred - the substances are disintegrated by ions. For example, HCl molecules in the aqueous medium are dissociated into hydrogen cations (H +, more precisely, H 3 O +) and chlorine anions (CL -). Sodium bromide (NaBr) is in an aqueous solution not in the form of molecules, but in the form of hydrated Na + and Br ions - (by the way, ions are also present in solid sodium bromide).

Recalling "ordinary" (molecular) equations, we do not take into account that no molecules come into the reaction, but ions. Here, for example, what does the reaction equation looks like between hydrochloric acid and sodium hydroxide:

HCl + NaOH \u003d NaCl + H 2 O. (1)

Of course, this scheme does not quite rightly describe the process. As we have already said, there are practically no HCl molecules in aqueous solution, and there are H + and CL ions. There are also cases with NaOH. It would be more correct to write the following:

H + + Cl - + Na + + OH - \u003d Na + + Cl - + H 2 O. (2)

That's what it is full ion equation. Instead of "virtual" molecules, we see particles that are actually present in the solution (cations and anions). Let's not stop on the question why H 2 O we recorded in molecular form. A little later, this will be explained. As you can see, there is nothing complicated: we replaced molecules by ions, which are formed during their dissociation.

However, even the complete ion equation is not immaculate. Indeed, we take a closer look: in the left, and in the right parts of equation (2) there are identical particles - Na + cations and clock anions. In the process of reaction, these ions do not change. Why are they generally needed? Remove them and get brief ion equation:

H + + OH - \u003d H 2 O. (3)

As you can see, everything comes down to the interaction of H + and OH ions - to the formation of water (neutralization reaction).

All, complete and short ion equations are recorded. If we solved the task of 31 on the exam in chemistry, you would get the maximum rating for it - 2 points.

So, once again about terminology:

HCl + NaOH \u003d NaCl + H 2 O is a molecular equation ("ordinary" equation, schematically reflecting the essence of the reaction);
H + + Cl - + Na + + OH - \u003d Na + + Cl - + H 2 O is the total ion equation (the real particles in the solution are visible);
H + + OH - \u003d H 2 O is a brief ion equation (we removed the whole "garbage" - particles that do not participate in the process).

Algorithm for writing ion equations

Make a molecular equation of reaction.
All particles dissociating in a solution are tangible, written in the form of ions; Substances that are not prone to dissociation, we leave "in the form of molecules."
We remove from the two parts of the equation t. N. Observer ions, i.e., particles that do not participate in the process.
We check the coefficients and we obtain the final response - a brief ion equation.

Example 1.. Make up complete and short ionic equations that describe the interaction of aqueous solutions of barium chloride and sodium sulfate.

Decision. We will act in accordance with the proposed algorithm. Make a molecular equation first. Chloride barium and sodium sulfate are two salts. Look into the section of the reference book "Properties of inorganic connections". We see that salts can interact with each other if a precipitate is formed during the reaction. Check:

Exercise 2. Complete equations of the following reactions:

KOH + H 2 SO 4 \u003d
H 3 PO 4 + Na 2 O \u003d
BA (OH) 2 + CO 2 \u003d
NaOH + CUBR 2 \u003d
K 2 S + HG (NO 3) 2 \u003d
Zn + FECL 2 \u003d

Exercise 3.. Write molecular equations of reactions (in aqueous solution) between: a) sodium carbonate and nitric acid, b) nickel (II) chloride and sodium hydroxide, c) orthophosphoric acid and calcium hydroxide, d) silver nitrate and potassium chloride, d) phosphorus oxide (V) and potassium hydroxide.

I sincerely hope that you have no problems with the performance of these three tasks. If this is not the case, it is necessary to return to the topic "Chemical properties of the main classes of inorganic connections."

How to turn the molecular equation into the full ion equation

Begins the most interesting. We must understand which substances should be recorded in the form of ions, and which - to leave in the "molecular form". We will have to remember the following.

In the form of ions write down:

soluble salts (emphasize only salts are well soluble in water);
alkali (remind you that alkalis are called soluble bases, but not NH 4 OH);
strong acids (H 2 SO 4, HNO 3, HCl, HBr, Hi, HCLO 4, HCLO 3, H 2 SEO 4, ...).

