The name for a rapidly cooled mixture of water and sand. Methods for separating mixtures and purifying substances. Physical methods for separating mixtures

1. Fill in the blanks in the text using the words “components”, “differences”, “two”, “physical”.

A mixture can be prepared by mixing at least two substances. Mixtures can be separated into individual components using physical methods based on differences physical properties components.

2. Complete the sentences.

a) The settling method is based on The fact is that the particles of the solid substance are quite large; they quickly settle to the bottom, and the liquid can be carefully drained from the sediment.

b) The centrifugation method is based on the action of centrifugal force - heavier particles settle, and light ones end up on top.

c) The filtering method is based on passing a solution of a solid through a filter where the solid particles are retained on the filter.

3. Fill in the missing word:

a) flour and granulated sugar - a sieve; sulfur and iron filings - magnet.

b) water and sunflower oil - separating funnel; water and river sand - filter.

c) air and dust - respirator; air and poisonous gas - absorbent.

4. Make a list of necessary filtration equipment.

a) paper filter
b) a glass with a solution
c) glass funnel
d) clean glass
d) glass rod
e) tripod with foot

5. Laboratory experience. Making regular and pleated filters from filter paper or paper napkin.

Which filter do you think the solution will pass through faster - a regular one or a folded one? Why?

Through folded - the filtration contact area is larger than that of a conventional filter.

6. Suggest ways to separate the mixtures shown in Table 16.

Methods for separating some mixtures

7. Home experience. Adsorption of Pepsi-Cola colorants by activated carbon.

Reagents and equipment: carbonated drink, activated carbon; pan, funnel, filter paper, electric (gas) stove.

Progress. Pour half a cup (100 ml) of carbonated drink into the pan. Add 5 tablets of activated carbon there. Heat the pan for 10 minutes on the stove. Filter the carbon. Explain the results of the experiment.

The solution became discolored due to the absorption of colorants by activated carbon.

8. Home experience. Adsorption of odorous vapors by corn sticks.

Reagents and equipment: corn sticks, perfume or cologne; 2 identical glass jars with lids.

Progress. Place a drop of perfume into two glass jars. Place 4-5 corn sticks in one of the jars. Cover both jars with lids. Shake the jar containing the corn sticks a little. For what?

To increase the rate of adsorption.

Open both jars. Explain the results of the experiment.

There is no smell in the jar where the corn sticks were, since it adsorbed the smell of perfume.

Abstract on the discipline: Chemistry

On the topic: Methods for separating mixtures

Riga - 2009

Introduction……………………………………………………………………………………..page 3

Types of mixtures…………………………………………………………………………………page 4

Methods for separating mixtures……………………………………………………..page 6

Conclusion………………………………………………………………………………….page 11

List of used literature……………………………………………………......page 12

Introduction

In nature, substances in their pure form are very rare. Most of the objects around us consist of a mixture of substances. In a chemistry laboratory, chemists work with pure substances. If the substance contains impurities, then any chemist can separate the substance needed for the experiment from the impurities. To study the properties of substances, it is necessary to purify this substance, i.e. divide into component parts. Separating a mixture is physical process. Physical methods for separating substances are widely used in chemical laboratories, in the production of food products, and in the production of metals and other substances.

Types of mixtures

There are no pure substances in nature. When examining boulders and granite, we are convinced that they consist of grains and veins of different colors; Milk contains fats, proteins, and water; oil and natural gas contain organic substances called hydrocarbons; air contains various gases; natural water is not a chemically pure substance. A mixture is a mixture of two or more dissimilar substances.

Mixtures can be divided into two large groups (ri

If the components of a mixture are visible to the naked eye, then such mixtures are called heterogeneous. For example, a mixture of wood and iron filings, a mixture of water and vegetable oil, a mixture of river sand and water, etc.

If the components of a mixture cannot be distinguished with the naked eye, then such mixtures are called homogeneous. Mixtures such as milk, oil, sugar solution in water, etc. are classified as homogeneous mixtures.

