Problems to determine the quantitative composition of a mixture. Chemical properties of metals

Problems involving mixtures are a very common type of problem in chemistry. They require a clear idea of ​​which substances enter into the reaction proposed in the problem and which do not.
We talk about a mixture when we have not one, but several substances (components) “poured” into one container. These substances should not interact with each other.

Typical misconceptions and mistakes that arise when solving problems using mixtures.

1. An attempt to write both substances into one reaction.
   It turns out something like this:
   “A mixture of calcium and barium oxides was dissolved in hydrochloric acid...”
   The reaction equation is written as follows:
   CaO + BaO + 4HCl = CaCl 2 + BaCl 2 + 2H 2 O.
   This is a mistake, because there can be any amount of each oxide in this mixture.
   And in the above equation it is assumed that their equal amount.
2. The assumption is that their molar ratio corresponds to the coefficients in the reaction equations.
   For example:
   Zn + 2HCl = ZnCl 2 + H 2
   2Al + 6HCl = 2AlCl3 + 3H2
   The amount of zinc is taken to be x, and the amount of aluminum is taken to be 2x (according to the coefficient in the reaction equation). This is also incorrect. These quantities can be any and they are not related to each other in any way.
3. Attempts to find the “amount of substance of a mixture” by dividing its mass by the sum of the molar masses of the components.
   This action makes no sense at all. Each molar mass can only refer to a single substance.

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

Necessary theoretical information.

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"] = m component / m mixture

• 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 = n component 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:

  N component A: n component B = 2: 3

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

  φ ["phi"] = V component / V mixture

Electrochemical voltage series of metals.

Reactions of metals with acids.

1. With mineral acids, which include all soluble acids ( except nitrogen and concentrated sulfur, whose interaction with metals occurs in a special way), react only metals, located in the electrochemical voltage series to (to the left) hydrogen.
2. In this case, metals that have several oxidation states (iron, chromium, manganese, cobalt) exhibit the minimum possible oxidation state - usually +2.
3. Interaction of metals with nitric acid leads to the formation, instead of hydrogen, of nitrogen reduction products, and with concentrated sulfuric acid- to the release of sulfur reduction products. Since a mixture of reduction products is actually formed, the problem often contains a direct reference to a specific substance.

Nitric acid reduction products.

Products of reduction of sulfuric acid.

Reactions of metals with water and alkalis.

1. In water at room temperature dissolve only metals that correspond to soluble bases (alkalis). These are alkali metals (Li, Na, K, Rb, Cs), as well as group IIA metals: Ca, Sr, Ba. This produces alkali and hydrogen. Boiling water can also dissolve magnesium.
2. Only amphoteric metals can dissolve in alkali: aluminum, zinc and tin. In this case, hydroxo complexes are formed and hydrogen is released.

Examples of problem solving.

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.s.) were released. Determine the mass fractions of metals in the mixture.

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

Example 1 solution.

1. Finding the amount of hydrogen:
   n = V / V m = 5.6 / 22.4 = 0.25 mol.
2. According to the reaction equation:

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

3. Now you can calculate the mass fractions of metals in the mixture:

   ω Fe = m Fe /m of the entire mixture = 14 / 20 = 0.7 = 70%

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 (n.s.) were released. Determine the mass fractions of metals in the mixture.

In the second example, they react 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. Finding the amount of hydrogen:
   n = V / V m = 8.96 / 22.4 = 0.4 mol.
2. Let the amount of aluminum be x mole, and the amount of iron be x mole. Then we can express the amount of hydrogen released in terms of x and y:

   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:

   (56 − 18)y = 11 − 7.2
   y = 3.8 / 38 = 0.1 mol (Fe)
   x = 0.2 mol (Al)

3. Next we find the masses of metals and their mass fractions in the mixture:

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

   respectively,
   ω Al = 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.s.) 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.

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

Example 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.s.) 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- an 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, the number of moles of gas is:
   n SO 2 = V / Vm = 5.6 / 22.4 = 0.25 mol
2. 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,
   n Al = 0.15 / 1.5 = 0.1 mol.
   Aluminum weight:
   m Al = n M = 0.1 27 = 2.7 g
3. The remainder is iron, weighing 3 g. You can find the mass of the mixture:
   m mixture = 16 + 2.7 + 3 = 21.7 g.
4. Mass fractions of metals:

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

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

Example 5.21.1 g of a mixture of zinc and aluminum was dissolved in 565 ml of nitric acid solution containing 20 wt. % НNO 3 and having a density of 1.115 g/ml. The volume of the released gas, which is a simple substance and the only product of the reduction of nitric acid, was 2.912 l (n.s.). 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:
   n N 2 = 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:

   M solution = ρ V = 1.115 565 = 630.3 g
   m HNO 3 = ω m solution = 0.2 630.3 = 126.06 g
   n HNO 3 = 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:

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

   Zn 0 − 2e = Zn 2+	|	5
   2N +5 + 10e = N 2		1

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

   X = 0.04, which means n Zn = 0.04 5 = 0.2 mol
   y = 0.03, which means n Al = 0.03 10 = 0.3 mol

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

4. 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):

   0,2	0,48	0,2	0,03
   5Zn	+ 12HNO3 =	5Zn(NO 3) 2	+N2+	6H2O

   0,3	1,08	0,3	0,09
   10Al	+ 36HNO3 =	10Al(NO 3) 3	+3N2+	18H2O

5. The next question is: 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:
   n HNO 3 = 0.48 + 1.08 = 1.56 mol,
   those. the acid was in excess and you can calculate its remainder in the solution:
   n HNO 3 rest. = 2 − 1.56 = 0.44 mol.
6. So, in final solution contains:

   Zinc nitrate in an amount of 0.2 mol:
   m Zn(NO 3) 2 = n M = 0.2 189 = 37.8 g
   aluminum nitrate in an amount of 0.3 mol:
   m Al(NO 3) 3 = n M = 0.3 213 = 63.9 g
   excess nitric acid in an amount of 0.44 mol:
   m HNO 3 rest. = n M = 0.44 63 = 27.72 g

