Pure substances and mixtures. Methods for separating mixtures. Topic: "Methods for separating mixtures" (Grade 8) 2 ways to separate a heterogeneous mixture

Topic: "Methods for separating mixtures" (Grade 8)

theoretical block.

The concept of "mixture" was defined in the 17th century. English scientist Robert Boyle: "A mixture is an integral system consisting of heterogeneous components."

Comparative characteristics of a mixture and a pure substance

Signs of comparison	pure substance	Mixture
Composition	Constant	fickle
Substances	Same	Various
Physical Properties	Permanent	Fickle
Energy change during formation	going on	Not happening
Separation	Via chemical reactions	Physical methods

Mixtures differ from each other in appearance.

The classification of mixtures is shown in the table:

Here are examples of suspensions (river sand + water), emulsions (vegetable oil + water) and solutions (air in a flask, salt + water, small change: aluminum + copper or nickel + copper).

Methods for separating mixtures

In nature, substances exist in the form of mixtures. For laboratory research, industrial production, for the needs of pharmacology and medicine, pure substances are needed.

Used to purify substances various ways separation of mixtures

Evaporation - the release of dissolved in liquid solids the way it is converted into steam.

Distillation - distillation, separation of substances contained in liquid mixtures according to boiling points, followed by cooling of the vapor.

In nature, water in its pure form (without salts) does not occur. Oceanic, sea, river, well and spring water are varieties of salt solutions in water. However, often people need clean water that does not contain salts (used in car engines; in chemical production to obtain various solutions and substances; in the manufacture of photographs). Such water is called distilled, and the method of obtaining it is called distillation.

Filtration is the filtering of liquids (gases) through a filter in order to purify them from solid impurities.

These methods are based on differences in physical properties mixture components.

Consider ways to separate heterogeneous and homogeneous mixtures.

Blend example	Separation method
Suspension - a mixture of river sand with water	settling Separation upholding based on different densities of substances. Heavier sand settles to the bottom. You can also separate the emulsion: to separate oil or vegetable oil from water. In the laboratory, this can be done using a separating funnel. Oil or vegetable oil forms the top, lighter layer. As a result of settling, dew falls out of the fog, soot is deposited from smoke, cream is settled in milk. Separation of a mixture of water and vegetable oil by settling
A mixture of sand and table salt in water	Filtration What is the basis for the separation of heterogeneous mixtures using filtering? On various solubility of substances in water and on various sizes of particles. Only particles of substances commensurate with them pass through the pores of the filter, while larger particles are retained on the filter. So you can separate a heterogeneous mixture of table salt and river sand. Various porous substances can be used as filters: cotton wool, coal, fired clay, pressed glass, and others. The filtering method is the basis for the operation of household appliances, such as vacuum cleaners. It is used by surgeons - gauze bandages; drillers and workers of elevators - respiratory masks. With the help of a tea strainer for filtering tea leaves, Ostap Bender, the hero of the work of Ilf and Petrov, managed to take one of the chairs from Ellochka Ogre (“The Twelve Chairs”). Separation of a mixture of starch and water by filtration
A mixture of iron powder and sulfur	Action by magnet or water Iron powder was attracted by a magnet, but sulfur powder was not. The non-wetting sulfur powder floated to the surface of the water, while the heavy wettable iron powder settled to the bottom. Separation of a mixture of sulfur and iron using a magnet and water
A solution of salt in water is a homogeneous mixture	Evaporation or crystallization The water evaporates and salt crystals remain in the porcelain cup. When water is evaporated from lakes Elton and Baskunchak, table salt is obtained. This separation method is based on the difference in the boiling points of the solvent and the solute. If a substance, such as sugar, decomposes when heated, then the water is not completely evaporated - the solution is evaporated, and then from saturated solution sugar crystals are precipitated. Sometimes it is required to remove impurities from solvents with a lower boiling point, for example, water from salt. In this case, the vapors of the substance must be collected and then condensed upon cooling. This method of separating a homogeneous mixture is called distillation or distillation. In special devices - distillers, distilled water is obtained, which is used for the needs of pharmacology, laboratories, and car cooling systems. At home, you can design such a distiller: If, however, a mixture of alcohol and water is separated, then the first to be distilled off (collected in a receiving test tube) is alcohol with t bp = 78 ° C, and water will remain in the test tube. Distillation is used to obtain gasoline, kerosene, gas oil from oil. Separation of homogeneous mixtures

A special method of separating components, based on their different absorption by a certain substance, is chromatography.

With the help of chromatography, the Russian botanist M. S. Tsvet was the first to isolate chlorophyll from the green parts of plants. In industry and laboratories, instead of filter paper for chromatography, starch, coal, limestone, and aluminum oxide are used. Are substances always required with the same degree of purification?

For different purposes, substances with different degrees of purification are needed. Cooking water is sufficiently settled to remove impurities and chlorine used to disinfect it. Drinking water must first be boiled. And in chemical laboratories for the preparation of solutions and experiments, in medicine, distilled water is needed, as purified as possible from the substances dissolved in it. Highly pure substances, the content of impurities in which does not exceed one millionth of a percent, are used in electronics, semiconductor, nuclear technology and other precision industries.

Methods for expressing the composition of mixtures.

Mass fraction of the component in the mixture- the ratio of the mass of the component to the mass of the entire mixture. Usually the mass fraction is expressed in %, but not necessarily.

ω ["omega"] = m component / m mixture

Mole fraction of a component in a mixture- the ratio of the number of moles (amount of substance) of the component to the total number of moles of all substances in the mixture. For example, if the mixture includes substances A, B and C, then:

χ [“chi”] component A \u003d n component A / (n (A) + n (B) + n (C))

Molar ratio of components. Sometimes in tasks for a mixture, the molar ratio of its components is indicated. For example:

n component A: n component B = 2: 3

Volume fraction of the component in the mixture (only for gases)- the ratio of the volume of substance A to the total volume of the entire gas mixture.

φ ["phi"] = V component / V mixture

Practice block.

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. Define mass fractions metals in the mixture.

In the first example, copper does not react with hydrochloric acid, that is, hydrogen is released during the reaction of acid 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.

Finding the amount of hydrogen:
n \u003d V / V m \u003d 5.6 / 22.4 \u003d 0.25 mol.
According to the reaction equation:
The amount of iron is also 0.25 mol. You can find its mass:
m Fe \u003d 0.25 56 \u003d 14 g.
Now you can calculate the mass fractions of metals in the mixture:
ω Fe \u003d m Fe / m of the whole mixture \u003d 14 / 20 \u003d 0.7 \u003d 70%

Answer: 70% iron, 30% 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 x - the number of moles of one of the metals, and for y - the amount of substance of the second.

Example 2 solution.