There are solid, liquid, gaseous substances. Substances can be mixed in any state of aggregation. The state of aggregation of a mixture is determined by the substance that is quantitatively superior to the others.

Heterogeneous mixtures are formed from substances of different states of aggregation, when the substances do not dissolve mutually and do not mix well (Table 1)

Types of heterogeneous mixtures
before mixing	Examples
Hard/solid	Minerals;
iron/sulfur	Solid/liquid
Lime mortar; wastewater	Solid/Gaseous
Smoke; dusty air	Liquid/solid
Pearl; minerals; water/ice	Liquid/liquid
Milk; vegetable oil/water	Liquid/Gaseous
Fog; clouds	Gaseous/solid
Styrofoam	Gaseous/liquid

Soap suds

Homogeneous mixtures are formed when substances dissolve well in each other and mix well (Table 2).
Types of homogeneous mixtures before mixing	Examples
Hard/solid	Physical state of the components
iron/sulfur	Alloy of gold and silver
Lime mortar; wastewater	Sugar/water
Smoke; dusty air	Iodine vapor in the air
Pearl; minerals; water/ice	Swollen gelatin
Milk; vegetable oil/water	Alcohol/water
Fog; clouds	Water/air
Styrofoam

Hydrogen in palladium When forming mixtures chemical transformations

usually does not occur, and the substances in the mixture retain their properties. Differences in the properties of substances are used to separate mixtures.

Methods for separating mixtures Mixtures, both heterogeneous and homogeneous, can be divided into component parts, i.e. for pure substances. Pure substances are substances that, using physical methods, cannot be separated into two or more other substances and do not change their physical properties. Exist various ways

1. separating mixtures, certain methods of separating mixtures are used depending on the composition of the mixture.
2. Screening;
3. Filtration;
4. Advocacy;
5. Decantation
6. Centrifugation;
7. Evaporation;
8. Evaporation;
9. Recrystallization;
10. Distillation (distillation);
11. Freezing;
12. Magnet action;
13. Chromatography;
14. Extraction;

Adsorption.

Let's get to know a few of them. It should be noted here that inhomogeneous mixtures are easier to separate than homogeneous ones. Below we give examples of separating substances from homogeneous and inhomogeneous mixtures.

Screening. Let's imagine that granulated sugar gets into the flour. Perhaps the simplest way to separate is screening . Using a sieve, you can easily separate small particles of flour from relatively large sugar crystals. IN agriculture

sifting is used to separate plant seeds from foreign debris. In construction, this is how gravel is separated from sand.

Filtration The solid component of the suspension is separated from the liquid using paper or fabric filters, cotton wool, a thin layer of fine sand. Let's imagine that we are given a mixture of table salt, sand and clay. It is necessary to separate table salt from the mixture. To do this, place the mixture in a beaker with water and shake. Table salt dissolves and the sand settles. The clay does not dissolve and does not settle to the bottom of the glass, so the water remains cloudy. To remove insoluble clay particles from the solution, the mixture is filtered. To do this, you need to assemble a small filtering device from a glass funnel, filter paper and a tripod. The salt solution is filtered. To do this, the filtered solution is carefully poured into a funnel with a tightly inserted filter. Sand and clay particles remain on the filter, and a clear salt solution passes through the filter. To isolate table salt dissolved in water, the method of recrystallization is used.

Recrystallization, evaporation

Recrystallization is a purification method in which a substance is first dissolved in water, then the solution of the substance in water is evaporated. As a result, the water evaporates and the substance is released in the form of crystals.
Let's give an example: You need to separate table salt from a solution.
Above we looked at an example when it was necessary to isolate table salt from a heterogeneous mixture. Now let's separate the table salt from the homogeneous mixture. The solution obtained by filtration is called filtrate. The filtrate should be poured into a porcelain cup. Place the cup with the solution on the tripod ring and heat the solution over the flame of an alcohol lamp. The water will begin to evaporate and the volume of the solution will decrease. This process is called by evaporation. As the water evaporates, the solution becomes more concentrated. When the solution reaches a state of saturation with table salt, crystals will appear on the walls of the cup. At this point, stop heating and cool the solution. Cooled table salt will separate out in the form of crystals. If necessary, salt crystals can be separated from the solution by filtration. The solution should not be evaporated until the water has completely evaporated, since other soluble impurities may also precipitate in the form of crystals and contaminate the table salt.