7. 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):

   Then for our task:

   M new solution = mass of acid solution + mass of metal alloy - mass of nitrogen
   m N 2 = n M = 28 (0.03 + 0.09) = 3.36 g
   m new solution = 630.3 + 21.1 − 3.36 = 648.04 g

8. Now you can calculate the mass fractions of substances in the resulting solution:

   ωZn(NO 3) 2 = m quantity / m solution = 37.8 / 648.04 = 0.0583
   ωAl(NO 3) 3 = m volume / m solution = 63.9 / 648.04 = 0.0986
   ω HNO 3 rest. = m water / m solution = 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.o.) were released, and when this mixture was exposed to the same mass of excess hydrochloric acid, 8.96 liters of gas (n.o.) 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 NO 2, and 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.s.) 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 hydrogen is released with a volume of 4.48 l (n.s.)?

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 (n.s.). 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.s.) 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.

2-3. When 27.2 g of a mixture of iron and iron (II) oxide was dissolved in sulfuric acid and the solution was evaporated to dryness, 111.2 g of iron sulfate—iron (II) sulfate heptahydrate—was formed. Determine the quantitative composition of the initial mixture.

2-4. When iron weighing 28 g reacted with chlorine, a mixture of iron (II) and (III) chlorides weighing 77.7 g was formed. Calculate the mass of iron (III) chloride in the resulting mixture.

2-5. What was the mass fraction of potassium in its mixture with lithium, if as a result of treating this mixture with excess chlorine, a mixture was formed in which the mass fraction of potassium chloride was 80%?

2-6. After treating a mixture of potassium and magnesium with a total mass of 10.2 g with excess bromine, the mass of the resulting mixture of solids turned out to be equal to 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. A mixture of lithium and sodium with a total mass of 7.6 g was oxidized with excess oxygen, a total of 3.92 l (n.s.) was consumed. The resulting mixture was dissolved in 80 g of 24.5% sulfuric acid solution. Calculate the mass fractions of substances in the resulting solution.

2-8. An aluminum-silver alloy was treated with an excess of a concentrated solution of nitric acid, and 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. Calculate the mass fractions of metals in the alloy.

3. Three metals and complex problems.

3-1. When 8.2 g of a mixture of copper, iron and aluminum was 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 (DS). Determine the composition of the initial mixture in mass percent.

3-2. 14.7 g of a mixture of iron, copper and aluminum, interacting with an excess of dilute sulfuric acid, releases 5.6 liters of hydrogen (n.s.). Determine the composition of the mixture in mass percent if chlorination of the same sample of the mixture requires 8.96 liters of chlorine (n.s.).

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

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

3-5. 13.8 g of a mixture consisting of silicon, aluminum and iron were treated with excess sodium hydroxide when heated, and 11.2 liters of gas (n.s.) were released. When such a mass of mixture is exposed to excess hydrochloric acid, 8.96 liters of gas (n.s.) are released. Determine the masses of substances in the original mixture.

3-6. When a mixture of zinc, copper and iron was treated with an excess of concentrated alkali solution, gas was released, and the mass of the undissolved residue turned out to be 2 times less than the mass of the original mixture. This residue was treated with an excess of hydrochloric acid, the volume of gas released turned out to be equal to the volume of gas released in the first case (the volumes were measured under the same conditions). Calculate the mass fractions of metals in the initial mixture.

3-7. There is a mixture of calcium, calcium oxide and calcium carbide with a molar ratio of components of 3:2:5 (in the order listed). What is the minimum volume of water that can enter chemical reaction 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 undissolved residue turned out to be 3.2 g. After separating the precipitate, the solution was treated with bromine in an alkaline medium, and at the end of the reaction it was treated with excess barium nitrate. The mass of the formed precipitate 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 the amphoteric metal, zinc, dissolves in alkali);

1-3. 37,5%;

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

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

3-3. 1.53% FeCl 3, 2.56% ZnCl 2, 1.88% AlCl 3 (iron reacts with chlorine to 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, turns into chromium (II) chloride, which, when exposed to bromine in an alkaline medium, turns into chromate; when barium salt is added, insoluble chromate is formed barium)

Problems to determine the quantitative composition of a mixture. Chemical properties metals

1. A mixture of aluminum and iron filings was treated with an excess of dilute hydrochloric acid, and 8.96 liters (n.s.) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 6.72 liters (n.s.) of hydrogen will be released. Calculate the mass fraction of iron in the initial mixture.

2. A mixture of magnesium and zinc filings was treated with an excess of dilute sulfuric acid, and 22.4 liters (n.s.) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 13.44 liters (n.s.) of hydrogen will be released. Calculate the mass fraction of magnesium in the initial mixture.

3. When a mixture of copper and copper(II) oxide was dissolved in concentrated nitric acid, 18.4 g of brown gas was released and 470 g of a solution with a mass fraction of salt of 20% was obtained. Determine the mass fraction of copper oxide in the initial mixture.

4. A mixture of aluminum sulfide and aluminum was treated with water, and 6.72 l (no.) of gas was released. If the same mixture is dissolved in an excess of sodium hydroxide solution, 3.36 liters (n.s.) of gas will be released. Determine the mass fraction of aluminum in the initial mixture.

5. If a mixture of potassium and calcium chlorides is added to a solution of sodium carbonate, 10 g of precipitate is formed. If the same mixture is added to a solution of silver nitrate, 57.4 g of precipitate is formed. Determine the mass fraction of potassium chloride in the initial mixture.

6. A mixture of copper and aluminum weighing 10 g was treated with 96% nitric acid, which released 4.48 liters of gas (n.s.). Determine the quantitative composition of the initial mixture and the mass fraction of aluminum in it.

