Basic level tasks.

1. Calculate the mass of barium sulfate that precipitates when pouring solutions, one of which contains 522 g of barium nitrate, and the second -500 g of potassium sulfate.

3. Calculate the mass of the precipitate in g obtained by reacting 20 g of sodium hydroxide with 32 g of honey (II) sulfate.

4. Find the mass of iron in kg obtained by the aluminothermic method from 400 kg of aluminum and 800 kg of iron oxide (III).

5. Into the reactor for catalytic oxidation ammonia, 180 liters of ammonia and 250 liters of oxygen were supplied. What volume of gas in liters is formed?

6. 5.6 g of iron was burned in 5.6 liters of chlorine (n.a.). Calculate the mass in g of the resulting product.

7. Calculate the volume of ammonia (l, n.a.) released during heating of a mixture consisting of 7.4 g of calcium hydroxide and 30 g of ammonium sulfate.

8. 40 g of copper(II) oxide was treated with a solution sulfuric acid containing 49 g of anhydrous substance. Find the mass of salt formed.

9. A solution containing 53 g of sodium carbonate was treated with a solution containing 49 g of sulfuric acid. Find the mass of salt formed.

10. 47 g of potassium oxide were treated with a solution containing 40 g of nitric acid. Find the mass of potassium nitrate formed.

11. Determine how many grams of ammonia can be obtained by heating a mixture of 20 g of ammonium chloride with 20 g of calcium hydroxide.

Combined tasks.

12. A 5% solution of phenol weighing 94 g was affected by 50 g of bromine. What is the mass of the formed precipitate in g?

13. Iron (II) sulfide weighing 17.6 g was treated with 7.3% hydrochloric acid weighing 150 g. Calculate the mass of the released gas.

15. 24 g of metallic magnesium was affected by 100 g of a 30% hydrochloric acid solution. Find the mass of magnesium chloride formed.

16. 40 g of alumina acted on 200 g of a 10% solution of sulfuric acid. Find the mass of water formed.

Tasks are difficult.

17. Mix 52.6 ml of 25% aluminum sulfate solution (pl. 1.3 g/ml) and 135 ml of 14% barium chloride solution (1.1 g/ml). Calculate the mass of precipitate formed.

18. In the interaction of 93 ml of a 10% solution of phosphoric acid (density 1.05 g / ml) with 50 g of a 30% solution of calcium hydroxide, ... g of a precipitate was formed.

19. K hydrochloric acid, containing 0.25 mol of hydrogen chloride, added 6 g of a 50% sodium hydroxide solution. What reaction of the medium (acidic, neutral, alkaline) will the resulting solution have?

20. 800 g of marble containing 2% non-carbonate impurities were treated with a 20% solution of nitric acid weighing 1260 g. Calculate the volume of carbon monoxide (II) in l obtained from this reaction (n.a.).

21. Chlorine, released during the interaction of 43.5 g of manganese (IV) oxide with a 36% solution of HCl with a volume of 500 ml and a density of 1.18 g / ml, was passed through a hot solution of potassium hydroxide weighing 600 g, with a mass fraction of 28% . Determine mass fraction potassium chlorate in the resulting solution.

22. Mixed 500 g of a 20% solution of copper (II) nitrate and 200 g of a 40% solution of potassium hydroxide. The precipitate formed was separated and calcined. How many grams of copper will be obtained by reducing the substance obtained by calcination with hydrogen?

Problem 518.
A solution containing 0.53 g of sodium carbonate in 200 g of water crystallizes at -0.13°C. Calculate the apparent degree of dissociation of the salt.
Decision:
M(Na 2 CO 3 ) = 106 g/mol.
Now we determine the decrease in the crystallization temperature t (cryst) of the solution without taking into account the dissociation of alkali (the cryoscopic constant for water is 1.86) according to the formula:

Here m 1 is the mass of the dissolved substance, m 2 is the mass of the solution, M - molar mass solute, K is the cryoscopic constant.

Comparing the found value of t (theoret) with the experimentally obtained value of t (expert), we calculate the isotonic coefficient using the formula:

Answer: 0,9.

