Methods for separating inhomogeneous mixtures. Topic: "Methods for separating mixtures" (grade 8) 2 methods for separating a heterogeneous mixture

Theoretical block.

The definition of the concept "mixture" was given in the 17th century. english scientist Robert Boyle: "A mixture is a complete system consisting of dissimilar components."

Comparative characteristics mixture and pure substance

The mixtures differ from each other in appearance.

The classification of mixtures is shown in the table:

We will give examples of suspensions (river sand + water), emulsions (vegetable oil + water) and solutions (air in a flask, table salt + water, loose 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 mixture separation are used to purify substances.

Evaporation - the separation of solids dissolved in a liquid by means of its transformation into steam.

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

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

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

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

Consider ways to split heterogeneousand homogeneous mixtures.

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

Using chromatography, a 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 cleaning?

For different purposes, substances with different degrees of purification are required. It is enough to settle water for cooking to remove impurities and chlorine used for its disinfection. Water for drinking must first be boiled. And in chemical laboratories for the preparation of solutions and conducting 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 per cent, are used in electronics, semiconductor, nuclear technology and other precision industries.

Ways 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 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 = n component A / (n (A) + n (B) + n (C))

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

ncomponent A: ncomponent B = 2: 3

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

φ ["phi"] = Vcomponent / Vmixture

Practical block.

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

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

Solution of 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 an excess of hydrochloric acid, 8.96 liters of gas (n.u.) were released. Determine the mass fraction of metals in the mixture.

In the second example, both metal. Here, hydrogen is already liberated from the acid in both reactions. Therefore, direct calculation cannot be used here. In such cases, it is convenient to solve with the help of 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.

Solution of example 2.

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

2. Let the amount of aluminum - x mole, and iron by mole. Then you can express the amount of released hydrogen in terms of x and y:

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 quantities of substances.
m = M n
Hence, the mass of aluminum
mAl = 27x,
iron mass
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. 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:

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.u.) were released and 5 g of the substance did not dissolve. Determine the mass fraction of metals in the mixture.

In the third example, two metals react and the third metal (copper) does not react. Therefore, the remainder of 5 g is the mass of copper. The amounts of the other two metals - zinc and aluminum (note that their total mass is 16 - 5 = 11 g) can be found using the system of equations, as in example # 2.

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

Example 4.The 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.u.) were released. The remaining mixture was treated with an excess of sodium hydroxide solution. Released 3.36 liters of gas and remained 3 g of undissolved residue. 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 does react with copper. This produces sulfur (IV) oxide.
With alkali reacts only aluminum- amphoteric metal (in addition to aluminum, zinc and tin also dissolve in alkalis, in hot concentrated alkali - beryllium can still be dissolved).

Solution example 4.

1.Only copper reacts with concentrated sulfuric acid, the number of moles of gas:
nSO2 = V / Vm = 5.6 / 22.4 = 0.25 mol

2. (do not forget that such reactions must be balanced using 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 = n M = 0.25 64 = 16 g.

4. Aluminum enters into a reaction with an alkali solution, thus forming a hydroxo complex of aluminum and hydrogen:
2Al + 2NaOH + 6H2O = 2Na + 3H2

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 a nitric acid solution containing 20 wt. % HNO3 and having a density of 1.115 g / ml. The volume of the evolved gas, which is a simple substance and the only product of the reduction of nitric acid, was 2.912 L (standard). 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 give hydrogen with metals, this is nitrogen. Both metals dissolved in acid.
The problem does not ask for the composition of the initial mixture of metals, but for the composition of the solution obtained after the reactions. This makes the task more difficult.

Solution 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 the substance of the dissolved HNO3:

m solution = ρ V = 1.115 565 = 630.3 g
mHNO3 = ω m solution = 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 - acid is definitely enough(these metals do not react with water). Accordingly, it will be necessary to check was there an excess of acid, and how much of it remained after the reaction in the resulting solution.

3. We compose the reaction equations ( do not forget about 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:

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, so 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 we turn to the composition of the solution. It will be convenient to rewrite the reactions again and write down the amounts of all reacted and formed substances (except for water) over the reactions:

10. The 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 has reacted is:
nHNO3 = 0.48 + 1.08 = 1.56 mol,
that is, the acid was in excess and you can calculate its remainder in solution:
nHNO3 = 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 the amount of 0.44 mol:
mHNO3 = 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.min. 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-pa = 37.8 / 648.04 = 0.0583
ωAl (NO3) 3 = mv-va / mr-pa = 63.9 / 648.04 = 0.0986
ωHNO3res. = 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 were treated with an excess of concentrated nitric acid, 4.48 liters of gas (n.u.) were released, and when this mixture was exposed to the same mass of excess hydrochloric acid, 8.96 liters of gas (n.a.) was released. y.). Determine the composition of the original mixture. (RCTU)

When solving this problem, one must remember, firstly, that concentrated nitric acid with an inactive metal (copper) gives NO2, and iron and aluminum do not react with it. In contrast, hydrochloric acid does not react with copper.