Finding the amount of hydrogen:
n \u003d V / V m \u003d 8.96 / 22.4 \u003d 0.4 mol.
Let the amount of aluminum be x mol, and iron y mol. Then we can express in terms of x and y the amount of hydrogen released:
It is much more convenient to solve such systems by the subtraction method, multiplying the first equation by 18:
27x + 18y = 7.2
and subtracting the first equation from the second:
(56 - 18)y \u003d 11 - 7.2
y \u003d 3.8 / 38 \u003d 0.1 mol (Fe)
x = 0.2 mol (Al)
Next, we find the masses of metals and their mass fractions in the mixture:

m Fe = n M = 0.1 56 = 5.6 g
m Al = 0.2 27 = 5.4 g
ω Fe = m Fe / m mixture = 5.6 / 11 = 0.50909 (50.91%),

respectively,

ω Al \u003d 100% - 50.91% \u003d 49.09%

Answer: 50.91% iron, 49.09% 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.

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 sulphuric 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 also dissolve in alkalis, and beryllium can still be dissolved in hot concentrated alkali).

Example 4 solution.

Only copper reacts with concentrated sulfuric acid, the number of moles of gas:
n SO2 \u003d V / Vm \u003d 5.6 / 22.4 \u003d 0.25 mol

0,25

0,25

Cu+

2H 2 SO 4 (conc.) = CuSO 4 +

SO 2 + 2H 2 O
(do not forget that such reactions must be equalized using an electronic balance)
Since the molar ratio of copper and sulfur dioxide is 1:1, then copper is also 0.25 mol. You can find the mass of copper:
m Cu \u003d n M \u003d 0.25 64 \u003d 16 g.
Aluminum reacts with an alkali solution, and an aluminum hydroxocomplex and hydrogen are formed:
2Al + 2NaOH + 6H 2 O = 2Na + 3H 2

Al 0 − 3e = Al 3+

2

2H + + 2e = H 2

3
Number of moles of hydrogen:
n H2 = 3.36 / 22.4 = 0.15 mol,
the molar ratio of aluminum and hydrogen is 2:3 and, therefore,
nAl \u003d 0.15 / 1.5 \u003d 0.1 mol.
Aluminum weight:
m Al \u003d n M \u003d 0.1 27 \u003d 2.7 g
The remainder is iron, weighing 3 g. You can find the mass of the mixture:
m mixture \u003d 16 + 2.7 + 3 \u003d 21.7 g.
Mass fractions of metals:

ω Cu \u003d m Cu / m mixture \u003d 16 / 21.7 \u003d 0.7373 (73.73%)
ω Al = 2.7 / 21.7 = 0.1244 (12.44%)
ω Fe = 13.83%

Answer: 73.73% copper, 12.44% aluminum, 13.83% 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 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 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.

Determine the amount of gas substance:
n N2 \u003d V / Vm \u003d 2.912 / 22.4 \u003d 0.13 mol.
We determine the mass of the nitric acid solution, the mass and amount of the dissolved HNO3 substance:

m solution \u003d ρ V \u003d 1.115 565 \u003d 630.3 g
m HNO3 \u003d ω m solution \u003d 0.2 630.3 \u003d 126.06 g
n HNO3 \u003d m / M \u003d 126.06 / 63 \u003d 2 mol

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.

We compose the reaction equations ( do not forget about the electronic balance) and, for convenience of calculations, we take for 5x - the amount of zinc, and for 10y - the amount of aluminum. Then, in accordance with the coefficients in the equations, nitrogen in the first reaction will be x mol, and in the second - 3y mol:

5x		x
5Zn	+ 12HNO 3 = 5Zn(NO 3) 2 +	N 2	+ 6H2O

Zn 0 − 2e = Zn 2+		5
2N+5+10e=N2		1

10y		3y
10Al	+ 36HNO 3 \u003d 10Al (NO 3) 3 +	3N2	+ 18H2O

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

x \u003d 0.04, which means n Zn \u003d 0.04 5 \u003d 0.2 mol
y \u003d 0.03, which means that n Al \u003d 0.03 10 \u003d 0.3 mol

Let's check the mass of the mixture:
0.2 65 + 0.3 27 \u003d 21.1 g.

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):

0,2	0,48	0,2	0,03
5Zn	+ 12HNO 3 =	5Zn(NO 3) 2	+N2+	6H2O

0,3	1,08	0,3	0,09
10Al	+ 36HNO 3 =	10Al(NO 3) 3	+ 3N 2 +	18H2O

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:
n HNO3 \u003d 0.48 + 1.08 \u003d 1.56 mol,
those. the acid was in excess and you can calculate its remainder in solution:
n HNO3 rest. \u003d 2 - 1.56 \u003d 0.44 mol.
So in final solution contains:

zinc nitrate in the amount of 0.2 mol:
m Zn(NO3)2 = n M = 0.2 189 = 37.8 g
aluminum nitrate in the amount of 0.3 mol:
m Al(NO3)3 = n M = 0.3 213 = 63.9 g
an excess of nitric acid in an amount of 0.44 mol:
m HNO3 rest. = n M = 0.44 63 = 27.72 g

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:

m new solution \u003d mass of acid solution + mass of metal alloy - mass of nitrogen
m N2 = n M = 28 (0.03 + 0.09) = 3.36 g
m new solution \u003d 630.3 + 21.1 - 3.36 \u003d 648.04 g
Now you can calculate the mass fractions of substances in the resulting solution:

ωZn (NO 3) 2 \u003d m in-va / m solution \u003d 37.8 / 648.04 \u003d 0.0583
ωAl (NO 3) 3 \u003d m in-va / m solution \u003d 63.9 / 648.04 \u003d 0.0986
ω HNO3 rest. \u003d m in-va / m solution \u003d 27.72 / 648.04 \u003d 0.0428

Answer: 5.83% zinc nitrate, 9.86% aluminum nitrate, 4.28% nitric acid.

Example 6When processing 17.4 g of a mixture of copper, iron and aluminum with an excess of concentrated nitric acid, 4.48 liters of gas (n.o.) were released, and when this mixture was exposed to the same mass of excess hydrochloric acid, 8.96 l of gas (n.o.). u.). 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 NO 2, and iron and aluminum do not react with it. Hydrochloric acid, on the other hand, does not react with copper.

Answer for example 6: 36.8% copper, 32.2% iron, 31% aluminum.

Tasks for independent solution.

1. Simple problems with two mixture components.

1-1. A mixture of copper and aluminum weighing 20 g was treated with a 96% solution of nitric acid, and 8.96 liters of gas (n.a.) were released. Determine the mass fraction of aluminum in the mixture.

1-2. A mixture of copper and zinc weighing 10 g was treated with a concentrated alkali solution. In this case, 2.24 liters of gas (n.y.) were released. Calculate the mass fraction of zinc in the initial mixture.

