Learning to solve problems on a mixture of organic substances

Generalization of the experience of teaching organic chemistry in specialized biological and chemical classes

One of the main criteria for mastering chemistry as an academic discipline is the ability of students to solve computational and qualitative problems. In the process of teaching in specialized classes with in-depth study of chemistry, this is of particular relevance, since all entrance exams in chemistry offer tasks of an increased level of complexity. The greatest difficulty in the study of organic chemistry is caused by the tasks of determining the quantitative composition of a multicomponent mixture of substances, the qualitative recognition of a mixture of substances, and the separation of mixtures. This is due to the fact that in order to solve such problems, it is necessary to deeply understand the chemical properties of the substances under study, be able to analyze, compare the properties of substances of different classes, and also have a good mathematical background. A very important point in learning is the generalization of information about the classes of organic substances. Let us consider the methodological methods of developing the ability of students to solve problems on a mixture of organic compounds.

hydrocarbons

• Where is what substance (qualitative composition)?
• How much substance is in the solution (quantitative composition)?
• How to separate the mixture?

STAGE 1. Summarizing knowledge about the chemical properties of hydrocarbons using a table(Table 1).

STEP 2. Solving quality problems.

Task 1. The gas mixture contains ethane, ethylene and acetylene. How to prove the presence of each of the gases in a given mixture? Write the equations for the necessary reactions.

Decision

Of the remaining gases, only ethylene will decolorize bromine water:

C 2 H 4 + Br 2 \u003d C 2 H 4 Br 2.

The third gas, ethane, burns:

2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O.

Table 1

Chemical properties of hydrocarbons

Reagent	Representatives of hydrocarbons
	CH 3 CH 3 ethane	CH 2 \u003d CH 2 ethylene	CHCH acetylene	C 6 H 6 benzene	C 6 H 5 CH 3 toluene	C 6 H 5 CH \u003d CH 2 styrene	C 6 H 10 cyclohexene
Br 2 (water)	–	+	+	–	–	+	+
KMnO 4	–	+	+	–	+	+	+
Ag2O (solution in NH 3 aq.)	–	–	+	–	–	–	–
Na	–	–	+	–	–	–	–
O2	+	+	+	+	+	+	+

Task 2. Isolate in pure form the components of the mixture consisting of acetylene, propene and propane. Write the equations for the necessary reactions.

Decision

When the mixture is passed through an ammonia solution of silver oxide, only acetylene is absorbed:

C 2 H 2 + Ag 2 O \u003d C 2 Ag 2 + HOH.

To regenerate acetylene, the resulting silver acetylenide is treated with hydrochloric acid:

C 2 Ag 2 + 2HCl \u003d C 2 H 2 + 2AgCl.

When the remaining gases are passed through bromine water, propene will be absorbed:

C 3 H 6 + Br 2 \u003d C 3 H 6 Br 2.

For propene regeneration, the resulting dibromopropane is treated with zinc dust:

C 3 H 6 Br 2 + Zn \u003d C 3 H 6 + ZnBr 2.

STAGE 3. Solution of calculation problems.

Task 3. It is known that 1.12 l (n.o.) of a mixture of acetylene with ethylene in the dark completely binds with 3.82 ml of bromine (= 3.14 g/ml). How many times will the volume of the mixture decrease after passing it through an ammonia solution of silver oxide?

Decision

Both components of the mixture react with bromine. Let's compose the reaction equations:

C 2 H 4 + Br 2 \u003d C 2 H 4 Br 2,

C 2 H 2 + 2Br 2 \u003d C 2 H 2 Br 4.

Let us denote the amount of ethylene substance through X mol, and the amount of acetylene substance through
y mol. It can be seen from the chemical equations that the amount of the reacting bromine substance will be in the first case X mole, and in the second - 2 y mol. Amount of gas mixture substance:

= V/V M \u003d 1.12 / 22.4 \u003d 0.05 mol,

and the amount of bromine substance:

(Br2) = V/M\u003d 3.82 3.14 / 160 \u003d 0.075 mol.

Let us compose a system of equations with two unknowns:

Solving the system, we get that the amount of ethylene substance in the mixture is equal to the amount of acetylene substance (0.025 mol each). Only acetylene reacts with an ammonia solution of silver, therefore, when the gas mixture is passed through a solution of Ag 2 O, the volume of the gas will decrease exactly by a factor of two.

Task 4. The gas released during the combustion of a mixture of benzene and cyclohexene was passed through an excess of barite water. This gave 35.5 g of sediment. Find the percentage composition of the initial mixture if the same amount of it can discolor 50 g of a solution of bromine in carbon tetrachloride with a mass fraction of bromine of 3.2%.

Decision

C 6 H 10 + Br 2 \u003d C 6 H 10 Br 2.

The amount of cyclohexene substance is equal to the amount of bromine substance:

(Br2) = m/M= 0.032 50/160 = 0.01 mol.

The mass of cyclohexene is 0.82 g.