Problem 519.
In equal amounts of water, 0.5 mol of sugar is dissolved in one case, and 0.2 mol of CaCI 2 in the other. The crystallization temperatures of both solutions are the same. Determine the apparent degree of dissociation CaCl 2 .
Decision:
From the equation t z \u003d KC M, where K is the cryoscopic constant of the solvent, which for water has a value of 1.86; C M is the molar concentration of the solution; tz - lowering the freezing point of the solution, we get:

t s (sugar) = 1.86 . 0.5 \u003d 0.930 0 C;
t s (CaCl 2) = 1.86. 0.2 \u003d 0.372 0 C

Now, taking into account that the crystallization temperatures of both solutions are the same and, comparing the found values ​​of t for sugar and calcium chloride, we find the isotonic coefficient (i), we get:

Calculate the apparent degree of dissociation () of the salt from the ratio:

Answer: 0,9.

Task 520.
At 100°C, the vapor pressure of a solution containing 0.05 mol of sodium sulfate in 450 g of water is 100.8 kPa (756.2 mmHg). Determine the apparent degree of dissociation of Na 2 SO 4 .
Decision:
The decrease in vapor pressure over the solution is calculated by the equation:

where P 0 is the pressure of saturated vapor over the solvent, 101.325 kPa; P is the pressure of saturated vapor over the solution, 100.800 kPa; N is the mole fraction of the dissolved substance; n 1, n 2 - the amount of the dissolved substance and solution. Find the amount of water in the solution:

We find the experimental value:

P (expert) \u003d P0 - P \u003d 101.325 - 100.800 \u003d 0.525 Pa.

Calculate the isotonic coefficient from the ratio:

Calculate the apparent degree of dissociation () of the salt from the ratio:

Here K is total number ions formed during the dissociation of the electrolyte.

Answer: 0,8.

Problem 521.
1 liter of 0.01M acetic acid solution contains 6.26 . 10 -21 of its molecules and ions. Determine the apparent degree of dissociation of acetic acid.
Decision:
1 mole of any substance contains 6.02 . 10-23 molecules. Let's determine the number of molecules contained in 0.01 mole of acetic acid from the proportion:

1: 6,02 . 10 -23 = 0.01:x; x = (0.01 . 6,02 . 10 -23)/1 = 6,02 . 10 -21

The isotonic coefficient is calculated from a comparison of the theoretical and practical quantities of acetic acid molecules:

Calculate the apparent degree of dissociation () of the acid from the ratio:

here K is the total number of ions formed during the dissociation of the electrolyte.

Ability to decide chemical problems- an important component of knowledge in the subject. According to state standard Chemistry education students graduating from high school should be able to solve more than a dozen types of standard problems. Among them are tasks for "surplus-deficiency".
I offer my own version of the presentation of material for solving such problems in the course of chemistry of the 9th grade.
I take 2–2.5 lessons to study this topic, depending on the level of abilities of the students in the class. Acquaintance with the algorithm for solving problems of this type occurs as part of the study of the topic “Theory electrolytic dissociation". However, if the class is strong, then as part of the experiment, this type of problem is sometimes studied at the end of the 8th grade in the chapter “Halogens”, and the freed time can be spent on studying organic chemistry in 9th grade.
In the first lesson, I analyze two types of problems for "excess-lack":
one of the two reacting substances is given in excess;
both reacted substances are spent on interaction with each other without a trace, i.e., they are given in stoichiometric quantities.
Two or three tasks similar to those studied in the lesson are necessarily offered as homework.
In the second lesson, I consolidate and deepen the studied material, introduce the concepts of “percentage concentration of solutions of substances” that have reacted, “density of solutions”. In addition, I complicate the tasks by introducing the “percentage of impurities in the original substance”, etc. This technique allows you to repeat the elements of already studied material, saving time. At the end of the second lesson of studying the topic or at the beginning of the third, I conduct a small independent work to consolidate the studied material, including one or two tasks, and independent work It is offered in three levels of difficulty, depending on the abilities of the student.

Lesson 1

Solving problems on "surplus-lack"

Goals.

• teach the algorithm for solving problems of a new type;
• consolidate oral counting skills;
• repeat the rules for calculating the relative molecular masses of substances;
• to fix the rules for the competent design of the conditions of the problem;
• to form the skills of chemical thinking, logic, and also contribute to the education of a harmonious, comprehensively developed personality.

DURING THE CLASSES

Consider the option when one of the reacted substances is given in excess, the other is in short supply.
When solving chemical problems, one should not forget about the rules for their competent design according to the scheme: given, find, solution, answer.