7. A mixture of magnesium and magnesium oxide weighing 6.4 g was treated with a sufficient amount of dilute sulfuric acid, and 2.24 liters of gas (n.s.) were released. Determine the quantitative composition of the initial mixture and the mass fraction of magnesium oxide in it.

8. A mixture of copper and zinc weighing 40 g was treated with a concentrated alkali solution. At the same time, a gas with a volume of 8.96 liters (n.s.) was released. Calculate the mass fraction of copper in the initial mixture.

On the topic: methodological developments, presentations and notes

Lesson development contains a detailed outline of the lesson, slides for the lesson, workbook on the topic being studied, instructional cards for conducting the experiment and other didactic material....

Task No. 1

When a mixture of copper and copper (II) oxide was dissolved in concentrated nitric acid, 18.4 g of brown gas was released and 470 g of a solution with a mass fraction of salt of 20% was obtained. Determine the mass fraction of copper oxide in the initial mixture.

Answer: 65.22%

Explanation:

When copper and copper (II) oxide are dissolved, the following reactions occur:

Cu + 4HNO 3(conc.) = Cu(NO 3) 2 + 2NO 2 + 2H 2 O (I)

CuO + 2HNO 3(conc.) = Cu(NO 3) 2 + H 2 O (II)

Brown NO 2 gas is released only when copper reacts with concentrated nitric acid. Let's find its quantity using the formula:

where m is the mass of the substance [g], M is the molar mass of the substance [g/mol].

M(NO 2) = 46 g/mol

ν (NO 2) = m(NO 2)/M(NO 2) = 18.4 g/46 g/mol = 0.4 mol.

According to reaction condition (I):

ν I (Cu(NO 3) 2) = ν(Cu) = 1/2ν(NO 2),

hence,

ν I (Cu(NO 3) 2) = ν(Cu) = 0.4 mol/2 = 0.2 mol.

Mass of copper nitrate formed due to the interaction of copper with concentrated nitric acid:

m I (Cu(NO 3) 2) = M(Cu(NO 3) 2) . ν(Cu(NO 3) 2) = 188 g/mol. 0.2 mol = 37.6 g.

The mass of reacted copper is:

m(Cu) = M(Cu) . ν(Cu) = 64 g/mol. 0.2 mol = 12.8 g.

The total mass of copper nitrate contained in the solution is equal to:

m total (Cu(NO 3) 2) = w(Cu(NO 3) 2) . m(p−ra)/100% = 20% . 470g / 100% = 94g.

Mass of copper nitrate formed by the reaction of copper oxide with concentrated nitric acid (II):

m II (Cu(NO 3) 2) = m(mixture) – m I (Cu(NO 3) 2) = 94 g – 37.6 g = 56.4 g.

The amount of copper nitrate formed by the reaction of copper oxide with concentrated nitric acid (II):

ν II (Cu(NO 3) 2) = m II (Cu(NO 3) 2) /M II (Cu(NO 3) 2) = 56.4 g/188 g/mol = 0.3 mol

According to the reaction condition:

(II) ν II (Cu(NO 3) 2) = ν(CuO) = 0.3 mol.

m(CuO) = M(CuO) . ν(CuO) = 80 g/mol. 0.3 mol = 24 g

m(mixtures) = m(CuO) + m(Cu) = 24 g + 12.8 g = 36.8 g

The mass fraction of copper oxide in the mixture is equal to:

w(CuO)% = m(CuO)/m(mixtures) . 100% = 24 g / 36.8 g. 100% = 65.22%

Task No. 2

Determine the mass fraction of sodium carbonate in the solution obtained by boiling 150 g of 8.4% sodium bicarbonate solution. What volume of a 15.6% barium chloride solution (density 1.11 g/ml) will react with the resulting sodium carbonate? The evaporation of water can be neglected.

Answer: 5.42%, 90 ml

Explanation:

The decomposition of sodium bicarbonate in solution is described by the reaction:

2NaHCO 3 → Na 2 CO 3 + CO 2 + H 2 O

The mass fraction of the dissolved substance is calculated by the formula:

w(in-in)% = m(in-in)/m(p−ra) . 100%,

where m(v-va) is the mass of the dissolved substance, m(p−ra) is the mass of the solution).

Let's calculate the mass of dissolved sodium bicarbonate (NaHCO 3):

M(NaHCO 3) = m(p−ra) . w(NaHCO 3)/100% = 150 g 8.4%/100% = 12.6 g

ν(NaHCO 3) = m(NaHCO 3)/M(NaHCO 3) = 12.6 g/84 g/mol = 0.15 mol

According to the reaction equation:

ν(Na 2 CO 3) = ν(CO 2) = 1/2ν(NaHCO 3) = 0.15 mol/2 = 0.075 mol

m(Na 2 CO 3) = M(Na 2 CO 3) . ν(Na 2 CO 3) = 106 g/mol. 0.075 mol = 7.95 g.

m(CO 2) = M(CO 2) . ν(CO 2) = 44 g/mol. 0.075 mol = 3.3 g.

Since the evaporation of water can be neglected, we find the mass of sodium bicarbonate formed after decomposition by subtracting the mass from the mass of the original solution carbon dioxide:

m(p−ra) = m(out.p−ra) - m(CO 2) = 150 g – 3.3 g = 146.7 g

w(Na 2 CO 3)% = m(Na 2 CO 3)/m(p−ra). 100% = 7.95 g/146.7 g 100% = 5.42%

The reaction between solutions of barium chloride and sodium carbonate is described by the equation:

BaCl 2 + Na 2 CO 3 → BaCO 3 ↓ + 2NaCl

According to the reaction equation:

ν(BaCl 2) = ν(Na 2 CO 3) = 0.075 mol, therefore

m(BaCl 2) = M(BaCl 2) . ν(BaCl 2) = 208 g/mol. 0.075 mol = 15.6 g

The mass of barium chloride solution is:

m(p−ra BaCl 2) = m(BaCl 2)/w(BaCl 2) . 100% = 15.6 g/15.6% . 100% = 100 g

The volume of the solution is calculated using the formula:

V(p−ra) = m(p−ra)/ρ(p−ra), where ρ(p−ra) is the density of the solution.