Let us write the equations for the reactions of hydrocarbon combustion:

C 6 H 6 + 7.5O 2 \u003d 6CO 2 + 3H 2 O,

C 6 H 10 + 8.5O 2 \u003d 6CO 2 + 5H 2 O.

0.01 mol of cyclohexene forms 0.06 mol of carbon dioxide when burned. The released carbon dioxide forms a precipitate with barite water according to the equation:

CO 2 + Ba (OH) 2 \u003d BaCO 3 + H 2 O.

The amount of substance of the sediment of barium carbonate (BaCO 3) \u003d m/M\u003d 35.5 / 197 \u003d 0.18 mol is equal to the amount of substance of all carbon dioxide.

The amount of carbon dioxide substance formed during the combustion of benzene is:

0.18 - 0.06 \u003d 0.12 mol.

Using the benzene combustion reaction equation, we calculate the amount of benzene substance - 0.02 mol. The mass of benzene is 1.56 g.

Weight of the whole mixture:

0.82 + 1.56 = 2.38 g

The mass fractions of benzene and cyclohexene are 65.5% and 34.5%, respectively.

Oxygen-containing
organic compounds

Solving problems on mixtures in the topic "Oxygen-containing organic compounds" occurs in a similar way.

STAGE 4. Compilation of a comparative summary table(Table 2).

STEP 5. Recognition of substances.

Task 5. Use qualitative reactions to prove the presence of phenol, formic acid and acetic acid in this mixture. Write the reaction equations, indicate the signs of their occurrence.

Decision

Of the components of the mixture, phenol reacts with bromine water to form a white precipitate:

C 6 H 5 OH + 3Br 2 \u003d C 6 H 2 Br 3 OH + 3HBr.

The presence of formic acid can be established using an ammonia solution of silver oxide:

HCOOH + 2Ag (NH 3) 2 OH \u003d 2Ag + NH 4 HCO 3 + 3NH 3 + HOH.

Silver is released in the form of a precipitate or a mirror coating on the walls of the test tube.

If, after adding an excess of ammonia solution of silver oxide, the mixture boils with a solution of baking soda, then it can be argued that acetic acid is present in the mixture:

CH 3 COOH + NaHCO 3 \u003d CH 3 COOHa + CO 2 + H 2 O.

table 2

Chemical properties of oxygen-containing
organic matter

Reagent	Representatives of oxygen-containing compounds
	CH 3 OH methanol	C 6 H 5 OH phenol	HCHO methanal	HCOOH formic acid	CH 3 CHO acet- aldehyde	HCOCH 3 methyl- formate	C 6 H 12 O 6 glucose
Na	+	+	–	+	–	–	+
NaOH	–	+	–	+	–	+	–
NaHCO3	–	–	–	+	–	–	–
Ba 2 (water)	–	+	+	+	+	+	+
Ag2O (solution in NH 3 aq.)	–	–	+	+	+	+	+

Task 6. Four unlabeled tubes contain ethanol, acetaldehyde, acetic acid, and formic acid. What reactions can be used to distinguish substances in test tubes? Write reaction equations.

Decision

Analyzing the features of the chemical properties of these substances, we come to the conclusion that to solve the problem, one should use a solution of sodium bicarbonate and an ammonia solution of silver oxide. Acetaldehyde reacts only with silver oxide, acetic acid only with sodium bicarbonate, and formic acid with both. A substance that does not react with any of the reagents is ethanol.

Reaction equations:

CH 3 CHO + 2Ag (NH 3) 2 OH \u003d CH 3 COOHNH 4 + 2Ag + 3NH 3 + HOH,

CH 3 COOH + NaHCO 3 \u003d CH 3 COOHa + CO 2 + HOH,

HCOOH + 2Ag (NH 3) 2 OH \u003d 2Ag + NH 4 HCO 3 + 3NH 3 + HOH,

HCOOH + NaHCO 3 \u003d HCOOHa + CO 2 + HOH.

STEP 6. Determination of the quantitative composition of the mixture.

Task 7. To neutralize 26.6 g of a mixture of acetic acid, acetaldehyde and ethanol, 44.8 g of a 25% potassium hydroxide solution was used. When the same amount of the mixture interacted with an excess of metallic sodium, 3.36 liters of gas were released at n.o. Calculate the mass fractions of substances in this mixture.

Decision

Acetic acid and ethanol will react with metallic Na, and only acetic acid will react with KOH. Let's compose the reaction equations:

CH 3 COOH + Na \u003d CH 3 COONa + 1 / 2H 2, (1)

C 2 H 5 OH + Na \u003d C 2 H 5 ONa + 1/2H 2, (2)

Task 8. A mixture of pyridine and aniline weighing 16.5 g was treated with 66.8 ml of 14% hydrochloric acid (= 1.07 g/ml). To neutralize the mixture, it was necessary to add 7.5 g of triethylamine. Calculate the mass fractions of salts in the resulting solution.