1-3. A mixture of magnesium and magnesium oxide weighing 6.4 g was treated with a sufficient amount of dilute sulfuric acid. At the same time, 2.24 liters of gas (n.o.) were released. Find the mass fraction of magnesium in the mixture.

1-4. A mixture of zinc and zinc oxide weighing 3.08 g was dissolved in dilute sulfuric acid. Zinc sulfate weighing 6.44 g was obtained. Calculate the mass fraction of zinc in the initial mixture.

1-5. Under the action of a mixture of iron and zinc powders weighing 9.3 g on an excess of copper (II) chloride solution, 9.6 g of copper was formed. Determine the composition of the initial mixture.

1-6. What mass of a 20% hydrochloric acid solution will be required to completely dissolve 20 g of a mixture of zinc with zinc oxide, if hydrogen is released in the amount of 4.48 liters (n.o.)?

1-7. When dissolved in dilute nitric acid, 3.04 g of a mixture of iron and copper releases nitric oxide (II) with a volume of 0.896 l (n.o.). Determine the composition of the initial mixture.

1-8. When dissolving 1.11 g of a mixture of iron and aluminum filings in a 16% hydrochloric acid solution (ρ = 1.09 g/ml), 0.672 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 18.8 g was calcined without access to air with an excess of graphite powder. The reaction product was treated with dilute hydrochloric acid, and 11.2 liters of gas (n.o.) were released. Determine the mass fractions of metals in the mixture.

2-2. To dissolve 1.26 g of an alloy of magnesium with aluminum, 35 ml of a 19.6% sulfuric acid solution (ρ = 1.1 g/ml) was used. The excess acid reacted with 28.6 ml of a 1.4 mol/L potassium hydrogen carbonate 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 27.2 g of a mixture of iron and iron (II) oxide in sulfuric acid and evaporating the solution to dryness, 111.2 g of ferrous sulfate, iron (II) sulfate heptahydrate, was formed. Determine the quantitative composition of the initial mixture.

2-4. The interaction of iron weighing 28 g with chlorine formed a mixture of iron (II) and (III) chlorides weighing 77.7 g. 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 80%?

2-6. After treatment with an excess of bromine of a mixture of potassium and magnesium with a total mass of 10.2 g, the mass of the resulting mixture of solids was 42.2 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.

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 8.2 g of a mixture of copper, iron and aluminum with an excess of concentrated nitric acid, 2.24 liters of gas were 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. 14.7 g of a mixture of iron, copper and aluminum, interacting with an excess of dilute sulfuric acid, releases 5.6 liters of hydrogen (n.o.). Determine the composition of the mixture in mass percent if chlorination of the same sample of the mixture requires 8.96 liters of chlorine (n.o.).

3-3. Iron, zinc and aluminum filings are mixed in a molar ratio of 2:4:3 (in the order listed). 4.53 g of this mixture was treated with an excess of chlorine. The resulting mixture of chlorides was dissolved in 200 ml of water. Determine the concentration of substances in the resulting solution.

3-4. An alloy of copper, iron and zinc weighing 6 g (the masses of all components are equal) was placed in an 18.25% solution of hydrochloric acid weighing 160 g. Calculate the mass fractions of the substances in the resulting solution.

3-5. 13.8 g of a mixture consisting of silicon, aluminum and iron, was treated with excess sodium hydroxide while heating, while 11.2 liters of gas (n.o.) were released. When exposed to such a mass of a mixture of excess hydrochloric acid, 8.96 liters of gas (n.o.) are 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 2 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 3:2:5 (in the order listed). What is the minimum volume of water that can enter into chemical interaction with such a mixture weighing 55.2 g?

3-8. A mixture of chromium, zinc and silver with a total weight of 7.1 g was treated with dilute hydrochloric acid, the mass of the undissolved residue was 3.2 g. The mass of the formed precipitate turned out to be 12.65 g. Calculate the mass fractions of metals in the initial mixture.

Answers and comments to tasks for independent solution.

1-1. 36% (aluminum does not react with concentrated nitric acid);

1-2. 65% (only amphoteric metal - zinc dissolves in alkali);

1-5. 30.1% Fe (iron, displacing copper, goes into the +2 oxidation state);

1-7. 36.84% Fe (iron in nitric acid goes to +3);

1-8. 75.68% Fe (iron reacts with hydrochloric acid to +2); 12.56 ml HCl solution.
2-1. 42.55% Ca (calcium and aluminum with graphite (carbon) form the carbides CaC 2 and Al 4 C 3; when they are hydrolyzed with water or HCl, acetylene C 2 H 2 and methane CH 4 are released, respectively);

2-3. 61.76% Fe (iron sulfate heptahydrate - FeSO 4 7H 2 O);

2-7. 5.9% Li 2 SO 4, 22.9% Na 2 SO 4, 5.47% H 2 O 2 (when lithium is oxidized with oxygen, its oxide is formed, and when sodium is oxidized, Na 2 O 2 peroxide is formed, which is hydrolyzed in water to hydrogen peroxide and alkali);

3-1. 39% Cu, 3.4% Al;

3-2. 38.1% Fe, 43.5% Cu;

3-3. 1.53% FeCl 3 , 2.56% ZnCl 2 , 1.88% AlCl 3 (iron reacts with chlorine to the +3 oxidation state);

3-4. 2.77% FeCl 2, 2.565% ZnCl 2, 14.86% HCl (do not forget that copper does not react with hydrochloric acid, so its mass is not included in the mass of the new solution);

3-5. 2.8 g Si, 5.4 g Al, 5.6 g Fe (silicon is a non-metal, it reacts with an alkali solution, forming sodium silicate and hydrogen; it does not react with hydrochloric acid);

3-6. 6.9% Cu, 43.1% Fe, 50% Zn;

3-8. 45.1% Ag, 36.6% Cr, 18.3% Zn barium)

Test block

Part A

1. Sand with salt refers to:

A. to simple substances

B. to chemical compounds

C. to homogeneous systems

D. to heterogeneous systems

2. Fog is:

A. aerosol

B. emulsion

C. solution

D. suspension

3. To obtain gasoline from natural oil, the following method is used:

A. Synthesis

B. sublimation

C. Filtration

D. distillation

4. Specify the best way to separate a mixture of gasoline and water:

A. filtering

B. distillation

C. sublimation

D. settling

5. Separation of a mixture of oil and water is based on:

A. on the difference in density of two liquids

B. on the solubility of one liquid in another

C. on color difference

D. on a similar state of aggregation of liquids

6. A mixture of copper and iron filings can be separated:

A. Filtration

B. magnetic action

C. Chromatography

D. distillation (distillation)

7. What is a pure substance, unlike a mixture:

And cast iron

In food mixture

From the air

D sea water

8. What applies to heterogeneous mixtures:

A mixture of oxygen and nitrogen

In muddy river water

With snowy crust

9.What is a solid mixture:

A glucose solution

With alcohol solution

D potassium sulfate solution

10. What is the name of the method of cleaning a heterogeneous mixture:

A distillation

In filtering

With evaporation

D jelly heating

Part B

1. Set the correct sequence for separating a mixture of table salt and river sand:

A) filter out

B) assemble the device for filtering

B) dissolve in water

D) evaporate the solution

D) assemble the device for evaporation

2. Choose the number of the pair of substances to be separated

1) evaporation

2) filtering

A) river sand and water

b) sugar and water

B) iron and sulfur

D) water and alcohol

3. Correlate the proposed examples of mixtures to one or another group (fog, smoke, fizzy drinks, river and sea silt, mortars, ointment, ink, lipstick, alloys, minerals), filling out the table:

Aggregate state of substances	Mixture examples
hard-hard
Solid-liquid
solid-gaseous
liquid-liquid
liquid-solid
liquid-gaseous
gaseous-gaseous
gaseous-liquid
gaseous-solid

Test-task block

one . Task 1. Fill in the table

Answer:

2. Solve the crossword

Answers in vertical columns - how the specified mixture is separated

Oil + water
Iodine + sugar
Water + river sand
Water + alcohol
Water + salt

							4
								5
	1

		2			3


R	BUT	W	D	E	L	E	H	AND	E

Answer:

3. Suggest several cleaning methods natural water in hiking conditions.

Answer:

4. Anagrams. Rearrange the letters in the words so that you get the main terms of this lesson. Write these terms in response

MIESSE, CONGREEPA, ZUPENSIAS, TAXOCHI, RIFOLIFANTE

Answer:

5. Divide the proposed concepts into 2 groups.

AIR, SEAWATER, ALCOHOL, OXYGEN, STEEL, IRON

Record your answer in the table. Give names to columns

???	???
1	1
2	2
3	3

Answer:

6. Fairy chemistry

In well-known fairy tales, the stepmother or other evil spirits forced the heroine to separate certain mixtures into separate components. Remember what these mixtures were and on the basis of what method they were separated? It is enough to remember 2-3 fairy tales.

Answer:

7. Answer briefly the questions

1. When ore is crushed at mining and processing plants, fragments of iron tools get into it. How can they be extracted from ore?

2. The vacuum cleaner sucks in air containing dust and releases clean air. Why?

3. Water after washing cars in large garages is contaminated with engine oil. What should be done before draining it into the sewer?

4. Flour is cleaned of bran by sifting. Why do they do it?

Answer:

A task

A mixture of lithium and sodium with a total mass of 7.6 g was oxidized with an excess of oxygen, a total of 3.92 liters (n.o.) was consumed. The resulting mixture was dissolved in 80 g of a 24.5% sulfuric acid solution. Calculate the mass fractions of substances in the resulting solution.

FROM mixture separation methods (both heterogeneous and homogeneous) are based on the fact that the substances that make up the mixture retain their individual properties. Heterogeneous mixtures may differ in composition and phase state, for example: gas + liquid; solid+liquid; two immiscible liquids, etc. The main methods for separating mixtures are shown in the diagram below. Let's consider each method separately.
Separation of heterogeneous mixtures
For separation of heterogeneous mixtures, which are solid-liquid or solid-gas systems, there are three main ways:

filtration,
settling (decanting,
magnetic separation

FILTRATION
a method based on the different solubility of substances and different particle sizes of the mixture components. Filtration separates a solid from a liquid or gas.

To filter liquids, filter paper can be used, which is usually folded into fours and inserted into a glass funnel. The funnel is placed in a beaker in which filtrate is the liquid that has passed through the filter.
The pore size in the filter paper is such that it allows water molecules and solute molecules to seep through unhindered. Particles larger than 0.01 mm are retained on the filter and do notpass through it, thus forming a layer of sediment.
Remember! With the help of filtration, it is impossible to separate true solutions of substances, that is, solutions in which dissolution occurred at the level of molecules or ions.
In addition to filter paper, chemical laboratories use special filters with

different pore sizes.
Filtration of gas mixtures is not fundamentally different from filtration of liquids. The only difference is that when filtering gases from particulate matter (SPM), filters of special designs (paper, coal) and pumps are used to force the gas mixture through the filter, for example, air filtration in a car interior or an exhaust hood over a stove.
Filtering can be divided:

cereals and water
chalk and water
sand and water, etc.
dust and air (various designs of vacuum cleaners)

SETTLEMENT
The method is based on different settling rates of solid particles with different weights (densities) in a liquid or air medium. The method is used to separate two or more solids. insoluble substances in water (or other solvent). A mixture of insoluble substances is placed in water, mixed thoroughly. After some time, substances with a density more than one settle to the bottom of the vessel, and substances with a density less than unity float. If a mixture contains several substances with different strength gravity, then in the lower layer more heavy substances and then lighter ones. These layers can also be separated. Previously, grains of gold were isolated from crushed gold-bearing rocks in this way. Gold-bearing sand was placed on an inclined chute, through which a stream of water was launched. The flow of water picked up and carried away the waste rock, and heavy grains of gold settled at the bottom of the gutter. In the case of gas mixtures, there is also the settling of solid particles on hard surfaces, such as dust settling on furniture or plant leaves.
Immiscible liquids can also be separated by this method. To do this, use a separating funnel.
For example, to separate gasoline and water, the mixture is placed in a separating funnel, waiting for the moment until a clear phase boundary appears. Then gently open the faucet and water flows into the glass.
Mixtures can be separated by settling:

river sand and clay
heavy crystalline precipitate from solution
oil and water
vegetable oil and water, etc.

MAGNETIC SEPARATION
The method is based on different magnetic properties of the solid components of the mixture. This method is used in the presence of ferromagnetic substances in the mixture, that is, substances that have magnetic properties such as iron.
All substances in relation to magnetic field can be roughly divided into three large groups:

feromagnetics: attracted by magnet - Fe, Co, Ni, Gd, Dy
paramagnets: weakly attracted-Al, Cr, Ti, V, W, Mo
diamagnets: repelled by magnet - Cu, Ag, Au, Bi, Sn, brass

Magnetic separation can separate b:

sulfur and iron powder
soot and iron, etc.

Separation of homogeneous mixtures
For separation of liquid homogeneous mixtures (true solutions) use the following methods:

evaporation (crystallization),
distillation (distillation),
chromatography.