TASK 1. 47 g of potassium oxide were treated with a solution containing 40 g of nitric acid. Find the mass of potassium nitrate formed.

Given :

m(K 2 O) \u003d 47 g,
m(HNO 3) = 40 g.

To find :

m(KNO 3).

Decision

M r(K 2 O) = 2 A r(K)+1 A r(O) = 239 + 116 = 94,

M r(HNO 3) = 1 A r(H)+1 A r(N) + 3 A r(O) = 1 1 + 1 14 + 3 16 = 63,

M r(KNO 3) = 1 A r(K)+1 A r(N) + 3 A r(O) = 1 39 + 1 14 + 3 16 = 101.

For the convenience of calculating x 1 let's take the mass of HNO 3 and find which of the substances that have entered into the reaction is given in excess, which is in short supply.

47/94 = x 1/126,x 1= 63 g.

Hence, Nitric acid given in short supply, because according to the condition it is 40 g, and according to the calculation, 63 g is needed, therefore, the calculation is carried out according to HNO 3:

40/126 = X/202, X= 64 g.

Answer . m(KNO 3) = 64 g.

TASK 2. 24 g of metallic magnesium were affected by 100 g of a 30% hydrochloric acid solution. Find the mass of magnesium chloride formed.

Given :

m(Mg) = 24 g,
m(solution HCl) = 100 g,
(HCl) = 30%.

To find :

m(MgCl2).

Decision

Calculate the relative molecular weights substances we are interested in:

M r(HCl) = 1 A r(H)+1 A r(Cl) \u003d 1 + 35.5 \u003d 36.5,

M r(MgCl 2) \u003d 1 A r(Mg) + 2 A r(Cl) \u003d 24 + 2 35.5 \u003d 95.

For the convenience of calculating x 1 let us take the mass of hydrochloric acid and find which of the substances that have entered into the reaction is given in excess, which is in short supply.

24/24 = x 1/73, x 1= 73 g.

It can be seen from the calculation that hydrochloric acid is given in short supply, since, according to the condition of the problem, it is given 30 g, and 73 g is required for the reaction. Therefore, we carry out the calculation for hydrochloric acid:

30/73 = X/95, X= 39 g.

Answer . m(MgCl 2) = 39 g.

Consider the option when both substances that have entered into the reaction are given in stoichiometric quantities, i.e., they react with each other without residue.

TASK 1. 64 g of sulfur acted on 36 g of aluminum. Find the mass of aluminum sulfide formed.

Given :

m(Al) = 36 g,
m(S) = 64 g.

To find :

m(Al 2 S 3).

Decision

We take the mass of Al as x 1 and we will find which of the substances that have entered into the reaction is given in excess, which is in short supply.

x 1/54 = 64/96, x 1= 36 g.

AT this case the substances that entered into the reaction are taken in stoichiometric quantities, so the calculation can be carried out according to any of them:

64/96 = X/150, X= 100 g.

Answer . m(Al 2 S 3) = 100 g.

TASK 2. A solution containing 53 g of sodium carbonate was treated with a solution containing 49 g of sulfuric acid. Find the mass of salt formed.

Given :

m(Na 2 CO 3) \u003d 53 g,
m(H 2 SO 4) = 49 g.

To find :

m(Na2SO4).

Decision

Calculate the relative molecular masses of the substances of interest to us:

M r ( Na 2 CO 3) = 2 A r(Na) + 1 A r(C)+3 A r(O) = 2 23 + 1 12 + 3 16 = 106.

M r(H 2 SO 4) \u003d 2 A r(H)+1 A r(S)+4 A r(O) = 2 1 + 1 32 + 4 16 = 98.

M r(Na 2 SO 4) \u003d 2 A r(Na) + 1 A r(S)+4 A r(O) = 2 23 + 1 32 + 4 16 = 142.

Let's take for x 1 a mass of sulfuric acid, in order to find out which substance is given in excess, which is in short supply.

53/106 = x 1/98, x 1= 49 g.

In this case, both substances are taken in stoichiometric quantities, so the calculation can be carried out using any of them:

49/98 = X/142, X= 71 g.

Answer . m(Na 2 SO 4) \u003d 71 g.