Decision

Let's compose the reaction equations:

C 5 H 5 N + HCl \u003d (C 5 H 5 NH) Cl,

C 6 H 5 NH 2 + HCl \u003d (C 6 H 5 NH 3) Cl,

(C 2 H 5) 3 N + Hcl \u003d ((C 2 H 5) 3 NH) Cl.

Calculate the amount of substances - participants in the reactions:

(HCl) = 0.274 mol,

((C 2 H 5) 3 N) = 0.074 mol.

0.074 mol of acid was also spent on the neutralization of triethylamine, and for the reaction with the mixture: 0.274 - 0.074 = 0.2 mol.

We use the same technique as in Problem 3. Denote X is the number of moles of pyridine and y is the number of aniline in the mixture. Let's make a system of equations:

Solving the system, we get that the amount of pyridine is 0.15 mol, and aniline is 0.05 mol. Let's calculate the amounts of substances of hydrochloric salts of pyridine, aniline and triethylamine, their masses and mass fractions. They are respectively 0.15 mol, 0.05 mol, 0.074 mol; 17.33 g, 6.48 g, 10.18 g; 18.15%, 6.79%, 10.66%.

LITERATURE

Kuzmenko N.E., Eremin V.V. Chemistry. 2400 tasks for schoolchildren and university applicants. Moscow: Bustard, 1999;
Ushkalova V.N., Ioanidis N.V.. Chemistry: competitive tasks and answers. Allowance for entering universities. M.: Education, 2000.

dissolving water. The solution obtained after passing gases through water had an acidic reaction. When this solution was treated with silver nitrate, 14.35 g of a white precipitate precipitated. Determine the quantitative and qualitative composition of the initial mixture of gases. Decision.

The gas that burns to form water is hydrogen, which is slightly soluble in water. React in sunlight with an explosion hydrogen with oxygen, hydrogen with chlorine. Obviously, there was chlorine in the mixture with hydrogen, because. the resulting HC1 is highly soluble in water and gives a white precipitate with AgNO3.

Thus, the mixture consists of gases H2 and C1:

1 mol 1 mol

HC1 + AgN03 -» AgCl 4- HN03.

x mol 14.35

When processing 1 mol of HC1, 1 mol of AgCl is formed, and when processing x mol, 14.35 g or 0.1 mol. Mr(AgCl) = 108 + 2 4- 35.5 = 143.5, M(AgCl) = 143.5 g/mol,

v= - = = 0.1 mol,

x = 0.1 mol of HC1 was contained in the solution. 1 mol 1 mol 2 mol H2 4-C12 2HC1 x mol y mol 0.1 mol

x \u003d y \u003d 0.05 mol (1.12 l) of hydrogen and chlorine reacted to form 0.1 mol

HC1. The mixture contained 1.12 liters of chlorine, and hydrogen 1.12 liters + 1.12 liters (excess) = 2.24 liters.

Example 6 A laboratory has a mixture of sodium chloride and iodide. 104.25 g of this mixture was dissolved in water and an excess of chlorine was passed through the resulting solution, then the solution was evaporated to dryness and the residue was calcined to constant weight at 300 °C.

The mass of dry matter turned out to be 58.5 g. Determine the composition of the initial mixture in percent.

Mr(NaCl) = 23 + 35.5 = 58.5, M(NaCl) = 58.5 g/mol, Mr(Nal) = 127 + 23 = 150 M(Nal) = 150 g/mol.

In the initial mixture: the mass of NaCl - x g, the mass of Nal - (104.25 - x) g.

When passing through a solution of chloride and sodium iodide, iodine is displaced by them. When passing the dry residue, the iodine evaporated. Thus, only NaCl can be a dry matter.

In the resulting substance: the mass of NaCl of the original x g, the mass of the obtained (58.5-x):

2 150 g 2 58.5 g

2NaI + C12 -> 2NaCl + 12

(104.25 - x)g (58.5 - x)g

2 150 (58.5 - x) = 2 58.5 (104.25 x)

x = - = 29.25 (g),

those. NaCl in the mixture was 29.25 g, and Nal - 104.25 - 29.25 = 75 (g).

Find the composition of the mixture (in percent):

w(Nal) = 100% = 71.9%,

©(NaCl) = 100% - 71.9% = 28.1%.

Example 7 68.3 g of a mixture of nitrate, iodide and potassium chloride are dissolved in water and treated with chlorine water. As a result, 25.4 g of iodine was released (neglect the solubility of which in water). The same solution was treated with silver nitrate. 75.7 g of sediment fell out. Determine the composition of the initial mixture.

Chlorine does not interact with potassium nitrate and potassium chloride:

2KI + C12 -» 2KS1 + 12,

2 mol - 332 g 1 mol - 254 g

Mg (K1) \u003d 127 + 39 - 166,

x = = 33.2 g (KI was in the mixture).

v(KI) - - = = 0.2 mol.

1 mol 1 mol

KI + AgN03 = Agl + KN03.

0.2 mol x mol

x = = 0.2 mol.

Mr(Agl) = 108 + 127 = 235,

m(Agl) = Mv = 235 0.2 = 47 (r),

then AgCl will be

75.7 g - 47 g = 28.7 g.