EVAPORATION. CRYSTALLIZATION.
The method is based on different boiling points of solvent and solute. Used to isolate soluble solids from solutions. Evaporation is usually carried out as follows: the solution is poured into a porcelain cup and heated while constantly stirring the solution. The water gradually evaporates and a solid remains at the bottom of the cup.
DEFINITION
Crystallization- phase transition of a substance from a gaseous (vaporous), liquid or solid amorphous state to a crystalline state.
In this case, the evaporated substance (water or solvent) can be collected by condensation on a colder surface. For example, if you place a cold glass slide over an evaporating dish, water droplets form on its surface. The distillation method is based on the same principle.
DISTILLATION. DISTILLATION.
If a substance, such as sugar, decomposes when heated, then the water is not completely evaporated - the solution is evaporated, and then sugar crystals are precipitated from a saturated solution. Sometimes it is required to purify solvents from impurities, for example, water from salt. In this case, the solvent should be evaporated, and then its vapors should be collected and condensed on cooling. This method of separating a homogeneous mixture is called distillation, or distillation.

In nature, water in its pure form (without salts) does not occur. Oceanic, sea, river, well and spring water are varieties of salt solutions in water. However, often people need clean water that does not contain salts (used in car engines; in chemical production to obtain various solutions and substances; in the manufacture of photographs). This water is called distilled it is used in the laboratory for chemical experiments.
Distillation can be divided:

water and alcohol
oil (for various fractions)
acetone and water, etc.

CHROMATOGRAPHY
Method for separation and analysis of mixtures of substances. Based on different rates of distribution of the test substance between two phases - stationary and mobile (eluent). The stationary phase, as a rule, is a sorbent (fine powder, such as aluminum oxide or zinc oxide or filter paper) with a developed surface, and the mobile phase is a gas or liquid flow. The flow of the mobile phase is filtered through the sorbent bed or moves along the sorbent bed, for example, on the surface of filter paper.

You can get a chromatogram yourself and see the essence of the method in practice. It is necessary to mix several inks and apply a drop of the resulting mixture on filter paper. Then, exactly in the middle of the colored spot, we will begin to pour clean water drop by drop. Each drop should be applied only after the previous one has been absorbed. Water plays the role of an eluent that transfers the test substance along the sorbent - porous paper. The substances that make up the mixture are retained by paper in different ways: some are well retained by it, while others are absorbed more slowly and continue to spread along with water for some time. Soon, a real colorful chromatogram will begin to spread across a sheet of paper: a spot of the same color in the center, surrounded by multi-colored concentric rings.
Thin-layer chromatography has become especially widespread in organic analysis. The advantage of thin layer chromatography is that it is possible to use the simplest and most sensitive detection method - visual control. Spots invisible to the eye can be developed using various reagents, as well as using ultraviolet light or autoradiography.
In the analysis of organic and inorganic substances using paper chromatography. Numerous methods have been developed for the separation of complex mixtures of ions, such as mixtures of rare earth elements, fission products of uranium, elements of the platinum group
MIXTURE SEPARATION METHODS USED IN INDUSTRY.
Methods for separating mixtures used in industry differ little from the laboratory methods described above.
Rectification (distillation) is most often used to separate oil. This process is described in more detail in the topic. "Oil refining".
The most common methods of purification and separation of substances in industry are settling, filtration, sorption and extraction. Filtration and settling methods are carried out similarly to the laboratory method, with the difference that settling tanks and large volume filters are used. Most often, these methods are used to clean Wastewater. Therefore, let's take a closer look at the methods extraction And sorption.
The term "extraction" can be applied to various phase equilibria (liquid-liquid, gas-liquid, liquid-solid, etc.), but more often it is applied to liquid-liquid systems, so the following definition can often be found:
DEFINITION
Extraction i - a method of separation, purification and isolation of substances, based on the process of distribution of a substance between two immiscible solvents.
One of the immiscible solvents is usually water, the other is an organic solvent, but this is not required. The extraction method is versatile; it is suitable for isolating almost all elements in various concentrations. Extraction allows you to separate complex multicomponent mixtures often more efficiently and faster than other methods. Performing an extraction separation or separation does not require complex and expensive equipment. The process can be automated, if necessary, it can be controlled remotely.
DEFINITION
Sorption- a method of isolation and purification of substances based on absorption solid(adsorption) or liquid-sorbent (absorption) of various substances (sorbates) from gas or liquid mixtures.
Most often in industry, absorption methods are used to clean gas-air emissions from dust particles or smoke, as well as toxic gaseous substances. In the case of absorption of gaseous substances, a chemical reaction can occur between the sorbent and the solute. For example, when absorbing gaseous ammoniaNH3a solution of nitric acid HNO 3 forms ammonium nitrate NH 4 NO 3(ammonium nitrate), which can be used as a highly effective nitrogen fertilizer.

12 16 ..

2.6. Processes for the separation of heterogeneous mixtures in food production

2.6.1. Classification of inhomogeneous systems and methods for their separation I

Heterogeneous systems are mixtures of at least two components that are in different phase states and separated by clear boundaries. In such systems, two phases of matter can be distinguished: a continuously distributed continuum of a phase called dispersion environment, and fragmented particles of various sizes and shapes located in it - dispersed phase. The particles of the dispersed phase have clear boundaries separating them from the dispersion medium. Inhomogeneous systems are also called heterogeneous or dispersed.The disperse medium of inhomogeneous systems can be in three states of aggregation. The dispersed phase can also be in these states. Theoretically, the existence of 9 inhomogeneous systems is possible. However, according to this classification, an inhomogeneous gas-gas (G-G) system does not exist, since the mixture of gases is a homogeneous system. In the above classification of inhomogeneous systems, it is also necessary to single out systems with solid phases T-Zh, T-G, T-T, which are not subject to separation and therefore cannot be considered heterogeneous.

Thus, dusts, fumes, mists, suspensions, emulsions and foams should be classified as heterogeneous systems.

Dust- an inhomogeneous system consisting of a gas and solid particles distributed in it with a size of 5 - 50 microns. It is formed mainly during crushing and transportation of solid materials.

Smoke- an inhomogeneous system consisting of a gas and solid particles distributed in it with a size of 0.3 - 5 microns. It is formed during the combustion of substances.

Fog- an inhomogeneous system consisting of a gas and liquid droplets 0.3 - 3 μm in size distributed in it, formed as a result of condensation.

Dusts, fumes, mists bear the common name aerosols.

Suspension- an inhomogeneous system consisting of a liquid and solid particles suspended in it. Depending on the size of the particles, suspensions are distinguished: rough with particles larger than 100 microns, thin with particles larger than 0.1 - 100 microns and colloidal solutions containing particles smaller than 0.1 µm.

Emulsion- an inhomogeneous system consisting of a liquid and drops of another liquid distributed in it, which does not dissolve in the first one. The size of the particles of the dispersed phase varies within a fairly wide range.