Settling, decanting

Used to separate insoluble substances from liquids upholding. If the solid particles are large enough, they quickly settle to the bottom and the liquid becomes clear. It can be carefully drained from the sediment, and this simple operation also has its own name - decanting. The smaller the size of the solid particles in the liquid, the longer the mixture will settle. You can also separate two liquids that do not mix with each other.

Centrifugation

If the particles of a heterogeneous mixture are very small, it cannot be separated either by settling or filtering. Examples of such mixtures include milk and toothpaste stirred in water. Such mixtures are separated centrifugation. Mixtures containing such liquid are placed in test tubes and rotated at high speed in special devices - centrifuges. As a result of centrifugation, the heavier particles are “pressed” to the bottom of the vessel, and the lighter ones end up on top. Milk is tiny particles of fat distributed in an aqueous solution of other substances - sugars, proteins. To separate such a mixture, a special centrifuge called a separator is used. When milk is separated, fats appear on the surface and are easy to separate. What remains is water with substances dissolved in it - this is skim milk.

Adsorption

In technology, the task often arises of purifying gases, such as air, from unwanted or harmful components. Many substances have one interesting property - they can “catch” to the surface of porous substances, like iron to a magnet. Adsorption is the ability of some solid substances to absorb gaseous or dissolved substances on their surface. Substances capable of adsorption are called adsorbents. Adsorbents are solid substances in which there are many internal channels, voids, pores, i.e. they have a very large total absorbing surface. Adsorbents are activated carbon, silica gel (in the box with new shoes you can find a small bag of white peas - this is silica gel), filter paper. Different substances “attach” to the surface of adsorbents differently: some are held firmly on the surface, others are held weaker. Activated carbon is capable of absorbing not only gaseous substances, but also substances dissolved in liquids. In case of poisoning, it is taken so that toxic substances are adsorbed on it.

Distillation (distillation)

Two liquids that form a homogeneous mixture, for example, ethyl alcohol and water, are separated by distillation or distillation. This method is based on the fact that the liquid is heated to boiling point and its vapor is discharged through a gas outlet tube into another vessel. As the steam cools, it condenses, leaving impurities in the distillation flask. The distillation device is shown in Fig. 2

The liquid is placed in a Wurtz flask (1), the neck of the Wurtz flask is tightly closed with a stopper with a thermometer inserted into it (2), and the reservoir with mercury should be at the level of the outlet tube opening. The end of the outlet tube is inserted through a tightly fitted plug into the Liebig refrigerator (3), at the other end of which the allonge (4) is strengthened. The narrowed end of the allonge is lowered into the receiver (5). The lower end of the refrigerator jacket is connected using a rubber hose to the water tap, and a drain is made from the upper end into the sink for draining. The refrigerator jacket should always be filled with water. The Wurtz flask and refrigerator are mounted in separate stands. The liquid is poured into the flask through a funnel with a long tube, filling the distillation flask to 2/3 of its volume. To ensure uniform boiling, place several boilers on the bottom of the flask - glass capillaries sealed at one end. After closing the flask, add water to the refrigerator and heat the liquid in the flask. Heating can be carried out on a gas burner, electric stove, water, sand or oil bath - depending on the boiling point of the liquid. Flammable and combustible liquids (alcohol, ether, acetone, etc.) should never be heated over an open fire to avoid accidents: only a water or other bath should be used. The liquid should not be completely evaporated: 10-15% of the initially taken volume should remain in the flask. A new portion of liquid can be poured only when the flask has cooled down a little.

Freezing

Substances that have different melting points are separated using the method freezing, cooling the solution. By freezing you can get very pure water at home. To do this, pour tap water into a jar or mug and place it in the freezer of the refrigerator (or take it out into the cold in winter). As soon as about half of the water turns into ice, the unfrozen part of it, where impurities accumulate, must be poured out and the ice allowed to melt.