74.5 g 143.5 g

KCl + AgN03 = AgCl + KN03

X \u003d 1 L_ \u003d 14.9 (KCl).

Therefore, the mixture contained: 68.3 - 33.2 - 14.9 = 20.2 g KN03.

Example 8. To neutralize 34.5 g of oleum, 74.5 ml of a 40% potassium hydroxide solution is consumed. How many moles of sulfur oxide (VI) account for 1 mole of sulfuric acid?

100% sulfuric acid dissolves sulfur oxide (VI) in any ratio. The composition expressed by the formula H2S04*xS03 is called oleum. Let's calculate how much potassium hydroxide is needed to neutralize H2SO4:

1 mol 2 mol

H2S04 + 2KOH -> K2S04 + 2H20 xl mol y mol

y - 2*x1 mole of KOH is used to neutralize SO3 in oleum. Let's calculate how much KOH is needed to neutralize 1 mol of SO3:

1 mol 2 mol

S03 4- 2KOH -> K2SO4 + H20 x2 mol z mol

z - 2 x2 mol of KOH goes to neutralize SOg in oleum. 74.5 ml of a 40% KOH solution is used to neutralize the oleum, i.e. 42 g or 0.75 mol KOH.

Therefore, 2 xl + 2x 2 \u003d 0.75,

98 xl + 80 x2 = 34.5 g,

xl = 0.25 mol H2SO4,

x2 = 0.125 mol SO3.

Example 9 There is a mixture of calcium carbonate, zinc sulfide and sodium chloride. If 40 g of this mixture is treated with an excess of hydrochloric acid, 6.72 liters of gases will be released, the interaction of which with an excess of sulfur oxide (IV) releases 9.6 g of sediment. Determine the composition of the mixture.

When exposed to a mixture of excess hydrochloric acid, carbon monoxide (IV) and hydrogen sulfide could be released. Only hydrogen sulfide interacts with sulfur oxide (IV), therefore, its volume can be calculated from the amount of precipitate released:

CaC03 + 2HC1 -> CaC12 + H20 + C02t(l)

100 g - 1 mol 22.4 l - 1 mol

ZnS + 2HC1 -> ZnCl2 + H2St (2)

97 g - 1 mol 22.4 l - 1 mol

44.8 l - 2 mol 3 mol

2H2S + S02 -» 3S + 2H20 (3)

xl l 9.6 g (0.3 mol)

xl = 4.48 L (0.2 mol) H2S; from equations (2 - 3) it can be seen that ZnS was 0.2 mol (19.4 g):

2H2S + S02 -> 3S + 2H20.

Obviously, the carbon monoxide (IV) in the mixture was:

6.72 l - 4.48 l \u003d 2.24 l (CO2).

Tasks for mixtures and alloys are a very common type of tasks for the exam in chemistry. They require a clear idea of ​​which of the substances enter into the reaction proposed in the problem and which do not.

O mixtures we say when we have not one, but several substances (components) “poured” into one container. These substances should not interact with each other.

Typical misconceptions and mistakes in solving problems on a mixture.

Often in such problems the reaction of metals with acids is used. To solve such problems, it is necessary to know exactly which metals interact with which acids and which do not.

Necessary theoretical information.

Methods for expressing the composition of mixtures.

Electrochemical series of voltages of metals.

Li Rb K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb H Sb Bi Cu Hg Ag Pd Pt Au

Reactions of metals with acids.

Nitric acid recovery products.

The more active the metal and the lower the acid concentration, the further nitrogen is reduced.
Nonmetals + conc. acid	Inactive metals (to the right of iron) + dil. acid	Active metals (alkaline, alkaline earth, zinc) + conc. acid	Active metals (alkali, alkaline earth, zinc) + medium dilution acid	Active metals (alkaline, alkaline earth, zinc) + very dil. acid
Passivation: do not react with cold concentrated nitric acid:
do not react with nitric acid at any concentration:

Sulfuric acid recovery products.

Inactive metals (to the right of iron) + conc. acid Nonmetals + conc. acid	Alkaline earth metals + conc. acid	Alkali metals and zinc + concentrated acid.	Dilute sulfuric acid behaves like a normal mineral acid (like hydrochloric acid)
Passivation: do not react with cold concentrated sulfuric acid:
do not react with sulfuric acid at any concentration:

Reactions of metals with water and alkalis.

Attention! Many mistakes in solving USE problems in chemistry are due to the fact that schoolchildren have a poor command of mathematics. Especially for you - material on how solve problems on percentages, alloys and mixtures.

Examples of problem solving.

Consider three examples of problems in which mixtures of metals react with hydrochloric acid:

Example 1When a mixture of copper and iron weighing 20 g was exposed to an excess of hydrochloric acid, 5.6 liters of gas (n.o.) were released. Determine the mass fractions of metals in the mixture.