Foam- an inhomogeneous system consisting of a liquid and gas bubbles distributed in it.

When the concentration of the dispersed phase changes, an inhomogeneous system can change its structure. This is accompanied by the so-called inversion phases. With inversion, the dispersion medium becomes a dispersed phase and vice versa. Thus, with an increase in the concentration of the solid phase in suspensions, a moment may come when the solid phase forms a continuous continuum (continuous medium) in which limited volumes of the liquid dispersed phase are distributed. In this case, it can be argued about the transition of the suspension into a plastic mass of class T-Zh.

Similar changes occur with foam if the liquid content in it increases; it passes into a supersaturated carbonated liquid, in which the dispersed phase of gas bubbles can be distinguished. Such a system is not sufficiently stable, although it can remain in this state for a relatively long time.

With an increase in the concentration of the solid dispersed phase, dust passes into a bulk product with specific properties, i.e. both solid and liquid media. Such a system has some elasticity and plasticity (the ability to maintain its shape under relatively small loads), however, it takes the form of a container into which it is filled; when poured onto a plane, it forms a cone with an angle of repose.

To separate inhomogeneous systems, methods and equipment are used that are distinguished by a wide variety of physical phenomena. The choice of the optimal equipment is determined by the choice of a sign according to which the dispersion medium and the dispersed phase differ significantly in their properties and according to which they should be separated. Such features are: density, strength, magnetic and electronic properties, etc. It is by the use of one or more of these features that the methods of separating these systems differ.

A sign consisting in the difference in densities that make up an inhomogeneous system is used in the following separation methods: deposition due to gravity, settling centrifugation (separation) and cyclone process.

In conservative force fields(forces of gravity, centrifugal forces, inertial forces), the particles of the dispersed phase acquire acceleration, which, according to Newton's second law, is proportional to the acting force and inversely proportional to the mass of the particles. In solution, the particles begin to move in the dispersion medium in the direction of the vector operating force. Their velocities eventually stabilize at a level corresponding to the balance of the driving force and the resistance forces of the medium. With a given speed, all "heavy" and denser than the dispersion medium particles settle on the hard surfaces of the equipment.

The sign, consisting in the difference in the magnetic properties that make up an inhomogeneous system, is used to isolate particles of metallomagnetic inclusions from a dispersion medium. In this case, under the action of magnetic forces, metal-magnetic particles are accelerated in the direction of their action, while the environment remains stationary. Due to this, phase separation occurs in space.

A sign based on the difference in electrical properties that make up an inhomogeneous system is used in electrostatic precipitators. Under the influence of high electrical voltage particles of the dispersed phase can be ionized and move in space to the filter electrodes.

The feature, which consists in the retention of particles of the dispersed phase on solid partitions, is used in processes filtering(due to pressure difference and centrifugal filtration).

A sign associated with the association of dispersed particles into larger complexes is used in the separation of dusty gas systems wet way.

It is also possible to combine methods for separating heterogeneous systems.

2.6.2. Material balances of separation processes

Consider an inhomogeneous system, for example, a suspension to be separated and consisting of a substance (continuous phase) and particles of the substance (dispersed phase) distributed in it.

Let's designate: - weights of the initial mix, the clarified liquid and the received deposit; - the content of the substance in the initial mixture, clarified liquid and sediment (mass fractions).

In the absence of losses in the separation process, the material balance equations have the form:

by total amount of substances

by dispersed phase (substance)

The joint solution of the equations makes it possible to determine the amount of clarified liquid and the amount of sediment obtained at a given content of the substance in the sediment and clarified liquid.

If dispersed particles are released slowly from the medium or it is necessary to pre-clarify an inhomogeneous system, methods such as flocculation, flotation, classification, coagulation, etc. are used.

Coagulation is the process of sticking together of particles in colloidal systems (emulsions or suspensions) with the formation of aggregates. Sticking occurs due to the collision of particles during Brownian motion. Coagulation refers to a spontaneous process that tends to move into a state that has a lower free energy. The coagulation threshold is the minimum concentration of an injected substance that causes coagulation. Artificial coagulation can be accelerated by adding special substances - coagulators to the colloidal system, as well as by adding to the system electric field(electrocoagulation), mechanical action (vibration, mixing), etc.

When coagulating, it is often added to the shared heterogeneous mixture coagulant chemicals that destroy the solvated shells, while reducing the diffusion part of the electrical double layer located near the surface of the particles. This facilitates the agglomeration of particles and the formation of aggregates. Thus, due to the formation of larger fractions of the dispersed phase, particle settling is accelerated. Salts of iron, aluminum or salts of other polyvalent metals are used as coagulants.

Peptization is the reverse process of coagulation, which is the breakdown of aggregates into primary particles. Peptization is carried out by adding peptizing substances to the dispersion medium. This process contributes to the disaggregation of substances into primary particles. Peptizing agents can be surface-active substances (surfactants) or electrolytes such as humic acids or ferric chloride. The peptization process is used to obtain liquid dispersion systems from pastes or powders.

In turn, flocculation is a kind of coagulation. In this process, small particles that are suspended in gas or liquid media form flocculent aggregates called floccules. Soluble polymers, such as polyelectrolytes, are used as flocculants. Flocculating substances can be easily removed by filtration or settling. Flocculation is used for water treatment and the separation of valuable substances from wastewater, as well as for mineral processing. In the case of water treatment, flocculants are used in low concentrations (from 0.1 to 5 mg/l).

In order to destroy aggregates in liquid systems, additives are used that induce charges on particles that prevent their convergence. This effect can also be achieved by changing the pH of the medium. This method is called deflocculation.

Flotation is the process of separating solid hydrophobic particles from a continuous liquid phase by selectively fixing them at the interface between the liquid and gaseous phases (the contact surface of liquid and gas or the surface of bubbles in the liquid phase). The resulting system of solid particles and gas inclusions is removed from the surface of the liquid phase. This process is used not only to remove particles of the dispersed phase, but also to separate different particles due to differences in their wettability. In this process, hydrophobic particles are fixed at the interface and separated from hydrophilic particles that settle to the bottom. The best flotation results occur when the particle size is between 0.1 and 0.04 mm.

There are several types of flotation: foam, oil, film, etc. The most common is froth flotation. This process allows the particles treated with reagents to be carried to the surface of the water with the help of air bubbles. This allows the formation of a foam layer, the stability of which is controlled by a foaming agent.

The classification is used in devices of variable cross section. With its help, it is possible to separate a certain amount of small particles from the main product, consisting of large particles. Classification is carried out using centrifuges and hydrocyclones due to the effect of centrifugal force.

The separation of suspensions using magnetic processing systems is a very promising method. Water that has been treated in a magnetic field retains changed properties for a long time, for example, reduced wetting ability. This process makes it possible to intensify the separation of suspensions.

theoretical block.