In industry and laboratory conditions They use methods for separating mixtures based on other different properties of the components of the mixture. For example, iron filings can be separated from a mixture magnet. The ability of substances to dissolve in various solvents is used when extraction– the method of separating solid or liquid mixtures by treating them with various solvents. For example, iodine from aqueous solution can be isolated by some organic solvent in which iodine dissolves better.

Conclusion

In laboratory practice and in Everyday life Very often it is necessary to isolate individual components from a mixture of substances. Note that mixtures include two or more substances and are divided into two large groups: homogeneous and heterogeneous. There are various ways to separate mixtures, such as filtering, evaporation, distillation (distillation) and others. Methods for separating mixtures mainly depend on the type and composition of the mixture.

List of used literature

1. S. Ozols, E. Lepiņš chemistry for primary school., 1996. P. 289

2. Information from the Internet

Theoretical block.

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

Comparative characteristics of the mixture and pure substance

Signs of comparison	Pure substance	Mixture
	Constant	Fickle
Substances	Same	Various
Physical properties	Permanent	Fickle
Energy change during formation	Happening	Not happening
Separation	By using chemical reactions	By physical methods

The mixtures differ from each other in appearance.

The classification of mixtures is shown in the table:

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

usually does not occur, and the substances in the mixture retain their properties. Differences in the properties of substances are used to separate 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 for separating mixtures are used to purify substances.

Evaporation is the separation of solids dissolved in a liquid by converting it into steam.

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

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

Filtration - straining liquids (gases) through a filter in order to clean them from solid impurities.

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

Consider separation methods heterogeneousand homogeneous mixtures.

Blend Example	Separation method
Suspension - a mixture of river sand and water	Advocacy Separation defending based on different densities of substances. Heavier sand settles to the bottom. You can also separate the emulsion: separate the oil or vegetable oil from the water. In the laboratory this can be done using a separatory funnel. Petroleum or vegetable oil forms the top, lighter layer. As a result of settling, dew falls out of the fog, soot settles out of the smoke, and cream settles in the 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 different solubility of substances in water and on different particle sizes. Only particles of substances comparable to them pass through the pores of the filter, while larger particles are retained on the filter. This way you can separate a heterogeneous mixture of table salt and river sand. Various porous substances can be used as filters: cotton wool, coal, baked clay, pressed glass and others. The filtration method is the basis for the operation of household appliances, such as vacuum cleaners. It is used by surgeons - gauze bandages; drillers and elevator workers - respiratory masks. Using a tea strainer to filter tea leaves, Ostap Bender, the hero of the work by Ilf and Petrov, managed to take one of the chairs from Ellochka the Ogress (“Twelve Chairs”). Separation of a mixture of starch and water by filtration
Mixture of iron and sulfur powder	Action by magnet or water Iron powder was attracted by a magnet, but sulfur powder was not. Non-wettable sulfur powder floated to the surface of the water, and heavy wettable iron powder settled to the bottom. Separating 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, leaving salt crystals 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 boiling points of the solvent and solute. If a substance, for example sugar, decomposes when heated, then the water is not completely evaporated - the solution is evaporated, and then saturated solution sugar crystals precipitate. Sometimes it is necessary to remove impurities from solvents with a lower boiling point, such as salt from water. 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 construct such a distiller: If you separate a mixture of alcohol and water, then the alcohol with boiling point = 78 °C will be distilled off first (collected in a receiving test tube), and water will remain in the test tube. Distillation is used to produce gasoline, kerosene, and gas oil from oil. Separation of homogeneous mixtures

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

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

For different purposes, substances with varying degrees of purification are required. Cooking water should be left to stand sufficiently to remove impurities and chlorine used to disinfect it. Water for drinking must first be boiled. And in chemical laboratories, for preparing solutions and conducting experiments, in medicine, distilled water is needed, purified as much as possible from substances dissolved in it. Particularly 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 of 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 as %, but not necessarily.