V(solution of BaCl 2) = m(solution of BaCl 2)/ρ(solution of BaCl 2)

V(solution BaCl 2) = 100 g/1.11 g/ml = 90 ml

Task No. 3

In what mass ratios should 10% solutions of sodium hydroxide and sulfuric acid be mixed to obtain a neutral solution of sodium sulfate? What is the mass fraction of salt in such a solution?

Answer: m(solution H 2 SO 4)/m(solution NaOH) = 1.225; w(Na 2 SO 4) = 8%Explanation:

The interaction of solutions of sodium hydroxide and sulfuric acid with the formation of an average salt (neutral solution) proceeds according to the following scheme:

2NaOH + H 2 SO 4 → Na 2 SO 4 + H 2 O

The masses of sodium hydroxide and sulfuric acid are equal:

m(NaOH) = 0.1m(solution NaOH); m(H 2 SO 4) = 0.1m(H 2 SO 4)

Accordingly, the amounts of sodium hydroxide and sulfuric acid are equal:

ν(NaOH) = m(NaOH)/M(NaOH) = 0.1m(NaOH solution)/40 g/mol

ν(H 2 SO 4) = m(H 2 SO 4)/M(H 2 SO 4) = 0.1m(solution H 2 SO 4)/98 g/mol

According to the reaction equation ν(H 2 SO 4) = 1/2ν(NaOH), therefore

0.1m(solution H 2 SO 4)/98 = ½. 0.1m(NaOH solution)/40

m(H 2 SO 4 solution)/m(NaOH solution) = 98/2/40 = 1.225

According to the reaction equation

ν(H 2 SO 4) = ν(Na 2 SO 4), therefore

ν(Na 2 SO 4) = 0.1m(H 2 SO 4 solution)/98 g/mol

m(Na 2 SO 4) = 0.1m(H 2 SO 4 solution)/98 g/mol. 142 g/mol = 0.145 m(H 2 SO 4 solution)

m(solution H 2 SO 4) = 1.225. m(NaOH solution)

The mass of the resulting sodium sulfate solution is:

m(Na 2 SO 4 solution) = m(H 2 SO 4 solution) + m(NaOH solution) = 2.225 m(NaOH solution) = 2.225(H 2 SO 4 solution)/ 1.225

w(Na 2 SO 4)% = 0.145 m(H 2 SO 4 solution). 1.225/2.225 m(H 2 SO 4 solution). 100% = 8.0%

Task No. 4

How many liters of chlorine (n.o.) will be released if 26.1 g of manganese (IV) oxide is added to 200 ml of 35% hydrochloric acid (density 1.17 g/ml) when heated? How many grams of sodium hydroxide in a cold solution will react with this amount of chlorine?

Answer: V(Cl 2) = 6.72 l; m(NaOH) = 24 g

Explanation:

When hydrochloric acid is added to manganese (IV) oxide, the reaction occurs:

MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2 O

Let's calculate the mass of hydrogen chloride and the mass of hydrochloric acid in solution:

m(solution of 35% HCl) = V(solution of HCl) . ρ(solution HCl) = 200 ml. 1.17 g/ml = 234 g

m(HCl) = M(HCl) . w(HCl)%/100% = 234 g. 35%/100% = 81.9 g

Hence the amount of HCl:

ν(HCl) = m(HCl)/M(HCl) = 81.9 g/36.5 g/mol = 2.244 mol

Amount of manganese(IV) oxide:

ν(MnO 2) = m(MnO 2)/M(MnO 2) = 26.1 g/87 g/mol = 0.3 mol

According to the reaction equation ν(MnO 2) = ¼ ν(HCl), and according to the condition ν(MnO 2)< ¼ν(HCl), следовательно, MnO 2 – вещество в недостатке, полностью прореагирует с соляной кислотой.

According to the reaction equation, ν(MnO 2) = ν(Cl 2) = 0.3 mol, therefore the volume released at no. chlorine is equal to:

V(Cl 2) = V m. ν(Cl 2) = 22.4 l/mol. 0.3 mol = 6.72 l

In a cold solution, alkali reacts with chlorine to form sodium hypochlorite (NaClO) and sodium chloride (NaCl) (disproportionation reaction):

Cl 2 + 2NaOH → NaClO + NaCl + H 2 O

According to the reaction equation, ν(Cl 2) = ½ ν(NaOH), therefore ν(NaOH) = 0.3 mol. 2 = 0.6 mol

The mass of sodium hydroxide reacted with chlorine in a cold solution is equal to:

m(NaOH) = M(NaOH) . ν(NaOH) = 40 g/mol. 0.6 mol = 24 g

Task No. 5

A mixture of copper and copper (II) oxide can react with 219 g of a 10% hydrochloric acid solution or 61.25 g of an 80% sulfuric acid solution. Determine the mass fraction of copper in the mixture.

Answer: 21%

Explanation:

Since copper is in the series of metal activities to the right of hydrogen, hydrochloric acid does not interact with it. HCl reacts only with copper(II) oxide to form salt and water:

Both copper and copper (II) oxide react with an 80% sulfuric acid solution:

Let's calculate the mass and amount of the HCl substance that reacts with CuO:

therefore, the amount of CuO reacting with HCl:

Let's calculate the mass and amount of the substance H 2 SO 4:

Since in the reaction of copper oxide with sulfuric acid

therefore, the reaction with Cu takes 0.2 mol of H 2 SO 4 and the amount of copper substance will be equal to:

The masses of substances are equal:

Therefore, the mass fraction of copper in the initial mixture is

Task No. 6

When 5.6 g of potassium hydroxide reacted with 5.0 g of ammonium chloride, ammonia was obtained. It was dissolved in 50 g of water. Determine the mass fraction of ammonia in the resulting solution. Determine the volume of a 10% nitric acid solution with a density of 1.06 g/ml that will be required to neutralize ammonia.