The concept of "mixture" was defined in the 17th century. English scientist Robert Boyle: "A mixture is an integral system consisting of heterogeneous components."

Comparative characteristics of a mixture and a pure substance

Signs of comparison	pure substance	Mixture
	Constant	fickle
Substances	Same	Various
Physical Properties	Permanent	Fickle
Energy change during formation	going on	Not happening
Separation	Through chemical reactions	Physical methods

Mixtures differ from each other in appearance.

The classification of mixtures is shown in the table:

Here are examples of suspensions (river sand + water), emulsions (vegetable oil + water) and solutions (air in a flask, salt + water, small change: aluminum + copper or nickel + copper).

Methods for separating mixtures

In nature, substances exist in the form of mixtures. For laboratory research, industrial production, for the needs of pharmacology and medicine, pure substances are needed.

Various methods of separation of mixtures are used to purify substances.

Evaporation - the separation of solids dissolved in a liquid by converting it into vapor.

Distillation- distillation, separation of substances contained in liquid mixtures according to boiling points, followed by cooling of the vapor.

Filtration is the filtering of liquids (gases) through a filter in order to purify them from solid impurities.

These methods are based on differences in the physical properties of the components of the mixture.

Consider ways to separate heterogeneousand homogeneous mixtures.

Blend example	Separation method
Suspension - a mixture of river sand with water	settling Separation upholding based on different densities of substances. Heavier sand settles to the bottom. You can also separate the emulsion: to separate oil or vegetable oil from water. In the laboratory, this can be done using a separating funnel. Oil or vegetable oil forms the top, lighter layer. As a result of settling, dew falls out of the fog, soot is deposited from smoke, cream is settled in milk. Separation of a mixture of water and vegetable oil by settling
A mixture of sand and table salt in water	Filtration What is the basis for the separation of heterogeneous mixtures using filtering? On various solubility of substances in water and on various sizes of particles. Only particles of substances commensurate with them pass through the pores of the filter, while larger particles are retained on the filter. So you can separate a heterogeneous mixture of table salt and river sand. Various porous substances can be used as filters: cotton wool, coal, fired clay, pressed glass, and others. The filtering method is the basis for the operation of household appliances, such as vacuum cleaners. It is used by surgeons - gauze bandages; drillers and workers of elevators - respiratory masks. With the help of a tea strainer for filtering tea leaves, Ostap Bender, the hero of the work of Ilf and Petrov, managed to take one of the chairs from Ellochka Ogre (“The Twelve Chairs”). Separation of a mixture of starch and water by filtration
A mixture of iron powder and sulfur	Action by magnet or water Iron powder was attracted by a magnet, but sulfur powder was not. The non-wetting sulfur powder floated to the surface of the water, while the heavy wettable iron powder settled to the bottom. Separation of a mixture of sulfur and iron using a magnet and water
A solution of salt in water is a homogeneous mixture	Evaporation or crystallization The water evaporates and salt crystals remain in the porcelain cup. When water is evaporated from lakes Elton and Baskunchak, table salt is obtained. This separation method is based on the difference in the boiling points of the solvent and the solute. If a substance, such as sugar, decomposes when heated, then the water is not completely evaporated - the solution is evaporated, and then sugar crystals are precipitated from a saturated solution. Sometimes it is required to remove impurities from solvents with a lower boiling point, for example, water from salt. In this case, the vapors of the substance must be collected and then condensed upon cooling. This method of separating a homogeneous mixture is called distillation or distillation. In special devices - distillers, distilled water is obtained, which is used for the needs of pharmacology, laboratories, and car cooling systems. At home, you can design such a distiller: If, however, a mixture of alcohol and water is separated, then the first to be distilled off (collected in a receiving test tube) is alcohol with tboil = 78 °C, and water will remain in the test tube. Distillation is used to obtain gasoline, kerosene, gas oil from oil. Separation of homogeneous mixtures

A special method of separating components, based on their different absorption by a certain substance, is chromatography.

Using chromatography, the Russian botanist was the first to isolate chlorophyll from the green parts of plants. In industry and laboratories, instead of filter paper for chromatography, starch, coal, limestone, and aluminum oxide are used. Are substances always required with the same degree of purification?

Methods for expressing the composition of mixtures.

· Mass fraction of the component in the mixture- the ratio of the mass of the component to the mass of the entire mixture. Usually the mass fraction is expressed in %, but not necessarily.

ω ["omega"] = mcomponent / mmixture

· Mole fraction of a component in a mixture- the ratio of the number of moles (amount of substance) of the component to the total number of moles of all substances in the mixture. For example, if the mixture includes substances A, B and C, then:

χ [“chi”] component A \u003d n component A / (n (A) + n (B) + n (C))

· Molar ratio of components. Sometimes in tasks for a mixture, the molar ratio of its components is indicated. For example:

ncomponent A: ncomponent B = 2: 3

· Volume fraction of the component in the mixture (only for gases)- the ratio of the volume of substance A to the total volume of the entire gas mixture.

φ ["phi"] = Vcomponent / Vmixture

Practice block.

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.a.) 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.

n \u003d V / Vm \u003d 5.6 / 22.4 \u003d 0.25 mol.

2. According to the reaction equation:

3. The amount of iron is also 0.25 mol. You can find its mass:
mFe = 0.25 56 = 14 g.

Answer: 70% iron, 30% 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.a.) were released. Determine the mass fractions of metals in the mixture.

Example 2 solution.

1. Find the amount of hydrogen:
n \u003d V / Vm \u003d 8.96 / 22.4 \u003d 0.4 mol.

2. Let the amount of aluminum be x mol, and iron y mol. Then we can express in terms of x and y the amount of hydrogen released:


	2HCl = FeCl2 +

4. We know the total amount of hydrogen: 0.4 mol. Means,
1.5x + y = 0.4 (this is the first equation in the system).

5. For a mixture of metals, you need to express masses through quantities of substances.
m = Mn
So the mass of aluminum
mAl = 27x,
mass of iron
mFe = 56y,
and the mass of the whole mixture
27x + 56y = 11 (this is the second equation in the system).

6. So, we have a system of two equations:

7. Solving such systems is much more convenient by subtracting by multiplying the first equation by 18:
27x + 18y = 7.2
and subtracting the first equation from the second:

8. (56 - 18)y \u003d 11 - 7.2
y \u003d 3.8 / 38 \u003d 0.1 mol (Fe)
x = 0.2 mol (Al)

mFe = n M = 0.1 56 = 5.6 g
mAl = 0.2 27 = 5.4 g
ωFe = mFe / mmixture = 5.6 / 11 = 0.50.91%),

respectively,
ωAl \u003d 100% - 50.91% \u003d 49.09%

Answer: 50.91% iron, 49.09% 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 liters of gas (n.a.) were released and 5 g of the substance did not dissolve. Determine the mass fractions of metals in the mixture.