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

Tasks for an independent solution.

1. Simple problems with two components of the mixture.

1-1. A mixture of copper and aluminum weighing 20 g was treated with a 96% solution of nitric acid, while 8.96 liters of gas (n.u.) 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. At the same time, 2.24 liters of gas (n.y.) were released. Calculate the mass fraction of zinc in the starting 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.u.) 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. Received zinc sulfate weighing 6.44, Calculate the mass fraction of zinc in the original 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 original mixture.

1-6. What mass of a 20% hydrochloric acid solution is required for the complete dissolution of 20 g of a mixture of zinc and zinc oxide, if 4.48 liters of hydrogen is released in this case?

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

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

2. The tasks are more complex.

2-1. A mixture of calcium and aluminum weighing 18.8 g was calcined in the absence of air with an excess of graphite powder. The reaction product was treated with dilute hydrochloric acid, while 11.2 liters of gas (n.u.) were evolved. Determine the mass fraction 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) were used. The excess acid reacted with 28.6 ml of a 1.4 mol / L potassium bicarbonate solution. Determine the mass fractions of metals in the alloy and the volume of gas (n.a.) released during the dissolution of the alloy.

Theoretical block.

The definition of the concept "mixture" was given in the 17th century. English scientist Robert Boyle: "A mixture is a complete system consisting of dissimilar components."

Comparative characteristics of the mixture and the pure substance

The classification of mixtures is shown in the table:

We will give examples of suspensions (river sand + water), emulsions (vegetable oil + water) and solutions (air in a flask, table salt + water, loose 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 mixture separation are used to purify substances.

Evaporation - the separation of solids dissolved in a liquid by means of its transformation into steam.

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

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

Filtration - filtering 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 ways to split heterogeneous and homogeneous mixtures.

Using chromatography, the Russian botanist MS 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 with the same degree of purification always required?

For different purposes, substances with different degrees of purification are required. It is enough to settle water for cooking to remove impurities and chlorine used for its disinfection. Water for drinking must first be boiled. And in chemical laboratories for the preparation of solutions and conducting 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 per cent, are used in electronics, semiconductor, nuclear technology and other precision industries.

Ways 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 in%, but not necessarily.

• 
  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:

• 
  The molar ratio of the components. Sometimes in problems for a mixture, the molar ratio of its components is indicated. For example:

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

Practical block.

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

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

Solution of example 1.

1. 
   Find the amount of hydrogen:
   n = V / V m = 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:
   m Fe = 0.25 56 = 14 g.
4. 
   Now you can calculate the mass fractions of metals in the mixture:
   ω Fe = m Fe / m of the whole mixture = 14/20 = 0.7 = 70%

Example 2.When a mixture of aluminum and iron weighing 11 g was exposed to an excess of hydrochloric acid, 8.96 liters of gas (NU) were released. Determine the mass fraction of metals in the mixture.

In the second example, both metal. Here, hydrogen is already liberated from the acid in both reactions. Therefore, direct calculation cannot be used here. In such cases, it is convenient to solve with the help of 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.

Solution of example 2.

1. 
   Find the amount of hydrogen:
   n = V / V m = 8.96 / 22.4 = 0.4 mol.
2. 
   Let the amount of aluminum - x mol, and iron - by mol. Then you can express the amount of released hydrogen in terms of x and y:
   
3. 
   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:
4. 
   (56 - 18) y = 11 - 7.2
   y = 3.8 / 38 = 0.1 mol (Fe)
   x = 0.2 mol (Al)
5. 
   Then we find the masses of metals and their mass fractions in the mixture:

respectively,

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.u.) were released and 5 g of the substance did not dissolve. Determine the mass fraction of metals in the mixture.

In the third example, two metals react and the third metal (copper) does not react. Therefore, the remainder of 5 g is the mass of copper. The amounts of the other two metals - zinc and aluminum (note that their total mass is 16 - 5 = 11 g) can be found using the system of equations, as in example # 2.

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

Example 4.The 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.u.) were released. The remaining mixture was treated with an excess of sodium hydroxide solution. Released 3.36 liters of gas and remained 3 g of undissolved residue. 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 does react with copper. This produces sulfur (IV) oxide.

Solution example 4.