Answer: ω(NH 3)% = 3.1%; V(solution HNO 3) = 55.5 ml

Explanation:

As a result of the exchange reaction of potassium hydroxide with ammonium chloride, ammonia is released and potassium chloride and water are formed:

KOH + NH 4 Cl → KCl + NH 3 + H 2 O

Let's calculate the amounts of reacting KOH and NH 4 Cl:

According to the reaction equation, KOH and NH 4 Cl react in equal quantities, and according to the conditions of the problem, ν(KOH) > ν(NH 4 Cl). Therefore, ammonium chloride reacts completely and 0.09346 mol of ammonia is formed.

The mass fraction of ammonia in the resulting solution is calculated by the formula:

Ammonia reacts with nitric acid according to the reaction:

HNO 3 + NH 3 → NH 4 NO 3

Consequently, an equal amount of nitric acid reacts with ammonia, i.e.

therefore, the mass of reacted nitric acid is equal to:

The mass of the nitric acid solution is equal to:

The volume of a 10% nitric acid solution is equal to:

Task No. 7

Phosphorus (V) chloride weighing 4.17 g completely reacted with water. What volume of potassium hydroxide solution with a mass fraction of 10% (density 1.07 g/ml) is required to completely neutralize the resulting solution?

Answer: V(KOH solution) = 84 ml

Explanation:

Phosphorus (V) chloride is completely hydrolyzed to form phosphoric acid and hydrogen chloride:

The resulting orthophosphoric and hydrochloric acids are neutralized with a solution of potassium hydroxide, and medium salts are formed:

Let's calculate the amount of phosphorus (V) chloride that reacts with water:

Consequently, (according to the reaction equation)

And (5 times more than ν(PCl 5)).

To neutralize orthophosphoric acid, 0.06 mol of KOH is required, and to neutralize hydrochloric acid, 0.1 mol of KOH is required.

Therefore, the total amount of alkali required to neutralize the acid solution is equal.

The mass of alkali is equal to:

The mass of the alkali solution is equal to:

The volume of a 10% alkali solution is equal to:

Task No. 8

When pouring 160 g of a 10% solution of barium nitrate and 50 g of an 11% solution of potassium chromate, a precipitate formed. Calculate the mass fraction of barium nitrate in the resulting solution.

Answer: ω(Ba(NO 3) 2) = 4.24%

Explanation:

In the exchange reaction, when barium nitrate interacts with potassium chromate, potassium nitrate is formed, and a precipitate precipitates - barium chromate:

Ba(NO 3) 2 + K 2 CrO 4 → BaCrO 4 ↓ + 2KNO 3

Let us calculate the masses and quantities of reacting barium nitrate and potassium chromate:

According to the reaction equation, salts Ba(NO 3) 2 and K 2 CrO 4 react 1 to 1, according to the conditions of the problem ν(Ba(NO 3) 2) > ν(K 2 CrO 4), therefore, potassium chromate is in short supply and reacts completely .

Let's calculate the amount and mass of barium nitrate remaining after the reaction:

Since barium chromate precipitates, the mass of the solution obtained by combining solutions of barium nitrate and potassium chromate is equal to:

Let us calculate the mass of barium chromate that precipitated:

Then the mass of the solution is:

Find the mass fraction of barium nitrate in the resulting solution:

Task No. 9

If a mixture of potassium and calcium chlorides is added to a solution of sodium carbonate, 10 g of precipitate is formed. If the same mixture is added to a solution of silver nitrate, 57.4 g of precipitate is formed. Determine the mass fraction of potassium chloride in the initial mixture.

Answer: ω(KCl) = 57.3%

Explanation:

Calcium chloride enters into an exchange reaction with sodium carbonate, resulting in the precipitation of calcium carbonate:

CaCl 2 + Na 2 CO 3 → CaCO 3 ↓ + 2NaCl (I)

Potassium chloride does not react with sodium carbonate.

Therefore, the mass of precipitated calcium carbonate is equal to 10 g. Let us calculate its amount of substance:

Both chlorides enter into an exchange reaction with silver nitrate:

CaCl 2 + 2AgNO 3 → 2AgCl↓ + Ca(NO 3) 2 (II)

KCl + AgNO 3 → AgCl↓ + KNO 3 (III)

Let's calculate the total amount of silver chloride substance:

According to reaction (II), therefore

According to reaction (III), therefore,

The mass of potassium chloride in the initial mixture is equal to:

Let's calculate the mass fraction of potassium chloride in the mixture:

Task No. 10

A mixture of sodium and sodium oxide was dissolved in water. In this case, 4.48 l (n.s.) of gas was released and 240 g of solution with a mass fraction of sodium hydroxide of 10% was formed. Determine the mass fraction of sodium in the initial mixture.

Answer: ω(KCl) = 59.74%

Explanation:

When sodium oxide reacts with water, an alkali is formed, and when sodium reacts with water, an alkali is formed and hydrogen is released:

2Na + 2H 2 O → 2NaOH + H 2 (I)

Na 2 O + H 2 O → 2NaOH (II)

Let's calculate the amount of hydrogen released when sodium interacts with water:

According to the equation of reaction (I) ν(Na) = ν(NaOH) = 2ν(H 2) = 0.4 mol, therefore, the mass of sodium reacting with water and the mass of alkali formed as a result of reaction (I) are equal to:

m(Na) = 23 g/mol 0.4 mol = 9.2 g and mI (NaOH) = 40 g/mol 0.4 mol = 16 g

Let's calculate the total mass of alkali in the solution:

The mass and amount of alkali substance formed by reaction (II) are equal to:

m II (NaOH) = 24 g – 16 g = 8 g

According to the reaction equation (II) ν(Na 2 O) = 1/2ν II (NaOH), therefore,

ν(Na 2 O) = 0.2 mol/2 = 0.1 mol

m(Na 2 O) = M(Na 2 O) ν(Na 2 O) = 62 g/mol 0.1 mol = 6.2 g

Let's calculate the mass of the initial mixture consisting of sodium and sodium oxide:

m(mixture) = m(Na) + m(Na 2 O) = 9.2 g + 6.2 g = 15.4 g

The mass fraction of sodium in the mixture is equal to:

Task No. 11

A mixture of sodium carbonate and sodium bicarbonate can react with 73 g of 20% hydrochloric acid solution and 80 g of 10% sodium hydroxide solution. Determine the mass fraction of sodium carbonate in the initial mixture.