In the first example, copper does not react with hydrochloric acid, that is, hydrogen is released when the acid reacts with iron. Thus, knowing the volume of hydrogen, we can immediately find the amount and mass of iron. And, accordingly, the mass fractions of substances in the mixture.

Example 1 solution.

Answer: iron, copper.

Example 2Under the action of an excess of hydrochloric acid on a mixture of aluminum and iron weighing 11 g, 8.96 liters of gas (n.o.) were released. Determine the mass fractions of metals in the mixture.

In the second example, the reaction is both metal. Here, hydrogen is already released from the acid in both reactions. Therefore, direct calculation cannot be used here. In such cases, it is convenient to solve using a very simple system of equations, taking for - the number of moles of one of the metals, and for - the amount of substance of the second.

Example 2 solution.

1. Find the amount of hydrogen: mol.
2. Let the amount of aluminum be a mole, and iron a mole. Then we can express through and the amount of hydrogen released:

   – molar ratio

3. We know the total amount of hydrogen: mol. So (this is the first equation in the system).
4. For a mixture of metals, you need to express masses through quantities of substances. So the mass of aluminum

   mass of iron

   and the mass of the whole mixture

   (this is the second equation in the system).

5. So we have a system of two equations:
   
   It is much more convenient to solve such systems by the subtraction method, multiplying the first equation by 18: and subtracting the first equation from the second:

   respectively,

Answer: iron, aluminum.

Example 316 g of a mixture of zinc, aluminum and copper was treated with an excess of hydrochloric acid solution. In this case, 5.6 l of gas (n.o.) was released and 5 g of the substance did not dissolve. Determine the mass fractions of metals in the mixture.

In the third example, two metals react, but the third metal (copper) does not react. Therefore, the remainder of 5 g is the mass of copper. The quantities of the remaining two metals - zinc and aluminum (note that their total mass is 16 - 5 = 11 g) can be found using a system of equations, as in example No. 2.

Answer to Example 3: 56.25% zinc, 12.5% ​​aluminum, 31.25% copper.

The next three examples of tasks (No. 4, 5, 6) contain the reactions of metals with nitric and sulfuric acids. The main thing in such tasks is to correctly determine which metal will dissolve in it, and which will not.

Example 4A mixture of iron, aluminum and copper was treated with an excess of cold concentrated sulfuric acid. At the same time, part of the mixture dissolved, and 5.6 liters of gas (n.o.) were released. The remaining mixture was treated with an excess of sodium hydroxide solution. 3.36 liters of gas evolved and 3 g of undissolved residue remained. Determine the mass and composition of the initial mixture of metals.

In this example, remember that cold concentrated sulfuric acid does not react with iron and aluminum (passivation), but reacts with copper. In this case, sulfur oxide (IV) is released.

With alkali reacts only aluminum- amphoteric metal (in addition to aluminum, zinc and tin are also dissolved in alkalis, and beryllium can still be dissolved in hot concentrated alkali).

Example 4 solution.

1. Only copper reacts with concentrated sulfuric acid, the number of moles of gas: mol

   (conc.)
   (do not forget that such reactions must be equalized using an electronic balance)

   Since the molar ratio of copper and sulfur dioxide, then copper is also a mole.
   You can find the mass of copper:

2. Aluminum reacts with an alkali solution, and an aluminum hydroxocomplex and hydrogen are formed:
3. Number of moles of hydrogen: mole, the mole ratio of aluminum and hydrogen and therefore

   Moth.

   Aluminum weight:

4. The remainder is iron, weighing 3 g. You can find the mass of the mixture: g.
5. Mass fractions of metals:

Answer: copper, aluminum, iron.

Example 521.1 g of a mixture of zinc and aluminum was dissolved in 565 ml of a nitric acid solution containing 20 wt. % HNO 3 and having a density of 1.115 g/ml. The volume of the released gas, which is a simple substance and the only product of the reduction of nitric acid, amounted to 2.912 l (n.o.). Determine the composition of the resulting solution in mass percent. (RCTU)

The text of this problem clearly indicates the product of nitrogen reduction - "simple substance". Since nitric acid does not produce hydrogen with metals, it is nitrogen. Both metals dissolved in acid.

The problem asks not the composition of the initial mixture of metals, but the composition of the solution obtained after the reactions. This makes the task more difficult.

Example 5 solution.

1. Determine the amount of gas substance: mol.
2. We determine the mass of the nitric acid solution, the mass and amount of the substance dissolved:

   mole

   Please note that since the metals have completely dissolved, it means - just enough acid(these metals do not react with water). Accordingly, it will be necessary to check Is there too much acid?, and how much of it remains after the reaction in the resulting solution.

3. We compose the reaction equations ( do not forget about the electronic balance) and, for convenience of calculations, we take as - the amount of zinc, and for - the amount of aluminum. Then, in accordance with the coefficients in the equations, nitrogen in the first reaction will be a mol, and in the second - a mol:
4. Then, taking into account that the mass of the mixture of metals is g, their molar masses are g / mol for zinc and g / mol for aluminum, we obtain the following system of equations:

   
   - amount of nitrogen
   is the mass of a mixture of two metals

   It is convenient to solve this system by multiplying the first equation by 90 and subtracting the first equation from the second.