1. 
   Only copper reacts with concentrated sulfuric acid, the number of moles of gas:
   n SO2 = V / Vm = 5.6 / 22.4 = 0.25 mol
   

   0,25		0,25
   Cu +	2H 2 SO 4 (conc.) = CuSO 4 +	SO 2 + 2H 2 O

2. 
   (do not forget that such reactions must be balanced using 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:
   m Cu = n M = 0.25 64 = 16 g.
4. 
   Aluminum reacts with an alkali solution, thus forming an aluminum hydroxo complex and hydrogen:
   2Al + 2NaOH + 6H 2 O = 2Na + 3H 2
   

   Al 0 - 3e = Al 3+		2
   2H + + 2e = H 2		3

5. 
   The 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,
   n Al = 0.15 / 1.5 = 0.1 mol.
   Aluminum weight:
   m Al = n M = 0.1 27 = 2.7 g
6. 
   The remainder is iron, weighing 3 g. You can find the mass of the mixture:
   m mixture = 16 + 2.7 + 3 = 21.7 g.
7. 
   Mass fractions of metals:

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 a nitric acid solution containing 20 wt. % НNО 3 and having a density of 1.115 g / ml. The volume of the evolved gas, which is a simple substance and the only product of the reduction of nitric acid, was 2.912 L (standard). 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 give hydrogen with metals, this is nitrogen. Both metals dissolved in acid.

Solution example 5.

1. 
   Determine the amount of gas substance:
   n N2 = V / Vm = 2.912 / 22.4 = 0.13 mol.
2. 
   We determine the mass of the nitric acid solution, the mass and amount of the substance of the dissolved HNO3:

Please note that since the metals have completely dissolved, it means - acid is definitely enough(these metals do not react with water). Accordingly, it will be necessary to check was there an excess of acid, and how much of it remained after the reaction in the resulting solution.

1. 
   We compose the reaction equations ( do not forget about 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:

1. 
   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 has reacted is:
   n HNO3 = 0.48 + 1.08 = 1.56 mol,
   those. acid was in excess and you can calculate its remainder in solution:
   n HNO3 rest. = 2 - 1.56 = 0.44 mol.
2. 
   So in final solution contains:

1. 
   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:

2. 
   m new solution = 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 = 630.3 + 21.1 - 3.36 = 648.04 g
3. 
   Now you can calculate the mass fractions of substances in the resulting solution:

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 were treated with an excess of concentrated nitric acid, 4.48 liters of gas (n.u.) were released, and when this mixture was exposed to the same mass of excess hydrochloric acid, 8.96 l of gas (n.o.) was released. y.). Determine the composition of the original mixture. (RCTU)

When solving this problem, one must remember, firstly, that concentrated nitric acid with an inactive metal (copper) gives NO 2, and iron and aluminum do not react with it. In contrast, hydrochloric acid does not react with copper.

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

Tasks for an independent solution.

1. Simple problems with two components of the mixture.

1-1. A mixture of copper and aluminum weighing 20 g was treated with a 96% solution of nitric acid, while 8.96 liters of gas (n.u.) 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. At the same time, 2.24 liters of gas (ny) were released. Calculate the mass fraction of zinc in the starting 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.u.) 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. Received zinc sulfate weighing 6.44, Calculate the mass fraction of zinc in the original 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 original mixture.

1-6. What mass of a 20% hydrochloric acid solution is required for the complete dissolution of 20 g of a mixture of zinc and zinc oxide, if 4.48 L of hydrogen is released in this case?

1-7. When 3.04 g of a mixture of iron and copper are dissolved in dilute nitric acid, nitric oxide (II) with a volume of 0.896 L (NU) is released. Determine the composition of the original mixture.

1-8. When dissolving 1.11 g of a mixture of iron and aluminum sawdust in a 16% solution of hydrochloric acid (ρ = 1.09 g / ml), 0.672 liters of hydrogen (NU) were released. Find the mass fraction of metals in the mixture and determine the volume of consumed hydrochloric acid.

2. The tasks are more complex.

2-1. A mixture of calcium and aluminum weighing 18.8 g was calcined in the absence of air with an excess of graphite powder. The reaction product was treated with dilute hydrochloric acid, while 11.2 liters of gas (NU) were evolved. Determine the mass fraction 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) were used. The excess acid reacted with 28.6 ml of a 1.4 mol / L potassium bicarbonate solution. Determine the mass fractions of metals in the alloy and the volume of gas (n.o.) released during the dissolution of the alloy.

2-3. When 27.2 g of a mixture of iron and iron (II) oxide were dissolved in sulfuric acid and the solution was evaporated to dryness, 111.2 g of ferrous sulfate - iron (II) sulfate heptahydrate - was formed. Determine the quantitative composition of the original 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 weight 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, a mixture of potassium and magnesium with a total weight 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. The aluminum-silver alloy 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 fraction of metals in the alloy.

3. Three metals and challenges.

3-1. When 8.2 g of a mixture of copper, iron and aluminum were treated with an excess of concentrated nitric acid, 2.24 liters of gas were released. The same volume of gas is released when the same mixture of the same mass is treated with an excess of dilute sulfuric acid (NU). 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 (NU). Determine the composition of the mixture in mass percent if 8.96 liters of chlorine (n.o.) is required for chlorination of the same weighed portion of the mixture.