However, the teacher, when selecting problems to solve in the classroom, must remember that in some cases (for example, if acid or acid oxide given in excess) the solution of the problem is not limited to the calculation of two proportions, since the reaction will proceed further with the formation of an acid salt. This will increase the complexity of the material. In the first lessons, when solving problems of this type, I do not include in the material problems for the passage of reactions with the formation of acidic or basic salts.

Homework

TASK 1. 40 g of alumina acted on 200 g of a 10% sulfuric acid solution. Find the mass of water formed.

Given :

M(solution H 2 SO 4) = 200 g,
(H 2 SO 4) = 10%,
m(Al 2 O 3) = 40 g.

To find:

m(H 2 O).

Decision

Calculate the relative molecular masses of the substances of interest to us:

M r(Al 2 O 3) \u003d 2 A r(Al) + 3 A r(O) = 227 + 316 = 102,

M r(H 2 SO 4) \u003d 2 A r(H)+1 A r(S)+4 A r(O) = 2 1 + 1 32 + 4 16 = 98,

M r(H 2 O) = 2 A r(H)+1 A r(O) = 2 1 + 1 16 = 18.

m(H 2 SO 4) \u003d 200 10/100 \u003d 20 g.

Let us find which of the substances that have entered into the reaction is given in excess, and which is in deficiency.

x 1/102 = 20/294, x 1= 6.94 g.

It can be seen from the calculation that Al 2 O 3 is given in excess, therefore, the calculation is carried out for acid:

20/294 = X/54, X= 3.67 g.

Answer .m(H 2 O) \u003d 3.67 g.

TASK 2. 40 g of copper(II) oxide was treated with a sulfuric acid solution containing 49 g of an anhydrous substance. Find the mass of salt formed.

Given :

m(CuO) = 40 g,
m(H 2 SO 4) = 49 g.

To find :

M(CuSO 4).

Decision

Let us find which of the substances that have entered into the reaction is given in excess, and which is in short supply.

x 1/80 = 49/98, x 1= 40 g.

According to the equation of this reaction, the substances are taken in stoichiometric quantities, so the calculation can be carried out using any of them:

40/80 = X/160, X= 80 g.

Answer . m(CuSO 4) = 80 g.

…nee-this is dangerous!

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Smoke! Smoke! Smoke!If you really want, For the heart to work tired, For the teeth to turn yellow,Eyes stop looking...To make you short of breath And a bear would overtake you, To make your lungs blacker, To take you to the hospital,And from there, a direct path to the tomb.

Smoke! Smoke! Smoke if you don't want to live If you want the heart

no longer beat in the side,Stones to strum in the kidneysSo that your lungs crackle. Health and mind were gone. You smoke more... smoke....

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faded faster, To cough more often than others, To make the voice faster

hoarse, So that the nerves more often

naughty,

So that the brain is worse than a fraction

decided

To make the cells faster

shared.

And quickly into cancer

have turned.

Think again, brothers!

Grade 9

Public lesson:Solving problems on

« excessflaw".

Evdokimova T.V.

Public lesson: Solving problems on

"excess - deficiency".

Goals:

To teach the algorithm for solving problems of a new type;

Strengthen oral counting skills;

Repeat the rules for calculating the relative molecular masses of substances;

To consolidate the rules of literacy for the formulation of the conditions of the problem;

To form the skills of chemical thinking, logic, and also contribute to the education of a harmonious, comprehensively developed personality.

During the classes:

Consider the option when one of the reacted substances is given in excess, the other - in

lack.

When solving chemical problems, one should not forget about the rules for their competent design according to the scheme:

given, find, solution, answer.

Task #1

47 g of potassium oxide were treated with a solution containing 40 g of nitric acid. Findmass of potassium nitrate formed.

Given: m (K 2 O) \u003d 47 g,

m(NJ) 3) =40g

Find: m (KNO 3 )

Decision:

Calculate the relative molecular masses of the substances of interest to us:Mg (K 2 O) \u003d 2Ar (K) + 1Ak (O) \u003d 2-39 + 1- 16 \u003d 94, Mr (HNO 3) \u003d lAr (H) + lAr (N) + 3 Ar (O) \u003d l 1+ 1 14+ 3 16 \u003d 63? Mr(KNO 3)=lAr(K)+lAr(N)+3Ar(O)=l 39+1 14 +3 16=101 K 2 + O + 2HNO3 \u003d 2KNO3 + H 2 O

y1mol 2mol2mol

m94 g/mol63 g/mol101 mol
m94g 126g 202g

For the convenience of calculating xi we take the mass of HNO 3 and find which of the substances that entered th

reaction, given in excess, which is in short supply.