   So mole

   So mole

   Let's check the mass of the mixture:

   G.
   

5. Now let's move on to the composition of the solution. It will be convenient to rewrite the reactions again and write down over the reactions the amounts of all reacted and formed substances (except water):
6. The next question is: did nitric acid remain in the solution and how much is left? According to the reaction equations, the amount of acid that reacted: mol,

   those. the acid was in excess and you can calculate its remainder in solution:

   Moth.

7. So in final solution contains:

   zinc nitrate in the amount of moles:

   aluminum nitrate in the amount of moles:

   excess nitric acid in the amount of moles:

8. What is the mass of the final solution? Recall that the mass of the final solution consists of those components that we mixed (solutions and substances) minus those reaction products that left the solution (precipitates and gases):
   

   Then for our task:

   Weight of acid solution + weight of metal alloy - weight of nitrogen

   Example 6When processing g of a mixture of copper, iron and aluminum with an excess of concentrated nitric acid, l gas (n.o.) was released, and when this mixture was exposed to the same mass of excess hydrochloric acid, l gas (n.o.). Determine the composition of the initial mixture. (RCTU)

   When solving this problem, we must remember, firstly, that concentrated nitric acid with an inactive metal (copper) gives, and iron and aluminum do not react with it. Hydrochloric acid, on the other hand, does not react with copper.

   Answer for example 6: copper, iron, aluminum.

   Tasks for independent solution.

   1. Simple problems with two mixture components.

   1-1. A mixture of copper and aluminum with a mass of g was treated with a solution of nitric acid, and l of gas (n.a.) was released. Determine the mass fraction of aluminum in the mixture.

   1-2. A mixture of copper and zinc weighing g was treated with a concentrated alkali solution. In this case, l of gas (n.y.) was released. Calculate the mass fraction of zinc in the initial mixture.

   1-3. A mixture of magnesium and magnesium oxide, weighing g, was treated with a sufficient amount of dilute sulfuric acid. At the same time, l of gas (n.o.) was released. Find the mass fraction of magnesium in the mixture.

   1-4. A mixture of zinc and zinc oxide weighing g was dissolved in dilute sulfuric acid. Zinc sulfate was obtained with a mass of g. Calculate the mass fraction of zinc in the initial mixture.

   1-5. Under the action of a mixture of powders of iron and zinc with a mass of g on an excess of a solution of copper (II) chloride, g of copper was formed. Determine the composition of the initial mixture.

   1-6. What mass solution of hydrochloric acid will be required to completely dissolve g of a mixture of zinc with zinc oxide, if hydrogen is released with a volume of l (n.o.)?

   1-7. When dissolved in dilute nitric acid, g of a mixture of iron and copper releases nitric oxide (II) with a volume of l (n.o.). Determine the composition of the initial mixture.

   1-8. When dissolving g of a mixture of iron and aluminum filings in a solution of hydrochloric acid (g / ml), l of hydrogen (n.o.) was released. Find the mass fractions of metals in the mixture and determine the volume of hydrochloric acid consumed.

   2. Tasks are more complex.

   2-1. A mixture of calcium and aluminum weighing g was calcined without access to air with an excess of graphite powder. The reaction product was treated with dilute hydrochloric acid, and l gas (n.o.) was released. Determine the mass fractions of metals in the mixture.

   2-2. To dissolve g of an alloy of magnesium with aluminum, a ml solution of sulfuric acid (g / ml) was used. The excess acid reacted with ml of mol/l potassium bicarbonate solution. Determine the mass fractions of metals in the alloy and the volume of gas (N.O.) released during the dissolution of the alloy.

   2-3. When dissolving g of a mixture of iron and iron oxide (II) in sulfuric acid and evaporating the solution to dryness, g of ferrous sulfate, iron sulfate heptahydrate (II), was formed. Determine the quantitative composition of the initial mixture.

   2-4. When iron (g) reacted with chlorine, a mixture of iron (II) and (III) chlorides (g) was formed. Calculate the mass of iron (III) chloride in the resulting mixture.

   2-5. What was the mass fraction of potassium in its mixture with lithium, if, as a result of the treatment of this mixture with an excess of chlorine, a mixture was formed in which the mass fraction of potassium chloride was ?

   2-6. After treatment with an excess of bromine of a mixture of potassium and magnesium with a total mass of g, the mass of the resulting mixture of solids was found to be g. This mixture was treated with an excess of sodium hydroxide solution, after which the precipitate was separated and calcined to constant weight. Calculate the mass of the resulting residue.

   2-7. A mixture of lithium and sodium with a total mass of g was oxidized with an excess of oxygen; in total, l (n.o.) was consumed. The resulting mixture was dissolved in the i-th solution of sulfuric acid. Calculate the mass fractions of substances in the resulting solution.