3-3. Iron, zinc and aluminum filings are mixed in a molar ratio of 2: 4: 3 (in the order of listing). 4.53 g of this mixture was treated with excess 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) were placed in an 18.25% solution of hydrochloric acid weighing 160 g. Calculate the mass fractions of substances in the resulting solution.

3-5. 13.8 g of a mixture consisting of silicon, aluminum and iron was treated by heating with excess sodium hydroxide, while 11.2 liters of gas (NU) were evolved. When an excess of hydrochloric acid acts on such a mass of a mixture, 8.96 liters of gas (n.u.) are released. Determine the mass of substances in the original 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 was 2 times less than the mass of the initial mixture. This residue was treated with an excess of hydrochloric acid, the volume of the evolved gas in this case turned out to be equal to the volume of the gas evolved in the first case (the volumes were measured under the same conditions). Calculate the mass fraction of metals in the original mixture.

3-7. There is a mixture of calcium, calcium oxide and calcium carbide with a molar ratio of the components of 3: 2: 5 (in the order of the listing). 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 mass of 7.1 g was treated with dilute hydrochloric acid, the mass of the insoluble residue was equal to 3.2 g. After separation of the precipitate, the solution was treated with bromine in an alkaline medium, and at the end of the reaction it was treated with an excess of barium nitrate. The mass of the sediment formed turned out to be equal to 12.65 g. Calculate the mass fractions of metals in the initial mixture.

Answers and comments to problems 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 in reaction with hydrochloric acid goes to +2); 12.56 ml HCl solution.
2-1. 42.55% Ca (calcium and aluminum with graphite (carbon) form carbides CaC 2 and Al 4 C 3; during their hydrolysis with water or HCl, acetylene C 2 H 2 and methane CH 4 are released, respectively);

2-3. 61.76% Fe (ferrous 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 by oxygen, its oxide is formed, and when sodium is oxidized, peroxide Na 2 O 2, which is hydrolyzed in water to hydrogen peroxide and alkali);

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

3-3. 1.53% FeCl 3, 2.56% ZnCl 2, 1.88% AlCl 3 (iron, in reaction with chlorine, goes into the oxidation state +3);

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 (chromium, when dissolved in hydrochloric acid, transforms into chromium (II) chloride, which, under the action of bromine in an alkaline medium, turns into chromate; when barium salt is added, an insoluble chromate is formed 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. filtering

D. distillation

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

A. filtration

B. distillation

C. sublimation

D. upholding

5. The 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 divided into:

A. filtration

B. by magnet action

C. chromatography

D. distillation (distillation)

7. What is a pure substance as opposed to a mixture:

And cast iron

From the air

8.What applies to inhomogeneous mixtures:

A mixture of oxygen and nitrogen

Into the muddy river water

With snow crust

9.What is the solid mixture:

A glucose solution

With alcohol solution

D potassium sulfate solution

10. What is the name of the method for cleaning an inhomogeneous mixture:

And distillation

In filtration

With evaporation

D hot water

Part B

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

A) filter out

B) assemble the filtering device

C) dissolve in water

D) evaporate the solution

E) 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

C) iron and sulfur

D) water and alcohol

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

In our article we will look at what pure substances and mixtures are, methods of separating mixtures. Each of us uses them in everyday life. Are pure substances found in nature at all? And how can you tell them apart from mixtures?

Pure substances and mixtures: methods for separating mixtures

Pure substances are those that contain only a certain kind of particles. Scientists believe that they practically do not exist in nature, since all of them, albeit in negligible proportions, contain impurities. Absolutely all substances are also water-soluble. Even if you immerse a silver ring in this liquid, for example, the ions of this metal will go into solution.

A sign of pure substances is the constancy of composition and physical properties. In the process of their formation, the amount of energy changes. Moreover, it can both increase and decrease. It is possible to divide a pure substance into separate components only with the help of chemical reaction... For example, only distilled water has a typical boiling and freezing point for this substance, no taste or smell. And its oxygen and hydrogen can only be decomposed by electrolysis.

And how do their totality differ from pure substances? Chemistry will help us to answer this question. The separation methods of mixtures are physical, since they do not lead to a change chemical composition substances. Unlike pure substances, mixtures have variable composition and properties, and they can be separated by physical methods.

What is a mixture

A mixture is a collection of individual substances. An example of this is sea water. Unlike disillated, it has a bitter or salty taste, boils at a higher temperature, and freezes at a lower temperature. The methods for separating mixtures of substances are physical. Thus, pure salt can be extracted from seawater by evaporation and subsequent crystallization.

Types of mixtures

If you add sugar to the water, after a while its particles will dissolve and become invisible. As a result, it will be impossible to distinguish them with the naked eye. Such mixtures are called homogeneous or homogeneous. They are also examples of air, gasoline, broth, perfume, sweet and salty water, an alloy of copper and aluminum. As you can see, they can be in different states of aggregation, but most often there are liquids. They are also called solutions.