47 g K 2 O-Xi g HNO 3 94 g K 2 O-126 rHNO 3 47/94= Xi /126, Xi = 63 g.

Consequently, nitric acid is given in short supply, tk. according to the condition of its 40 g, and according to the calculation it is necessary

63 g, so we calculate according to HNO3:

40 g HNO 3 - x g KNO 3 126 rHNO 3 -202 rKNO 3 40/126= x/202, x = 64g. Answer: m (KNO 3) \u003d 64 g.

Task: 2.24 g of metallic magnesium was affected by 100 g of a 30% solution of seededacids. Find the mass of magnesium chloride formed.

Given:

m (Mg) \u003d 24 g, m (p - pHCI) \u003d 100 r, w (HCI) \u003d 30% /

Find: M (MGCI)

Decision. Calculate the relative molecular masses of the substances of interest to us:

Mg (HC1) \u003d 1 Ag (H) + 1 Ag (C1) \u003d 1 + 35.5 \u003d 36.5. to Mr(MGCI) \u003d l Ar (MG) + 2Ar (CI) \u003d 24 \u003d 235.5 \u003d 95.

24 30(xr)xr

Mg + 2HCI === MgCI 2 + H 2

v 1mol2 mol 1mol

M 24 g/mol 36.5 g/mol 95 g/mol
m 24 g 73 g 95 g

For the convenience of calculating xi let's take the mass of hydrochloric acid and find which substances,reacted is given in excess, which is in short supply.

24g Mg --- x 1 rHCI

23r Mg --- 73r HCI

24/24 \u003d x 1 / 73x 1 \u003d 73 r

It can be seen from the calculation that hydrochloric acid is given in deficiency, because according to the condition of the problem, it is given 30 g, and 73 g is required for the reaction. Therefore, we carry out the calculation for hydrochloric acid:

30 g HCI - xrMgCI 2

73 g HCI - 95 g MgCI 2

30/73 \u003d x / 95, x \u003d 39 g.

Answer: m (MgCI 2 )=39 u /

Let us consider the option when both reacting substances are given in stoichiometricquantities, i.e., they react with each other without a trace.

Task 1.64 g of sulfur acted on 36 g of aluminum. Find the mass of sulfide formed aluminum.

Given:

M (A1)=36g
M (S) = 64 g/______________________________________________________________________________

Find: M (AI 2 S 3 ).

Decision.

36g(x1g)64g x rt

2 AI +3 S == AI 2 S 3 /

v 2 mol3 mol1 mol

М 27 g/mol32g/mol150 g/mol

t54g 96 g 150 g *

Let's take the mass AI as xi and find which of the substances that have reacted is given in excess,which one is in short supply.

X ir AI-64rS

54r AI-96g S

xi /54 = 64/96, xj =36 r .

In this case, the reactants are taken in stoichiometric quantities,Therefore, the calculation can be carried out on any of them:

64 g S- x r AI 2 S 3

96 g S-150rAI 2 S 3\

64/96=x/150, x=100g. Answer: m (AI 2 S 3) \u003d 100 g.

Task 2.A solution containing 53 g of sodium carbonate was treated with a solutioncontaining 49 g of sulfuric acid. Find the mass of salt formed.

Given: m (Na 2 CO 3) \u003d 53 g

m ( H 2 SO 4 )=49 r_______________________ _______________________________________________

To find:

M (Na 2 SO 4 )

Decision. Calculate the relative molecular weights of the substances of interest to us:

Mr(Na 2 CO 3) \u003d 2 Ag (Na) + 1 Ag © + 3Ar (O) \u003d 223 + 112 + 3 16 \u003d 106

Mr (H 2 SO 4) \u003d 2Ar (H) + lAr (S) + 4Ar (O) \u003d 21 + 132 + 416 = 98 Mr(Na 2 SO 4) \u003d 2Ar (Na) + 1 Ar (S) + 4 Ar (O) \u003d 223 + 132 + 416 \u003d 142 /

53r 49r(xlr)xr

Na 2 CO 3 + H 2 SO 4 \u003d Na 2 SO 4 + H 2 O +CO Z

v 1mol 1 mol 1 mol