ω ["omega"] = mcomponent / mmixture

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

χ ["chi"] component A = ncomponent A / (n(A) + n(B) + n(C))

· Molar ratio of components. Sometimes problems for a mixture indicate the molar ratio of its components. 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

Practical block.

Let's look at three examples of problems in which mixtures of metals react with salt acid:

Example 1.When a mixture of copper and iron weighing 20 g was exposed to excess hydrochloric acid, 5.6 liters of gas (n.e.) 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 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.

Solution to example 1.

n = V / Vm = 5.6 / 22.4 = 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 2.When a mixture of aluminum and iron weighing 11 g was exposed to excess hydrochloric acid, 8.96 liters of gas (no.) 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 x to be the number of moles of one of the metals, and y to be the amount of substance of the second.

Solution to example 2.

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

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

	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 the amount of substances.
m = Mn
So, the mass of aluminum
mAl = 27x,
mass of iron
mFe = 56у,
and the mass of the entire mixture
27x + 56y = 11 (this is the second equation in the system).

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

7. It is much more convenient to solve such systems using the subtraction method, multiplying the first equation by 18:
27x + 18y = 7.2
and subtracting the first equation from the second:

8. (56 − 18)y = 11 − 7.2
y = 3.8 / 38 = 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 = 100% − 50.91% = 49.09%

Answer: 50.91% iron, 49.09% aluminum.

Example 3.16 g of a mixture of zinc, aluminum and copper were treated with an excess of hydrochloric acid solution. In this case, 5.6 liters of gas (n.o.) were 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 amounts 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 4.A mixture of iron, aluminum and copper was treated with an excess of cold concentrated sulfuric acid. In this case, 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 were released and 3 g of undissolved residue remained. Determine the mass and composition of the initial mixture of metals.

In this example, we must remember that cold concentrated sulfuric acid does not react with iron and aluminum (passivation), but does react with copper. This releases sulfur (IV) oxide.
With alkali reacts only aluminum- amphoteric metal (in addition to aluminum, zinc and tin also dissolve in alkalis, and beryllium can also be dissolved in hot concentrated alkali).

Solution to example 4.

1. Only copper reacts with concentrated sulfuric acid, 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 a mass of copper:
mCu = n M = 0.25 64 = 16 g.

4. Aluminum reacts with an alkali solution, resulting in the formation of a hydroxo complex of aluminum and hydrogen:
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 = n M = 0.1 27= 2.7 g

6. The remainder is iron, weighing 3 g. You can find the mass of the mixture:
mmixture = 16 + 2.7 + 3 = 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 5.21.1 g of a mixture of zinc and aluminum was dissolved in 565 ml of nitric acid solution containing 20 wt. % НNO3 and having a density of 1.115 g/ml. The volume of gas released, which is simple substance and the only product of the reduction of nitric acid, amounted to 2.912 l (no.). Determine the composition of the resulting solution in mass percent. (RHTU)

The text of this problem clearly indicates the product of nitrogen reduction - a “simple substance”. Since nitric acid with metals does not produce hydrogen, it is nitrogen. Both metals dissolved in the acid.
The problem asks not the composition of the initial mixture of metals, but the composition of the resulting solution after the reactions. This makes the task more difficult.

Solution to example 5.

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 dissolved HNO3:

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

Please note that since the metals have completely dissolved, it means - there was definitely 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 reaction equations ( don't forget about your electronic balance) and, for the convenience of calculations, we take 5x to be the amount of zinc, and 10y to be 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, taking into account 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 = 0.04, which means nZn = 0.04 5 = 0.2 mol
y = 0.03, which means nAl = 0.03 10 = 0.3 mol

8. Check the mass of the mixture:
0.2 65 + 0.3 27 = 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 above the reactions the amounts of all reacted and formed substances (except water):

10. Next question: is there any nitric acid left 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. = 2 − 1.56 = 0.44 mol.