Answer: ω(Na 2 CO 3) = 38.7%

Explanation:

The interaction of sodium carbonate and bicarbonate with hydrochloric acid proceeds according to the reactions:

Na 2 CO 3 + 2HCl → 2NaCl + CO 2 + H 2 O (I)

NaHCO 3 + HCl → NaCl + CO 2 + H 2 O (II)

Sodium bicarbonate reacts with sodium hydroxide to form a medium salt:

NaHCO 3 + NaOH → Na 2 CO 3 + H 2 O (III)

Let's calculate the mass and amount of NaOH substance reacting according to reaction (III):

m(NaOH)=80 g*0.1 = 8 g

Therefore, according to reaction (III), 0.2 mol of NaHCO 3 reacts, since ν III (NaHCO 3) = ν (NaOH).

Let's calculate the total mass and amount of HCl substance reacting according to reactions (I) and (II):

v(HCl) = m(HCl)/M(HCl) = 14.6 g/36.5 g/mol = 0.4 mol

According to the reaction equation (II) ν II (NaHCO 3) = ν II (HCl), therefore, 0.2 mol of HCl reacts with sodium bicarbonate. Then it interacts with sodium carbonate according to reaction (I).

ν I (HCl) = 0.4 mol – 0.2 mol = 0.2 mol.

According to the reaction equation (I) ν(Na 2 CO 3) = 1/2ν(HCl) = 0.2 mol/2 = 0.1 mol.

Let's calculate the masses of bicarbonate and sodium carbonate in the initial mixture:

m(NaHCO 3) = M(NaHCO 3) ν(NaHCO 3) = 84 g/mol 0.2 mol = 16.8 g

m(Na 2 CO 3) = M(Na 2 CO 3) ν(Na 2 CO 3) = 106 g/mol 0.1 mol = 10.6 g

The mass of the initial mixture of salts is:

m(mixture) = m(NaHCO 3) + m(Na 2 CO 3) = 16.8 g + 10.6 g = 27.4 g

The mass fraction of sodium carbonate is equal to:

Task No. 12

A mixture of aluminum sulfide and aluminum was treated with water, and 6.72 l (no.) of gas was released. If the same mixture is dissolved in an excess of sodium hydroxide solution, 3.36 liters (n.s.) of gas will be released. Determine the mass fraction of aluminum in the initial mixture.

Answer: ω(Al) = 15.25%

Explanation:

When treating a mixture of aluminum sulfide and aluminum with water, only aluminum sulfide reacts:

Al 2 S 3 + 6H 2 O → 2Al(OH) 3 ↓ + 3H 2 S (I)

As a result of this interaction, hydrogen sulfide is formed, the amount of substance of which is:

n(H 2 S) = V(H 2 S)/V m = 6.72/22.4 = 0.3 mol,

The amount of hydrogen sulfide released depends only on the mass of aluminum sulfide (it does not depend on the amount of aluminum metal). Therefore, based on equation I, we can conclude that:

n(Al 2 S 3) = n(H 2 S) ⋅ 1/3, where 1 is the coefficient of Al 2 S 3, and 3 is the coefficient of H 2 S. Then:

n(Al 2 S 3) = 0.3 ⋅ 1/3 = 0.1 mol.

Hence:

m(Al 2 S 3) = n(Al 2 S 3) ⋅ M(Al 2 S 3) = 0.1 ⋅ 150 g = 15 g;

When the same mixture reacts in an excess of sodium hydroxide solution, gas is released only when sodium hydroxide reacts with aluminum:

Al 2 S 3 + 8NaOH → 2Na + 3Na 2 S (II)

Let's calculate the amount of hydrogen released during the interaction of aluminum with a NaOH solution:

n III (H 2) = V(H 2)/V m = 3.36/22.4 = 0.15 mol

According to the reaction equation (III) n(Al) = nIII (H 2) ⋅ 2/3, therefore, n(Al) = 0.1 mol

The mass of aluminum is:

m(Al) = M(Al) n(Al) = 27 g/mol 0.1 mol = 2.7 g

Therefore, the mass of the initial mixture is:

m(mixtures) = m(Al 2 S 3) + m(Al) = 15 g + 2.7 g = 17.7 g.

The mass fraction of aluminum in the initial mixture is equal to:

ω(Al) = 100% ⋅ 2.7/17.7 = 15.25%

Task No. 13

For the complete combustion of a mixture of carbon and silicon dioxide, oxygen weighing 22.4 g was consumed. What volume of a 20% solution of potassium hydroxide (ρ = 1.173 g/ml) can react with the initial mixture if it is known that the mass fraction of carbon in it is 70 %?

Answer: 28.6 ml

Explanation:

Silicon dioxide does not react with oxygen. When carbon is burned, carbon dioxide is produced:

C + O 2 → CO 2

Let's calculate the amount of carbon involved in combustion:

According to the reaction equation ν(O 2) = ν(C), therefore, ν(C) = 0.7 mol

Let's calculate the mass of burned carbon:

m(C) = M(C) ν(C) = 12 g/mol 0.7 mol = 8.4 g

Let's calculate the mass of the initial mixture of carbon and silicon dioxide:

Let's calculate the mass and amount of silicon dioxide:

m(SiO 2) = m(mixtures) – m(C) = 12 g – 8.4 g = 3.6 g

Only silicon dioxide reacts with alkali:

2KOH + SiO 2 → K 2 SiO 3 + H 2 O

As a result of this reaction, alkali is consumed in the amount of:

ν(KOH) = 2 0.06 mol = 0.12 mol

m(KOH) = M(KOH) ν(KOH) = 56 g/mol 0.12 mol = 6.72 g

Let's calculate the mass and volume of the KOH alkali solution:

Task No. 14

A mixture of bicarbonate and potassium carbonate with a mass fraction of carbonate in it of 73.4% can react with 40 g of a 14% solution of potassium hydroxide. The initial mixture was treated with an excess of sulfuric acid solution. What volume (n.s.) of gas is released in this case?