   2-8. An alloy of aluminum and silver was treated with an excess of a concentrated solution of nitric acid, the residue was dissolved in acetic acid. The volumes of gases released in both reactions, measured under the same conditions, turned out to be equal to each other. Calculate the mass fractions of metals in the alloy.

   3. Three metals and complex tasks.

   3-1. When processing g of a mixture of copper, iron and aluminum with an excess of concentrated nitric acid, l of gas was released. The same volume of gas is also released when the same mixture of the same mass is treated with an excess of dilute sulfuric acid (N.O.). Determine the composition of the initial mixture in mass percent.

   3-2. g of a mixture of iron, copper and aluminum, interacting with an excess of dilute sulfuric acid, releases l of hydrogen (n.o.). Determine the composition of the mixture in mass percent if chlorination of the same sample of the mixture requires l chlorine (n.o.).

   3-3. Iron, zinc and aluminum filings are mixed in a molar ratio (in the order listed). g of this mixture was treated with an excess of chlorine. The resulting mixture of chlorides was dissolved in ml of water. Determine the concentration of substances in the resulting solution.

   3-4. An alloy of copper, iron and zinc with a mass of g (the masses of all components are equal) was placed in a solution of hydrochloric acid with a mass of g. Calculate the mass fractions of the substances in the resulting solution.

   3-5. g of a mixture consisting of silicon, aluminum and iron, was treated with heating with an excess of sodium hydroxide, while 1 l of gas (n.o.) was released. Under the action of a mixture of excess hydrochloric acid on such a mass, l of gas (n.o.) is released. Determine the masses of substances in the initial mixture.

   3-6. When a mixture of zinc, copper, and iron was treated with an excess of a concentrated alkali solution, gas was released, and the mass of the undissolved residue turned out to be several times less than the mass of the initial mixture. This residue was treated with an excess of hydrochloric acid, and the volume of the released gas turned out to be equal to the volume of the gas released in the first case (the volumes were measured under the same conditions). Calculate the mass fractions of metals in the initial mixture.

   3-7. There is a mixture of calcium, calcium oxide and calcium carbide with a molar ratio of components (in the order listed). What is the minimum volume of water that can enter into chemical interaction with such a mixture of mass r?

   3-8. A mixture of chromium, zinc and silver with a total mass of g was treated with dilute hydrochloric acid, the mass of the undissolved residue was equal to g. The mass of the precipitate formed turned out to be equal to g. Calculate the mass fractions of metals in the initial mixture.

   Answers and comments to tasks for independent solution.

   1-1. (aluminum does not react with concentrated nitric acid); and; (chromium, when dissolved in hydrochloric acid, turns into chromium (II) chloride, which, under the action of bromine in an alkaline medium, turns into chromate; when a barium salt is added, insoluble barium chromate is formed)

   
   

   The composition of the initial mixture for the production of artificial stone. (Photo gallery "Our technologies" on the page of the same name. What is included in the composition of artificial facing stone produced on flexible elastic molds. In essence, the decorative facing stone we are talking about is a typical Portland cement-based sand concrete, made by vibrocasting into special flexible elastic matrices - shaped and specially colored.Consider the main components of the concrete mixture for the production of artificial facing stone by vibrocasting.The binder is the basis of any artificial facing stone.In this case, it is Portland cement grade M-400 or M-500.To ensure that the quality of concrete always remains consistently high, we recommend using only "fresh" cement (as you know, it quickly loses its properties over time and from improper storage) of the same manufacturer with a good reputation.For the production of decorative facing stone, both ordinary, gray cement and and white cement. In nature, there are a number of colors and shades that can only be replicated on white cement. In other cases, gray portland is used (for reasons of economic feasibility).

   Many domestic manufacturers of artificial facing stone have recently been actively using gypsum as a binder. At the same time, they claim that their products are expanded clay concrete. And, as a rule, expanded clay concrete is really presented at the stands of companies. But there is one point that determines the behavior of manufacturers of artificial facing stone. The cost of flexible elastic injection molds that allow you to accurately repeat the texture of the stone is very high.

   And if the technology is followed, the turnover of injection molds, that is, the time from the moment the concrete is poured to the moment the product is stripped, is 10-12 hours, versus 30 minutes on plaster. This is what drives companies to use gypsum as a binder. And the price of gypsum is at least five times lower than the price of white cement. All this provides companies with super profits. But the issue price for the end user is very high! The extremely low frost resistance and strength of such products will not allow you to enjoy the view of the facades for a long time.

   In the presented photographs, the plaster product is one year after installation. Multiple cracks and fractures are clearly visible. Therefore, the use of this material on an industrial scale is difficult. Based on the tasks we are facing, we prefer to produce artificial facing stone - a material that is close to natural stone in terms of hardness and abrasion properties, suitable for both external and internal cladding, rather than decorations that are fragile and whimsical to water. Filler. Depending on the type of fillers used, cement-based artificial facing stone can be "heavy" (2-2.4 g/cm3) or "light" (about 1.6 g/cm3). Ideally, heavy concrete is used for the production of paving stone, decorative paving slabs, curbs, plinth frames, and interior stone. For the production of artificial facing stone used for exterior decoration, lightweight concrete is used.