11. So, in final solution contains:

zinc nitrate in an amount of 0.2 mol:
mZn(NO3)2 = n M = 0.2 189 = 37.8 g
aluminum nitrate in an amount of 0.3 mol:
mAl(NO3)3 = n M = 0.3 213 = 63.9 g
excess nitric acid in an amount of 0.44 mol:
mHNO3rest. = n M = 0.44 63 = 27.72 g

12. What is the mass of the final solution?
Let us remember 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. mnew solution = mass of acid solution + mass of metal alloy - mass of nitrogen
mN2 = n M = 28 (0.03 + 0.09) = 3.36 g
mnew solution = 630.3 + 21.1 − 3.36 = 648.04 g

ωZn(NO3)2 = mv-va / mr-ra = 37.8 / 648.04 = 0.0583
ωAl(NO3)3 = mv-va / mr-ra = 63.9 / 648.04 = 0.0986
ωHNO3rest. = mv-va / mr-ra = 27.72 / 648.04 = 0.0428

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

Example 6.When 17.4 g of a mixture of copper, iron and aluminum was treated with an excess of concentrated nitric acid, 4.48 liters of gas (n.e.) were released, and when this mixture was exposed to the same mass of excess hydrochloric acid, 8.96 liters of gas (n.e.) were released. y.). Determine the composition of the initial mixture. (RHTU)

When solving this problem, we must remember, firstly, that concentrated nitric acid with an inactive metal (copper) produces NO2, but iron and aluminum do not react with it. Hydrochloric acid, on the contrary, does not react with copper.

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

Problems 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.e.) 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. In this case, 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. We obtained zinc sulfate weighing 6.44 g. Calculate the mass fraction of zinc in the original mixture.

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

1-6. What mass of a 20% solution of hydrochloric acid will be required to completely dissolve 20 g of a mixture of zinc and zinc oxide, if at the same time hydrogen is released with a volume of 4.48 l (no.)?

1-7. When 3.04 g of a mixture of iron and copper is dissolved in dilute nitric acid, nitrogen oxide (II) is released with a volume of 0.896 l (no.). Determine the composition of the initial mixture.

1-8. When 1.11 g of a mixture of iron and aluminum filings was dissolved in a 16% solution of hydrochloric acid (ρ = 1.09 g/ml), 0.672 liters of hydrogen (n.e.) were released. Find the mass fractions of metals in the mixture and determine the volume of hydrochloric acid consumed.

2. The tasks are more complex.

2-1. A mixture of calcium and aluminum weighing 18.8 g was calcined without 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 magnesium-aluminium alloy, 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 potassium bicarbonate solution with a concentration of 1.4 mol/l. Determine the mass fractions of metals in the alloy and the volume of gas (no.) released during the dissolution of the alloy.

Each component of a mixture retains a set of its characteristics, so different substances can be isolated from the mixture.

Advocacy- This is a method based on different densities of substances.

For example, a mixture of vegetable oil and water can be separated into oil and water by simply letting the mixture sit.

Filtration- This is a method based on the different ability of the filter to pass the substances that make up the mixture. For example, a filter can be used to separate solid impurities from a liquid.

Evaporation- This is the separation of non-volatile solids from solution in a volatile solvent - particularly water. For example, to isolate salt dissolved in water, you simply evaporate the water. The water will evaporate, but the salt will remain.

Preparation of solutions.

Chemical pollution environment and its consequences.

A14. Qualitative reactions, indicators, gases.

A14. Determining the nature of the solution environment of acids and alkalis using indicators. Qualitative reactions to ions in solution (chloride, sulfate, carbonate ions, ammonium ion). Obtaining gaseous substances. Qualitative reactions to gaseous substances (oxygen, hydrogen, carbon dioxide, ammonia).

1) Table of indicator color changes in different environments:

2) Qualitative reactions to ions in solution.