In inhomogeneous or heterogeneous mixtures, particles of individual substances can be distinguished. Iron and wood filings, sand and table salt are typical examples. Inhomogeneous mixtures are also called suspensions. Among them, suspensions and emulsions are distinguished. The former include a liquid and a solid. So, the emulsion is a mixture of water and sand. An emulsion is a combination of two liquids with different densities.

There are heterogeneous mixtures with special names. So, an example of foam is polystyrene, and aerosols include fog, smoke, deodorants, air fresheners, antistatic agents.

Methods for separating mixtures

Of course, many mixtures have more valuable properties than the individual individual substances that are included in their composition. But even in everyday life situations arise when they need to be separated. And in industry, entire industries are based on this process. For example, petrol, gas oil, kerosene, fuel oil, diesel and engine oil, rocket fuel, acetylene and benzene are obtained from oil as a result of its processing. Agree, it is more profitable to use these products than to mindlessly burn oil.

Now let's see if there is such a thing as chemical methods for separating mixtures. Let's say from aqueous solution salt, we need to get pure substances. For this, the mixture must be heated. As a result, the water turns into steam and the salt crystallizes. But this will not happen the transformation of some substances into others. This means that the basis of this process is physical phenomena.

Methods for separating mixtures depend on the state of aggregation, ability to solubility, difference in boiling point, density and composition of its components. Let's consider each of them in more detail with specific examples.

Filtration

This separation method is suitable for mixtures containing a liquid and an insoluble solid. For example, water and river sand. This mixture must be passed through a filter. As a result, clean water will freely pass through it, and the sand will remain.

Upholding

Some methods of separating mixtures are based on the action of gravity. Thus, suspensions and emulsions can be decomposed. If vegetable oil gets into the water, this mixture must first be shaken. Then leave it for a while. As a result, the water will be at the bottom of the vessel, and the oil in the form of a film will cover it.

In laboratory conditions, it is used for settling. As a result of its work, a denser liquid is poured into a vessel, and a light one remains.

Deposition is characterized by a low rate of the process. It takes a certain amount of time for the sediment to form. Under industrial conditions, this method is carried out in special structures called sedimentation tanks.

Magnet action

If the mixture contains metal, then it can be separated using a magnet. For example, to separate iron and. But do all metals have such properties? Not at all. For this method, only mixtures containing ferromagnets are suitable. In addition to iron, these include nickel, cobalt, gadolinium, terbium, dysprosium, holmium, erbium.

Distillation

This name is translated from Latin means "dripping". Distillation is a method for separating mixtures based on the difference in boiling points of substances. Thus, even at home, alcohol and water can be separated. The first substance begins to evaporate already at a temperature of 78 degrees Celsius. Touching the cold surface, alcohol vapors condense, turning into liquid state.

In industry, in this way, oil refined products, fragrances, and pure metals are obtained.

Evaporation and crystallization

These separation methods are suitable for liquid solutions. The substances that make up their composition differ in boiling point. Thus, it is possible to obtain crystals of salt or sugar from the water in which they are dissolved. For this, the solutions are heated and evaporated until saturated. In this case, crystals are deposited. If it is necessary to obtain pure water, then the solution is brought to a boil, followed by condensation of vapors on a colder surface.

Methods for separating gas mixtures

Gaseous mixtures are separated by laboratory and industrial methods, since this process requires special equipment. The raw materials of natural origin are air, coke oven gas, generator gas, associated gas and natural gas, which is a combination of hydrocarbons.

Physical methods of separation of mixtures in gaseous state the following:

• Condensation is the process of gradual cooling of a mixture, during which condensation of its constituents occurs. In this case, first of all, high-boiling substances that are collected in separators pass into the liquid state. Thus, hydrogen is obtained from and also ammonia is separated from the unreacted part of the mixture.
• Sorption is the absorption of some substances by others. This process has opposite components, between which equilibrium is established in the course of the reaction. For the forward and reverse process, different conditions are required. In the first case, it is a combination of high pressure and low temperature. This process is called sorption. Otherwise use opposite conditions: low pressure at high temperature.
• Membrane separation is a method in which the property of semi-permeable partitions is used to selectively pass molecules various substances.
• Reflux is the process of condensation of high-boiling parts of mixtures as a result of their cooling. In this case, the temperature of the transition to the liquid state of individual components should differ significantly.

Chromatography

The name of this method can be translated as "writing in color". Imagine adding ink to the water. If you dip the end of the filter paper into this mixture, it will begin to absorb. In this case, water will be absorbed faster than ink, which is associated with a different degree of sorption of these substances. Chromatography is not only a method for separating mixtures, but also a method for studying such properties of substances as diffusion and solubility.