   Approximately this is what manufacturers working on American technology do. In the regions, unfortunately, heavy concrete is used predominantly. Of course, it is much easier to make a decorative stone on sand, but a light stone will always be preferable for the consumer. It's just a matter of choice. For the production of heavy artificial facing stone, coarse quartz sand of a fraction of 0.63-1.5 mm is used (the use of fine sand worsens the strength characteristics of concrete) and, when appropriate, fine gravel, such as marble, of a fraction of 5-10 mm. "Light" facing stone is made using expanded clay sand. But in the production of artificial facing stone on expanded clay, the following factor should be taken into account. In July 2001, we received information from customers about the appearance of "shots" on the surface of products (lightweight concrete) (dotted swelling of the white material). As a result of consultations with specialists, it was found that the “shots” appear as a result of the decay of lime inclusions in expanded clay.

   When free calcium interacts with moisture (water or its vapor), a chemical reaction occurs accompanied by an increase in the volume of free calcium grains, resulting in the so-called “shot” effect. CaO + H2O \u003d Ca (OH) 2 + CO2 \u003d CaCO3 The peculiarity of this chemical reaction is that it takes a very long time - up to 6 months. Expanded clay manufacturers produce products in accordance with GOST, which allows the presence of lime grains up to 3% of the total mass. The effect of "shots" reduces the consumer properties of products, so the task was to find a new filler for the production of lightweight concrete.

   It has been observed that the lime disintegration reaction causes degradation of the surface of the product ONLY in interior decoration. When using products for finishing plinths and facades of buildings, visible damage to the finishing material is not observed. According to the statements of an employee of NIIZhB, lime decay is leveled when using products for exterior decoration of buildings. In connection with the identification of this pattern, since August 2001, products for interior work have been produced not on expanded clay, but on another (heavier) aggregate. To switch to a single filler, we offer the following solutions to this problem: 1. Use crushed expanded clay with a fraction of at least 2 cm as a filler. 2. Create expanded clay dumps with exposure in an open area for at least 6-9 months.

   3. Creation of a non-uniform filler from quartz sand and a lighter artificial filler. 4. Use of slag pumice. however, the bulk density of the finished product will increase to 1800-2000 kg/m3. Lightweight aggregate must meet the following requirements. bulk weight is about 600 kg/m3. sand of a fraction of 0-0.5 cm or 0-1 cm (the presence of a fine fraction of 15% by volume. Compressive strength of 18 kg / cm (expanded clay index. Water absorption up to 25% (expanded clay index. In the production of artificial facing stone, decorative paving slabs) , small architectural products on flexible elastic molds, the following fillers can be used: Slag pumice, Granulated slag, Crushed stone and slag sand, Foam glass, Expanded perlite sand, Rigidly expanded perlite, Expanded vermoculite, Expanded polystyrene, Enriched quartz sand, Marble chips, Building sand (white ), Molding sand, Volcanic pumice. Pigments and dyes. The most important component of a decorative facing stone is the pigments (dyes) used. Skillful or inept use of colorants directly affects the appearance of the final product. In experienced hands, ordinary concrete turns into something completely indistinguishable from natural "wild" stone. How to achieve this? Mineral inorganic pigments (titanium, iron, chromium oxides) and special light and weather resistant dyes are used for coloring cement. Experienced manufacturers usually choose dyes from companies such as Bayer, Du Pont, Kemira and others no less reputable. This is due not only to the consistently high quality of their products, but also to their wide range of products. So, Bayer offers several dozen iron oxide pigments. By combining them with each other, you can choose almost any desired shade of color. So, Portland cement, expanded clay sand and pigments are the main composition of artificial facing stone. Many manufacturers of architectural concrete products limit themselves to this, despite the fact that there are a huge number of various additives in cements to improve certain characteristics. In any major city, you can find suppliers of domestic and imported additives for concrete. These are various superplasticizers that improve workability and increase the strength of concrete; polymer-latex additives that have a beneficial effect on the durability of concrete; concrete hardening accelerators and air-entraining additives; volumetric water repellents, many times reducing water absorption (useful for facade, basement and paving stone); chemical fibers for dispersed reinforcement, which dramatically increases crack resistance and much more. To use any of these additives or not - decide for yourself, we just want to recommend the use of protective impregnating compounds for surface treatment of decorative facing stone. Properly selected water repellent for concrete will achieve the following results. will increase the aesthetics of the perception of the stone and eliminate the "dustiness" - a characteristic feature of any cement concrete. will increase the service life of the facade stone (the point here is that the process of destruction of decorative concrete primarily affects the color saturation long before the first signs of destruction appear, the reason for which is the exposure of aggregate particles on the front surface of the stone. It will sharply reduce the risk of efflorescence on the surface of the stone, which are a real disaster for cementitious decorative concretes, which is why they should be given the closest attention.