Answer: V(CO 2) = 6.72 l

Explanation:

Only an acidic salt, potassium bicarbonate, can react with alkali:

KHCO 3 + KOH → K 2 CO 3 + H 2 O

Let's calculate the mass and amount of potassium hydroxide:

According to the reaction equation ν(KOH) = ν(KHCO 3), therefore, ν(KHCO 3) = 0.1 mol

The mass of potassium bicarbonate in the mixture is equal to:

m(KHCO 3) = M(KHCO 3) ν(KHCO 3) = 100 g/mol 0.1 mol = 10 g

Mass fraction of bicarbonate in a mixture of salts

ω(KHCO 3) = 100% − ω(K 2 CO 3) = 100% − 73.4% = 26.6%

Let's calculate the mass of the salt mixture:

Let's calculate the mass and amount of sodium carbonate:

When the salts of potassium carbonate and bicarbonate interact with excess sulfuric acid, an acid salt is formed - potassium hydrogen sulfate:

KHCO 3 + H 2 SO 4 → KHSO 4 + CO 2 + H 2 O

K 2 CO 3 + 2H 2 SO 4 → 2KHSO 4 + CO 2 + H 2 O

The total amount and volume of carbon dioxide released are equal to:

ν(CO 2) = 0.2 mol + 0.1 mol = 0.3 mol

V(CO 2) = V m ν(CO 2) = 22.4 l/mol 0.3 mol = 6.72 l

Task No. 15

A mixture of magnesium and zinc filings was treated with an excess of dilute sulfuric acid, and 22.4 liters (n.s.) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 13.44 liters (n.s.) of hydrogen will be released. Calculate the mass fraction of magnesium in the initial mixture.

Answer: 19.75%

Explanation:

When treating a mixture of magnesium and zinc filings with dilute sulfuric acid, hydrogen is released:

Zn + H 2 SO 4 (diluted) → ZnSO 4 + H 2 (I)

Mg + H 2 SO 4 (dil.) → MgSO 4 + H 2 (II)

Only Zn (an amphoteric metal) reacts with an excess NaOH solution:

Zn + 2NaOH + 2H 2 O → Na 2 + H 2 (III)

Let's calculate the amount of hydrogen released by reaction (III):

According to the equation of reaction (III) ν III (H 2) = ν(Zn), therefore, ν(Zn) = 0.6 mol

The mass of zinc is:

m(Zn) = M(Zn) ν(Zn) = 65 g/mol 0.6 mol = 39 g

According to reaction equations (I) and (II) ν I (H 2) = ν (Zn) and ν II (H 2) = ν (Mg), the total amount of hydrogen released is:

Therefore, the amount of hydrogen released by reaction (II):

ν II (H 2) = ν(Mg) = 1 mol – 0.6 mol = 0.4 mol

The mass of magnesium is:

m(Mg) = M(Mg) ν(Mg) = 24 g/mol 0.4 mol = 9.6 g

Let's calculate the mass of magnesium and zinc:

m(mixtures) = m(Mg) + m(Zn) = 9.6 g + 39 g = 48.6 g

The mass fraction of magnesium in the initial mixture is equal to:

Task No. 16

8 g of sulfur was burned in excess oxygen. The resulting gas was passed through 200 g of an 8% sodium hydroxide solution. Determine the mass fractions of salts in the resulting solution.

Answer: ω(NaHSO 3) ≈ 4.81%; ω(Na 2 SO 3) ≈ 8.75%

Explanation:

When sulfur burns in excess oxygen, sulfur dioxide is formed:

S + O 2 → SO 2 (I)

Let's calculate the amount of burnt sulfur:

According to the reaction equation, ν(S) = ν(SO 2), therefore, ν(SO 2) = 0.25 mol

The mass of released sulfur dioxide is equal to:

m(SO 2) = M(SO 2) ν(SO 2) = 64 g/mol 0.25 mol = 16 g

Let's calculate the mass and amount of sodium hydroxide:

The reaction between alkali and NaOH proceeds in stages: first an acidic salt is formed, then it turns into an intermediate one:

NaOH + SO 2 → NaHSO 3 (II)

NaHSO 3 + NaOH → Na 2 SO 3 + H 2 O (III)

Since ν(NaOH) > ν(SO 2) and ν(NaOH)< 2ν(SO 2), следовательно, образуются средняя и кислая соли.

As a result of reaction (II), ν(NaHSO 3) = 0.25 mol is formed, and ν(NaOH) = 0.4 mol – 0.25 mol = 0.15 mol remains.

When sodium hydrogen sulfate and alkali interact (reaction (III)), sodium sulfite is formed with an amount of substance ν(Na 2 SO 3) = 0.15 mol, and what remains is ν(NaHSO 3) = 0.25 mol – 0.15 mol = 0, 1 mol.

Let's calculate the masses of hydrosulfite and sodium sulfite:

m(NaHSO 3) = M(NaHSO 3) ν(NaHSO 3) = 104 g/mol 0.1 mol = 10.4 g

m(Na 2 SO 3) = M(Na 2 SO 3) ν(Na 2 SO 3) = 126 g/mol 0.15 mol = 18.9 g

The mass of the solution obtained by passing sulfur dioxide through an alkali solution is equal to:

m(p−ra) = m(p−ra NaOH) + m(SO 2) = 200 g + 16 g = 216 g

The mass fractions of salts in the resulting solution are equal to:

Task No. 17

A mixture of aluminum and iron filings was treated with an excess of dilute hydrochloric acid, and 8.96 liters (n.s.) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 6.72 liters (n.s.) of hydrogen will be released. Calculate the mass fraction of iron in the initial mixture.