As you can see, remember this list is completely simple: it includes strong acids and bases and all soluble salts. By the way, especially vigilant young chemists, which may be outraged that strong electrolytes (insoluble salts) did not enter this list, I can report the following: Inclusion of insoluble salts in this list does not reject the fact that they are strong electrolytes.

All other substances must be present in the ionic equations in the form of molecules. Those demanding readers who are not satisfied with the blurry term "all other substances", and which, following the example of the hero of the famous film, require "announce the full list" give the following information.

In the form of molecules write:

all insoluble salts;
all weak bases (including insoluble hydroxides, NH 4 OH and substances similar to it);
all weak acids (H 2 CO 3, HNO 2, H 2 S, H 2 SiO 3, HCN, HCLO, almost all organic acids ...);
in general, all weak electrolytes (including water !!!);
oxides (all types);
all gaseous compounds (in particular, H 2, CO 2, SO 2, H 2 S, CO);
simple substances (metals and non-metals);
almost all organic compounds (exception - water soluble salts of organic acids).

UV-F seems to have forgotten anything! Although it is easier to, in my opinion, still remember the list of N 1. From a fundamentally important thing in the list N 2, I will once again note the water.

Let's train!

Example 2.. Make a complete ion equation that describe the interaction of copper (II) hydroxide and hydrochloric acid.

Decision. Let's start naturally with a molecular equation. Copper hydroxide (II) is an insoluble base. All insoluble bases react with severe acids to form salt and water:

Cu (OH) 2 + 2HCl \u003d CUCL 2 + 2H 2 O.

And now we find out what substances to record in the form of ions, and which - in the form of molecules. We will be helped by the above lists. Copper hydroxide (II) is an insoluble base (see solubility table), weak electrolyte. Insoluble bases are recorded in molecular form. HCl - severe acid, in the solution almost completely dissociates to ions. CUCL 2 - soluble salt. We write in ion form. Water - only in the form of molecules! We get a complete ion equation:

Cu (OH) 2 + 2H + + 2Cl - \u003d Cu 2+ + 2Cl - + 2H 2 O.

Example 3.. Make a complete ionic equation of carbon dioxide reaction with an aqueous solution of NaOH.

Decision. Carbon dioxide is a typical acidic oxide, NaOH - alkali. In the interaction of acidic oxides with aqueous solutions, alkalis and water are formed. We make a molecular equation of reaction (do not forget, by the way, about the coefficients):

CO 2 + 2NAOH \u003d Na 2 CO 3 + H 2 O.

CO 2 - oxide, gaseous compound; We maintain a molecular form. NaOH - a strong base (alkali); We write in the form of ions. Na 2 CO 3 - soluble salt; We write in the form of ions. Water - weak electrolyte, practically does not dissociate; We leave in molecular form. We get the following:

CO 2 + 2NA + + 2OH - \u003d Na 2+ + CO 3 2- + H 2 O.

Example 4.. Sulfide sodium in an aqueous solution reacts with zinc chloride to form a precipitate. Make a complete ion equation of this reaction.

Decision. Sodium sulphide and zinc chloride are salts. With the interaction of these salts, zinc sulfide precipitate falls:

Na 2 S + ZnCl 2 \u003d ZNS ↓ + 2NACL.

I will immediately write a complete ion equation, and you independently analyze it:

2NA + + S 2- + Zn 2+ + 2Cl - \u003d zns ↓ + 2NA + + 2CL -.

I offer you several tasks for independent work and a small test.

Exercise 4.. Make molecular and complete ion equations of the following reactions:

NaOH + HNO 3 \u003d
H 2 SO 4 + MGO \u003d
Ca (NO 3) 2 + Na 3 PO 4 \u003d
COBR 2 + CA (OH) 2 \u003d

Exercise 5.. Write complete ion equations describing the interaction: a) nitrogen oxide (V) with an aqueous solution of barium hydroxide, b) cesium hydroxide solution with hydrogenic acid, c) aqueous solutions of copper sulfate and potassium sulfide, d) calcium hydroxide and an aqueous iron nitrate solution ( Iii).