So, we got acquainted with such concepts as "pure substances" and "mixtures". The former are elements or compounds consisting only of particles of a certain type. Examples are salt, sugar, distilled water. Mixtures are a collection of individual substances. A number of methods are used to separate them. The way they are separated depends on the physical properties of its constituents. The main ones are settling, evaporation, crystallization, filtration, distillation, magnetism and chromatography.

Separation methods for mixtures

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

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

After settling or centrifugation, the supernatant can be separated from the sediment by the method decantation- by carefully draining the liquid from the sediment.

It is possible to separate a mixture of two liquids insoluble in each other (after settling) using a separating funnel, the principle of which is clear from the following illustration:

To separate mixtures of substances in different states of aggregation, in addition to settling and centrifugation, filtration is also widely used. The method consists in the fact that the filter has different throughput in relation to the components of the mixture. Most often this is due to different particle sizes, but it can also be due to the fact that individual components of the mixture interact more strongly with the filter surface ( 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 below it:

In some cases, heterogeneous mixtures can be separated due to the different magnetic properties of the components. For example, a mixture of powders of sulfur and metallic iron 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 are called magnetic separation.

If the mixture is a solution of a refractory solid in any liquid, this substance can be isolated 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 separating not only volatile components from non-volatile 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 vapor of the lighter-boiling component is the first to volatilize. Vapors of this substance, after passing through the refrigerator, condense and drain into the receiver. The distillation method is widely used in the oil industry in the primary processing of oil to separate oil into fractions (gasoline, kerosene, diesel, etc.).

By the same method of distillation, water is obtained purified from impurities (primarily salts). Distilled water is called distilled water.

Lesson type. Learning new material.

Lesson objectives. Educational- to study the concepts of "pure substance" and "mixture", homogeneous (homogeneous) and heterogeneous (heterogeneous) mixtures, consider ways to separate mixtures, teach students to separate mixtures into components.

Developing- to develop the intellectual and cognitive skills of students: to highlight essential features and properties, establish causal relationships, classify, analyze, draw conclusions, perform experiments, observe, draw up observations in the form of tables, diagrams.

Educational- to promote the education of students' organization, accuracy during the experiment, the ability to organize mutual assistance when working in pairs, the spirit of competition when performing exercises.

Teaching methods. Methods of organizing educational and cognitive activities- verbal (heuristic conversation), visual (tables, figures, demonstrations of experiments), practical ( laboratory works, exercise).

Methods to stimulate interest in learning – cognitive games, educational discussions.

Control methods- oral control, written control, experimental control.

Equipment and reagents.On student tables- Sheets of paper, spoons for substances, glass rods, glasses of water, magnets, sulfur and iron powders.

On the teacher's desk- spoons, test tubes, test tube holder, spirit lamp, magnet, water, beakers, rack with a ring, rack with a foot, funnel, glass rods, filters, porcelain cup, separating funnel, test tube with gas outlet tube, receiver tube, beaker -fridge "with water, filter paper tape (2x10 cm), red ink, flask, sieve, iron and sulfur powders in a mass ratio of 7: 4, river sand, table salt, vegetable oil, copper sulfate solution, semolina, buckwheat.

DURING THE CLASSES

Organizing time

Tag missing, explain the purpose of the lesson, and familiarize students with the lesson plan.

P l a n u ro k a

1. Pure substances and mixtures. Distinctive features.

2. Homogeneous and inhomogeneous mixtures.

3. Methods for separating mixtures.

Conversation on the topic "Substances and their properties"

Teacher. Remember what chemistry studies.

Student. Substances, properties of substances, changes occurring with substances, i.e. transformation of substances.

Teacher. What is called a substance?

Student. Substance is what the physical body is made of.

Teacher. You know that substances are simple and complex. Which substances are called simple and which are called complex?

Student. Simple substances consist of atoms of one chemical element, complex - from atoms of various chemical elements.

Teacher. What are the physical properties of substances?

Student. Physical state, melting and boiling points, electrical and thermal conductivity, solubility in water, etc..

Explanation of the new material

Pure substances and mixtures.
Distinctive features

Teacher. Only pure substances have constant physical properties. Only pure distilled water has melting point = 0 ° С, boiling point = 100 ° С, has no taste. Sea water freezes at a lower temperature and boils at a higher temperature; it tastes bitter-salty. The Black Sea water freezes at a lower temperature and boils at a higher temperature than the Baltic Sea water. Why? The point is that in sea ​​water contains other substances, such as dissolved salts, i.e. it is a mixture of various substances, the composition of which varies within wide limits, while the properties of the mixture are not constant. The definition of the concept "mixture" was given in the 17th century. by the English scientist Robert Boyle: "A mixture is a complete system consisting of dissimilar components."

Consider the distinctive features of the mixture and the pure substance. To do this, we will do the following experiments.

Experience 1. Using the instructions for the experiment, study the essential physical properties of iron and sulfur powders, prepare a mixture of these powders and determine whether these substances retain their properties in the mixture.

Discussion with students of the results of the experiment.