Answer: ω(Fe) ≈ 50.91%

Explanation:

Reactions of aluminum and iron filings in an excess of dilute hydrochloric acid proceed with the release of hydrogen and the formation of salts:

Fe + 2HCl → FeCl 2 + H 2 (I)

2Al + 6HCl → 2AlCl 3 + 3H 2 (II)

Only aluminum filings react with an alkali solution, resulting in the formation of a complex salt and hydrogen:

2Al + 2NaOH + 6H 2 O → 2Na + 3H 2 (III)

Let us calculate the amount of hydrogen released by reaction (III):

Let's calculate the amount of hydrogen released by reactions (I) + (II):

According to the reaction equation (III) ν III (Al) = 2/3ν III (H 2), therefore, ν III (Al) = 2/3 · 0.3 mol = 0.2 mol.

The mass of aluminum filings is:

m(Al) = M(Al) ν(Al) = 27 g/mol 0.2 mol = 5.4 g

According to the reaction equation (II) ν II (Al) = 2/3ν II (H 2), therefore, ν II (H 2) = 3/2 0.2 mol = 0.3 mol

The amount of hydrogen released by reaction (I) is equal to:

ν I (H 2) = ν I + II (H 2) – ν II (H 2) = 0.4 mol – 0.3 mol = 0.1 mol.

According to the reaction equation (I) ν I (Fe) = ν I (H 2), therefore, ν I (Fe) = 0.1 mol

The mass of iron filings is:

m(Fe) = M(Fe) ν(Fe) = 56 g/mol 0.1 mol = 5.6 g

The mass of iron and aluminum filings is equal to:

m(sawdust) = m(Al) + m(Fe) = 5.4 g + 5.6 g = 11 g

The mass fraction of iron in the initial mixture is equal.

Problems on mixtures (USE-2017, No. 33)

Free-response questions are worth a maximum of 4 points

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.s.) were released. Find the mass fraction of magnesium in the mixture.

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 0.986 l (n.s.). Determine the composition of the initial mixture.

When iron weighing 28 g reacted with chlorine, a mixture of iron (II) and (III) chlorides weighing 77.7 g was formed. Calculate the mass of iron (III) chloride in the resulting mixture.

When 8.2 g of a mixture of copper, iron and aluminum was 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 (DS). Determine the composition of the initial mixture in mass percent.

When a mixture of copper and copper oxide 2 was dissolved in concentrated nitric acid, 18.4 g of brown gas was released and 470 solutions were obtained with a mass fraction of salt of 20%.
Determine the mass fraction of copper oxide in the initial mixture.

Multiple-choice assignments are worth a maximum of 2 points (Unified State Examination 2017)

1. (Unified State Exam No. 5) Match the formula of a substance with the class/group to which this substance belongs

FORMULA OF SUBSTANCE CLASS/GROUP

A) CaH 2 1) salt-forming oxide

B) NaH 2 PO 4 2) non-salt-forming oxide

B) H 3 N 3) medium salt

D) SeO 3 4) acid

5) sour salt

6) Binary connection

2. (Unified State Exam No. 7) What properties can the following substances exhibit?

FORMULA OF SUBSTANCE PROPERTIES

A) HNO 3 1) properties of bases

B) NaOH 2) properties of salts

B) Fe(OH) 2 3) properties of acids

D) Zn(NO 3) 2 4) properties of acids and salts

5) properties of salts and bases

3. (USE No. 7) Establish a correspondence between a solid substance and the products of its interaction with water:

A) BaBr 2 1) Ba(OH) 2 + HBr

B) Al 2 S 3 2) Ba(OH) 2 + NH 3

B) KH 2 3) Ba 2+ + 2Br -

D) Ba 3 N 2 4) H 2 + KOH

5) Al(OH) 3 + H 2 S

6) does not interact

4. (USE No. 5) Establish a correspondence between the oxide and its corresponding hydroxide

A) N 2 O 3 1) HPO 3

B) SeO 2 2) CuOH

B) Cu 2 O 3) H 2 SeO 3

D) P 2 O 3 4) H 2 SeO 4

5. (Unified State Exam No. 9, No. 17) The following scheme of substance transformations is specified:

Sr === X ==== NH 3 === Y

Determine which of the indicated substances are substances X and Y.

Write down the numbers of the selected substances under the corresponding letters in the table.

6. (USE No. 22) Establish a correspondence between the formula of the salt and the products of electrolysis of an aqueous solution of this salt, which were released on the inert electrodes: for each position indicated by a letter, select the corresponding position indicated by a number.

SALT FORMULA ELECTROLYSIS PRODUCTS

A) Na 3 PO4 1) O 2, H 2

B) KF 2) O 2, Mg

B) MgBr 2 3) H 2, Mg

D) Mg(NO 3) 2 4) Na, O 2

Write down the selected numbers in the table under the corresponding

7. (USE No. 27) Calculate mass copper sulfate(CuSO 4 * 5H 2 O), which must be dissolved in water to obtain 240 g of a 10% copper sulfate solution.

8. (Unified State Exam No. 27) Two solutions were mixed, the mass of the first solution was 80 g, with a mass fraction of sodium sulfate of 5%, the mass of the second solution was 40 g, with a mass fraction of sodium sulfate of 16%. Determine the mass fraction of sodium sulfate in the newly obtained solution.

Answer: ___________________ (Write down the number to the nearest tenth.)

9. (USE 35 (5)) Using the electron balance method, create the reaction equation:

CH 3 -CH = CH-CH 3 + KMnO 4 + H 2 O == CH3-CH(OH)-CH(OH)-CH 3 + MnO 2 + KOH