For cations:

Cation	Reagent	Observed reaction
Li+	flame
Na+	flame	Yellow staining
K+	flame	Purple staining
Ca 2+	flame	Brick red coloring
Sr 2+	flame	Carmine red coloring
Va 2+	flame SO 4 2-	Yellow-green coloration Precipitation of a white precipitate, insoluble in acids: Ba 2+ + SO 4 2- → BaSO 4
Cu 2+	HE -	Precipitation of blue color: Cu 2+ + 2OH - → Cu(OH) 2
Pb 2+	S 2-	Black precipitate: Pb 2+ + S 2- → PbS
Ag+	Cl-	Precipitation of white precipitate; insoluble in HNO 3, but soluble in conc. NH 3 H 2 O: Ag + +Cl - →AgCl
Fe 2+	1) OH - 2) potassium hexacyanoferrate (III) (red blood salt), K 3	1) Precipitation of a light green precipitate: Fe 2+ + 2OH - → Fe (OH) 2 2) Precipitation of a blue precipitate: K + + Fe 2+ + 3- → KFe 4
Fe 3+	1) OH - 2) potassium hexacyanoferrate (II) (yellow blood salt) K 4 3) thiocyanide ion SCN -	1) Precipitation of a brown precipitate: Fe 3+ + 3OH - → Fe (OH) 3 2) Precipitation of a blue precipitate: K + + Fe 3+ + 4- → KFe 3) The appearance of a bright red color due to the formation of complex ions Fe(SCN ) 2+ , Fe(SCN) + 2
Al 3+	alkali ( amphoteric properties hydroxide)	Precipitation of aluminum hydroxide when adding the first portions of alkali and its dissolution during further addition
NH4+	lye, heating	Ammonia smell: NH 4 + + OH - → NH 3 + H 2 O
H+ (acidic environment)	indicators: litmus, methyl orange	red coloring red coloring

For anions:

Anion	Reagent	Observed reaction
SO 4 2-	Va 2+	Precipitation of a white precipitate, insoluble in acids: Ba 2+ + SO 4 2- BaSO 4
NO 3 -	Add conc. H 2 SO 4 and Si, heat	Formation of a blue solution containing Cu 2+ ions, release of a brown gas (NO 2)
RO 4 3-	Ag+ ions	Precipitation of a light yellow precipitate in a neutral environment: ZAg+ + PO 4 3- Ag 3 PO 4
S 2-	Pb 2+ ions	Black precipitate: Pb 2+ + S 2- PbS
CO 3 2-	Ca 2+ ions	Precipitation of a white precipitate, soluble in acids: Ca 2+ + CO 3 2- = CaCO3
CO2	lime water Ca(OH) 2	Ca(OH) 2 + CO 2 CaCO 3 + H 2 O, CaCO 3 + CO 2 + H 2 O Ca(HCO 3) 2 Precipitation of a white precipitate and its dissolution when passing CO 2
SO 3 2-	H+ ions	The appearance of the characteristic odor SO 2: 2H + + SO 3 2- H 2 O + SO 2
F-	Ca 2+ ions	Precipitation of a white precipitate: Ca 2+ + 2F - CaF 2
Cl-	Ag + ions	The formation of a white precipitate, insoluble in HNO 3, but soluble in conc. NH 3 H 2 O: Ag + +CI - AgCl AgCI + 2(NH 3 H 2 O) + + CI - +2H 2 O
Br-	Ag + ions	Precipitation of a light yellow precipitate, insoluble in HNO 3: Ag + + Br - = AgBr precipitate darkens in the light
I -	Ag + ions	Precipitation of a yellow precipitate, insoluble in HNO 3 and NH 3 conc.: Ag + + I - AgI precipitate darkens in the light
OH - (alkaline environment)	indicators: litmus phenolphthalein	blue coloring crimson coloring

3) Obtaining gaseous substances. Qualitative reactions to gaseous substances (oxygen, hydrogen, carbon dioxide, ammonia).