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

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.a.) 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 also dissolve 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:
nSO2 = V / Vm = 5.6 / 22.4 = 0.25 mol


	2H2SO4 (conc.) = CuSO4 +

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

3. Since the molar ratio of copper and sulfur dioxide is 1:1, then copper is also 0.25 mol. You can find the mass of copper:
mCu \u003d n M \u003d 0.25 64 \u003d 16 g.

4. Aluminum reacts with an alkali solution, and an aluminum hydroxocomplex and hydrogen are formed:
2Al + 2NaOH + 6H2O = 2Na + 3H2

Al0 − 3e = Al3+

5. Number of moles of hydrogen:
nH2 = 3.36 / 22.4 = 0.15 mol,
the molar ratio of aluminum and hydrogen is 2:3 and, therefore,
nAl = 0.15 / 1.5 = 0.1 mol.
Aluminum weight:
mAl \u003d n M \u003d 0.1 27 \u003d 2.7 g

6. The remainder is iron, weighing 3 g. You can find the mass of the mixture:
mmix \u003d 16 + 2.7 + 3 \u003d 21.7 g.

7. Mass fractions of metals:

ωCu = mCu / mmixture = 16 / 21.7 = 0.7.73%)
ωAl = 2.7 / 21.7 = 0.1.44%)
ωFe = 13.83%

Answer: 73.73% copper, 12.44% aluminum, 13.83% 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. % HNO3 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.a.). 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:
nN2 = V / Vm = 2.912 / 22.4 = 0.13 mol.

2. Determine the mass of the nitric acid solution, the mass and amount of the dissolved HNO3 substance:

msolution \u003d ρ V \u003d 1.115 565 \u003d 630.3 g
mHNO3 = ω msolution = 0.2 630.3 = 126.06 g
nHNO3 = m / M = 126.06 / 63 = 2 mol

3. Compose reaction equations ( do not forget about the electronic balance) and, for convenience of calculations, we take for 5x - the amount of zinc, and for 10y - the amount of aluminum. Then, in accordance with the coefficients in the equations, nitrogen in the first reaction will be x mol, and in the second - 3y mol:


	12HNO3 = 5Zn(NO3)2 +

Zn0 − 2e = Zn2+


	36HNO3 = 10Al(NO3)3 +

Al0 − 3e = Al3+

5. Then, given that the mass of the mixture of metals is 21.1 g, their molar masses are 65 g/mol for zinc and 27 g/mol for aluminum, we obtain the following system of equations:

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

7. x \u003d 0.04, which means nZn \u003d 0.04 5 \u003d 0.2 mol
y \u003d 0.03, which means that nAl \u003d 0.03 10 \u003d 0.3 mol

8. Check the mass of the mixture:
0.2 65 + 0.3 27 \u003d 21.1 g.

9. 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):

10. 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:
nHNO3 = 0.48 + 1.08 = 1.56 mol,
i.e. the acid was in excess and you can calculate its remainder in the solution:
nHNO3res. \u003d 2 - 1.56 \u003d 0.44 mol.

11. So, in final solution contains:

zinc nitrate in the amount of 0.2 mol:
mZn(NO3)2 = n M = 0.2 189 = 37.8 g
aluminum nitrate in the amount of 0.3 mol:
mAl(NO3)3 = n M = 0.3 213 = 63.9 g
an excess of nitric acid in an amount of 0.44 mol:
mHNO3res. = n M = 0.44 63 = 27.72 g

12. 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):

13.
Then for our task:

14. new solution \u003d mass of acid solution + mass of metal alloy - mass of nitrogen
mN2 = n M = 28 (0.03 + 0.09) = 3.36 g
new solution \u003d 630.3 + 21.1 - 3.36 \u003d 648.04 g

ωZn(NO3)2 \u003d mv-va / mr-ra \u003d 37.8 / 648.04 \u003d 0.0583
ωAl(NO3)3 \u003d mv-va / mr-ra \u003d 63.9 / 648.04 \u003d 0.0986
ωHNO3res. \u003d mv-va / mr-ra \u003d 27.72 / 648.04 \u003d 0.0428

Answer: 5.83% zinc nitrate, 9.86% aluminum nitrate, 4.28% nitric acid.

Example 6When processing 17.4 g of a mixture of copper, iron and aluminum with an excess of concentrated nitric acid, 4.48 liters of gas (n.a.) were released, and when this mixture was exposed to the same mass of excess hydrochloric acid, 8.96 l of gas (n.a.). u.). 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 NO2, while iron and aluminum do not react with it. Hydrochloric acid, on the other hand, does not react with copper.

Answer for example 6: 36.8% copper, 32.2% iron, 31% aluminum.

Tasks for independent solution.

1. Simple problems with two mixture components.

1-2. A mixture of copper and zinc weighing 10 g was treated with a concentrated alkali solution. In this case, 2.24 liters of gas (n. y.) were released. Calculate the mass fraction of zinc in the initial mixture.

1-3. A mixture of magnesium and magnesium oxide weighing 6.4 g was treated with a sufficient amount of dilute sulfuric acid. At the same time, 2.24 liters of gas (n.a.) were released. Find the mass fraction of magnesium in the mixture.

1-4. A mixture of zinc and zinc oxide weighing 3.08 g was dissolved in dilute sulfuric acid. Zinc sulfate weighing 6.44 g was obtained. Calculate the mass fraction of zinc in the initial mixture.

1-6. What mass of a 20% hydrochloric acid solution will be required to completely dissolve 20 g of a mixture of zinc with zinc oxide, if hydrogen is released in the amount of 4.48 liters (n.a.)?

1-7. When dissolved in dilute nitric acid, 3.04 g of a mixture of iron and copper releases nitric oxide (II) with a volume of 0.896 l (n.a.). Determine the composition of the initial mixture.

1-8. When dissolving 1.11 g of a mixture of iron and aluminum filings in a 16% hydrochloric acid solution (ρ = 1.09 g / ml), 0.672 liters of hydrogen (n.a.) were 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 18.8 g was calcined without access to air with an excess of graphite powder. The reaction product was treated with dilute hydrochloric acid, and 11.2 liters of gas (n.a.) were released. Determine the mass fractions of metals in the mixture.

2-2. To dissolve 1.26 g of an alloy of magnesium with aluminum, 35 ml of a 19.6% sulfuric acid solution (ρ = 1.1 g/ml) was used. The excess acid reacted with 28.6 ml of a 1.4 mol/L potassium hydrogen carbonate solution. Determine the mass fractions of metals in the alloy and the volume of gas (n.a.) released during the dissolution of the alloy.