Teacher. Describe state of aggregation and the color of gray.

Student. Sulfur is a yellow solid.

Teacher. What is the state of aggregation and color of iron in powder form?

Student. Iron is a solid gray matter.

Teacher. How do these substances relate: a) to a magnet; b) to the water?

Student. Iron is attracted by a magnet, but sulfur is not; iron powder sinks in water, because iron is heavier than water, and sulfur powder floats to the surface of the water, since it is not wetted by water.

Teacher. What can you say about the ratio of iron and sulfur in the mixture?

Student. The ratio of iron and sulfur in the mixture can be different, i.e. fickle.

Teacher. Are the properties of iron and sulfur preserved in the mixture?

Student. Yes, the properties of each substance in the mixture are preserved.

Teacher. How can a mixture of sulfur and iron be separated?

Student. This can be done by physical methods: magnet or water.

Teacher . Experience 2. Now I will show the reaction between sulfur and iron. Your task is to carefully observe this experiment and determine whether iron and sulfur retain their properties in the iron (II) sulfide obtained as a result of the reaction and whether iron and sulfur can be isolated from it by physical methods.

I thoroughly mix powders of iron and sulfur in a mass ratio of 7: 4:

m (Fe ): m ( S ) = А r ( Fe ): А r ( S ) = 56: 32 = 7: 4,

I put the mixture in a test tube, heat it up in the flame of an alcohol lamp, heat it up strongly in one place and stop heating when a violent exothermic reaction begins. After the test tube has cooled, I gently break it, wrap it in a towel, and remove the contents. Take a close look at the resulting substance - iron (II) sulfide. Are gray iron powder and yellow sulfur powder visible in it separately?

Student. No, the resulting substance is dark gray in color.

Teacher. Then I test the resulting substance with a magnet. Are iron and sulfur separate?

Student. No, the resulting substance is not magnetized.

Teacher. I put iron (II) sulfide in water. What do you observe while doing this?

Student. Iron (II) sulfide sinks in water.

Teacher. Do sulfur and iron retain their properties as part of iron (II) sulfide?

Student. No, the new substance has properties that are different from the properties of the substances taken for the reaction.

Teacher. Is it possible to separate iron (II) sulfide by physical methods into simple substances?

Student. No, neither a magnet nor water can separate iron (II) sulfide into iron and sulfur.

Teacher. Does the energy change during the formation chemical?

Student. Yes, for example, when iron and sulfur interact, energy is released.

Teacher. Let's enter the results of the discussion of the experiments in the table.

table

Comparative characteristics of the mixture and the pure substance

To consolidate this part of the lesson, complete the exercise: determine where in the picture(see page 34) depicts a simple substance, complex substance or mixture.

Homogeneous and inhomogeneous mixtures

Teacher. Let's find out if the mixtures differ in appearance from each other.

The teacher demonstrates examples of suspensions (river sand + water), emulsions (vegetable oil + water) and solutions (air in a flask, table salt + water, loose change: aluminum + copper or nickel + copper).

Teacher. In suspensions, particles of a solid are visible, in emulsions - droplets of liquid, such mixtures are called inhomogeneous (heterogeneous), and in solutions the components are indistinguishable, they are homogeneous (homogeneous) mixtures. Consider the classification scheme for mixtures(diagram 1).

Scheme 1

Give examples of each type of mixture: suspensions, emulsions and solutions.

Methods for separating mixtures

Teacher. 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 mixture separation are used to purify substances (Scheme 2).

Scheme 2

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

Consider ways to split heterogeneous mixtures.

How can you separate a suspension - a mixture of river sand with water, that is, to clear the water from sand?

Student. By settling and then filtering.

Teacher. Right. Separation upholding based on different densities of substances. The heavier sand settles to the bottom. It is also possible to 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 upper, lighter layer... (The teacher demonstrates the relevant experiences.)

As a result of sedimentation, dew falls out of the fog, soot is deposited from the smoke, cream is settled in milk.

And what is the basis for the separation of heterogeneous mixtures using filtering?

Student. On different solubility of substances in water and on different particle sizes.

Teacher. It is true that 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.

Pupil shows an experience: pours water into a mixture of sand and salt, mixes, and then passes the suspension (suspension) through the filter - a solution of salt in water passes through the filter, and large particles of water-insoluble sand remain on the filter.

Teacher. What substances can be used as filters?

Student. Various porous substances can be used as filters: cotton wool, coal, fired clay, pressed glass and others.

Teacher. What examples of filtering applications in human life can you give?

Student. 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 elevator workers - respiratory masks. With the help of a tea strainer for filtering tea leaves, Ostap Bender, the hero of the work by Ilf and Petrov, managed to take one of the chairs from Ellochka the Cannibal ("The Twelve Chairs").

Teacher. And now, having got acquainted with these methods of separating the mixture, let's help the heroine of the Russian folk tale"Vasilisa the Beautiful".