Gas ( a brief description of)	Obtaining (reaction equations)	Collecting	Recognition
Hydrogen (H 2) is the lightest, colorless, odorless.	By replacing hydrogen with metals from acid solutions: Zn + 2HCl = ZnCl 2 + H 2.	In a test tube turned upside down.	When brought close to the flame, a “popping” or “barking” sound is heard.
Oxygen (O2) is odorless and colorless, heavier than air, slightly soluble in water.	1. Decomposition of potassium permanganate: 2KMnO 4 = K 2 MnO 4 + MnO 2 + O 2 2. Decomposition of hydrogen peroxide (MnO 2 catalyst): 2H 2 O 2 = 2H 2 O + O 2	1. Air displacement.	The flashing of a smoldering splinter brought into a vessel with oxygen.
Carbon dioxide– carbon monoxide (IV) – CO 2. Colorless, odorless, non-flammable, heavier than air. Let's dissolve in water.	1.In industry: CaCO 3 = CaO + CO 2 2.In the laboratory: CaCO 3 + 2HCl = CaCl 2 + H 2 O + CO 2

heterogeneous (heterogeneous)	homogeneous (homogeneous)
Heterogeneous mixtures are those in which the interface between the original components can be identified either with the naked eye or under a magnifying glass or microscope:	Substances in such mixtures are mixed with each other as much as possible, one might say, at the molecular level. In such mixtures, it is impossible to detect the interface between the original components even under a microscope:
Examples
Suspension (solid + liquid) Emulsion (liquid + liquid) Smoke (solid + gas) Solid powder mixture (solid+solid)	True solutions (for example, a solution of table salt in water, a solution of alcohol in water) Solid solutions (metal alloys, crystalline salt hydrates) Gas solutions (a mixture of gases that do not react with each other)

Methods for separating mixtures

Heterogeneous mixtures of the gas-liquid, liquid-solid, gas-solid types are unstable in time under the influence of gravity. In such mixtures, components with a lower density gradually rise upward (float), and with a higher density, they sink down (settle). This process of spontaneous separation of mixtures over time is called defending. For example, a mixture of fine sand and water quite quickly spontaneously divides into two parts:

To speed up the process of deposition of substances with a higher density from a liquid in laboratory conditions, they often resort to a more advanced version of the settling method - centrifugation. The role of gravity in centrifuges is played by centrifugal force, which always occurs during rotation. Since centrifugal force directly depends on the speed of rotation, it can be made many times greater than the force of gravity simply by increasing the number of revolutions of the centrifuge per unit time. Thanks to this, a much faster separation of the mixture is achieved compared to settling.

After settling or centrifugation, the supernatant can be separated from the sediment using the method decanting— by carefully draining the liquid from the sediment.

You can separate a mixture of two liquids that are insoluble in each other (after settling) using a separating funnel, the principle of operation of which is clear from the following illustration:

To separate mixtures of substances located in different states of aggregation In addition to sedimentation and centrifugation, filtration is also widely used. The method consists in the fact that the filter has a different throughput in relation to the components of the mixture. Most often this is due to different particle sizes, but it may also be due to the fact that individual components of the mixture interact more strongly with the filter surface ( are adsorbed them).

For example, a suspension of a solid insoluble powder with water can be separated using a porous paper filter. The solid remains on the filter, and the water passes through it and is collected in a container located underneath it:

In some cases, heterogeneous mixtures can be separated due to the different magnetic properties of the components. For example, a mixture of sulfur and metallic iron powders can be separated using a magnet. Iron particles, unlike sulfur particles, are attracted and held by a magnet:

Separation of mixture components using magnetic field called magnetic separation.

If the mixture is a solution of a refractory solid in a liquid, this substance can be separated from the liquid by evaporating the solution:

To separate liquid homogeneous mixtures, a method called distillation, or distillation. This method has a principle of operation similar to evaporation, but allows you to separate not only volatile components from nonvolatile ones, but also substances with relatively close boiling points. One of the simplest options for distillation apparatus is shown in the figure below:

The meaning of the distillation process is that when a mixture of liquids boils, the vapors of the lighter-boiling component evaporate first. The vapors of this substance, after passing through the refrigerator, condense and flow into the receiver. The distillation method is widely used in the oil industry during primary oil refining to separate oil into fractions (gasoline, kerosene, diesel, etc.).

The distillation method also produces water purified from impurities (primarily salts). Water that has been purified by distillation is called distilled water.