Student. In this tale, Baba Yaga ordered Vasilisa to separate the rye from the nigella and the poppy from the ground. The heroine of the tale was helped by pigeons. Now we can separate the cereals by filtering through a sieve if the grains are of different sizes, or by shaking with water if the particles have different densities or different wettability with water. Take, as an example, a mixture of grains of various sizes: a mixture of semolina and buckwheat.(The student shows how semolina with smaller particles passes through the sieve, but buckwheat remains on it.)

Teacher. But today you already got acquainted with a mixture of substances that have different wettability with water. What mixture am I talking about?

Student. It is a mixture of iron and sulfur powders. We carried out a laboratory experiment with this mixture..

Teacher. Remember how you split such a mixture.

Student. By settling in water and using a magnet.

Teacher. What did you observe while separating a mixture of iron and sulfur powders with water?

Student. Non-wettable sulfur powder floated to the surface of the water, and heavy wettable iron powder settled to the bottom.

Teacher. And how was the separation of this mixture with the help of a magnet?

Student. Iron powder was attracted by a magnet, but sulfur powder was not..

Teacher. So, we got acquainted with three methods of separating heterogeneous mixtures: settling, filtration and magnet action. Now let's look at ways to separate homogeneous (homogeneous) mixtures... Remember, after separating the sand by filtration, we got a solution of salt in water - a homogeneous mixture. How to isolate pure salt from the solution?

Student. Evaporation or crystallization.

The teacher demonstrates the experience: the water evaporates, and salt crystals remain in the porcelain cup.

Teacher. By evaporation of water 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, for example sugar, decomposes on heating, then the water is evaporated incompletely - the solution is evaporated, and then sugar crystals are precipitated from the 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.

The teacher shows the distillation of a solution of copper sulfate, water evaporates when t bale = 100 ° C, then the vapors are condensed in a receiving tube cooled with water in a glass.

Teacher. Distilled water is obtained in special devices - distillers, which is used for the needs of pharmacology, laboratories, and car cooling systems.

A student demonstrates a drawing of a "device" designed by him for distilling water.

Teacher. If you separate a mixture of alcohol and water, then the first will be distilled off (collected in the receiving tube) alcohol with t bip = 78 ° C, and water will remain in the test tube. Distillation is used to obtain gasoline, kerosene, gas oil from oil.

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

The teacher demonstrates experience. He hangs a strip of filter paper over a container of red ink, immersing only the end of the strip in it. The solution is absorbed by the paper and rises along it. But the border of the rise of the paint lags behind the border of the rise of the water. This is how the two substances are separated: water and the dye in the ink.

Teacher. Using 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 with the same degree of purification always required?

Student. For different purposes, substances with different degrees of purification are required. It is enough to settle water for cooking to remove impurities and chlorine used for its disinfection. Water for drinking must first be boiled. And in chemical laboratories for the preparation of solutions and conducting 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.

Teacher. Listen to L. Martynov's poem "Distilled Water":

Water
Favored
To pour!
She
Shone
So pure
No matter what to get drunk
Not to wash.
And it was not without reason.
She missed
Willows, tala
And the bitterness of flowering vines,
She lacked algae
And fish, oily from dragonflies.
She didn't have enough to be wavy
She missed flowing everywhere.
She lacked life
Clean -
Distilled water!

To consolidate and check the assimilation of the material, students answer the following questions.

1. When grinding ore at ore processing plants, fragments of iron tools fall into it. How can they be extracted from the ore?

2. Dispose of iron objects before recycling household waste and waste paper. What's the easiest way to do this?

3. The vacuum cleaner draws in dusty air and releases clean air. Why?

4. Car wash water in large garages becomes contaminated with engine oil. What should you do before draining it down the drain?

5. Flour is purified from bran by sifting. Why is this done?

6. How to separate tooth powder and table salt? Gasoline and water? Alcohol and water?

REFERENCE

Alikberova L.Yu. Entertaining chemistry. M .: AST-Press, 1999; Gabrielyan O.S., Voskoboinikova N.P., Yashukova A.V. Handbook of the teacher. Chemistry. 8th grade. M .: Bustard, 2002; Gabrielyan O.S. Chemistry.
8th grade. M .: Bustard, 2000; Guzei L.S., Sorokin V.V., Surovtseva R.P. Chemistry. 8th grade. M .: Bustard, 1995; Ilf I.A., Petrov E.P. Twelve Chairs. M .: Education, 1987; Kuznetsova N.E., Titova I.M., Gara N.N., Zhegin A.Yu. Chemistry. A textbook for students of the 8th grade of educational institutions. M .: Ventana-Graf, 1997; Rudzitis G.E., Feldman F.G. Chemistry. Textbook for grade 8 educational institutions. M .: Education, 2000; Tyldsepp A.A., Cork V.A.... We study chemistry. M .: Education, 1998.