By dissolving a mixture of copper. Problems for mixtures and alloys of metals. Typical misconceptions and errors that arise when solving problems on a mixture
Mix tasks (USE-2017, No. 33)
Tasks with a free answer are estimated at 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. At the same time, 2.24 liters of gas (n.u.) were released. Find the mass fraction of magnesium in the mixture.
When 3.04 g of a mixture of iron and copper are dissolved in dilute nitric acid, nitric oxide (II) with a volume of 0.986 L (NU) is released. Determine the composition of the original mixture.
The interaction of iron weighing 28 g with chlorine formed a mixture of iron (II) and (III) chlorides weighing 77.7 g. Calculate the mass of iron (III) chloride in the resulting mixture.
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.
When dissolving a mixture of copper and copper oxide 2 in concentrated nitric acid, 18.4 g of brown gas were released and 470 solution was obtained with a mass fraction of salt of 20%.
Determine the mass fraction of copper oxide in the initial mixture.
Tasks with a choice of answers, estimated at a maximum of 2 points (USE 2017)
1. (Unified State Exam No. 5) Establish a correspondence between the formula of a substance and 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) acidic salt
6) Binary compound
2. (Unified State Exam № 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. (Unified State Exam No. 7) Establish a correspondence between a solid 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. (Unified State Exam No. 5) Establish a correspondence between the oxide and the corresponding hydroxide
A) N 2 O 3 1) HPO 3
B) SeO 2 2) CuOH
C) Cu 2 O 3) H 2 SeO 3
D) P 2 O 3 4) H 2 SeO 4
5. (USE No. 9, No. 17) The following scheme of transformations of substances is given:
Sr === X ==== NH 3 === Y
Determine which of the specified substances are substances X and Y.
Write down the numbers of the selected substances in the table under the appropriate letters.
6. (Unified State Exam No. 22) Establish a correspondence between the salt formula and electrolysis products aqueous solution of this salt, which were released on inert electrodes: for each position marked with a letter, select the corresponding position marked with a number.
SALT FORMULA ELECTROLYSIS PRODUCTS
A) Na 3 PO4 1) O 2, H 2
B) KF 2) O 2, Mg
C) 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 the mass of copper sulfate (CuSO 4 * 5H 2 O), which must be dissolved in water to obtain 240 g of a 10% copper sulfate solution.
8. (USE No. 27) Mixed two solutions, the mass of the first solution is 80 g, with a mass fraction of sodium sulfate 5%, the mass of the second solution weighing 40 g, with a mass fraction of sodium sulfate 16%. Determine the mass fraction of sodium sulfate in the newly obtained solution.
Answer: ___________________ g. (Write down the number to the nearest tenth.)
9. (USE 35 (5)) Using the electronic 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
Tasks to determine the quantitative composition of the mixture. Chemical properties metals.
1. A mixture of aluminum and iron filings was treated with an excess of dilute hydrochloric acid, while 8.96 L (NU) hydrogen was released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 6.72 liters of hydrogen will be released. Calculate the mass fraction of iron in the original mixture.
2. A mixture of magnesium and zinc sawdust was treated with an excess of dilute sulfuric acid, while 22.4 liters (NU) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 13.44 liters (NU) of hydrogen will be released. Calculate the mass fraction of magnesium in the original mixture.
3. When a mixture of copper and copper (II) oxide was dissolved in concentrated nitric acid, 18.4 g of brown gas were 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, while 6.72 L (NL) of gas was evolved. If the same mixture is dissolved in an excess of sodium hydroxide solution, 3.36 L (NU) gas will be released. Determine the mass fraction of aluminum in the original mixture.
5. If a mixture of potassium and calcium chlorides is added to a sodium carbonate solution, 10 g of precipitate is formed. If the same mixture is added to the silver nitrate solution, 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, while 4.48 liters of gas (NU) were released. 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, while 2.24 liters of gas (NU) 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 solution of alkali. At the same time, a gas with a volume of 8.96 liters (NU) was released. Calculate the mass fraction of copper in the initial mixture.
On the subject: methodological developments, presentations and notes
Lesson development contains a detailed lesson outline, slides for the lesson, workbook on the topic under study, instructional cards for conducting the experiment and other didactic material ...
Mix tasks are a very common task in chemistry. They require a clear idea of which of the 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 with a mixture.
- 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 composed as follows:
CaO + BaO + 4HCl = CaCl 2 + BaCl 2 + 2H 2 O.
This is a mistake, because this mixture can contain any amount of each oxide.
And in the above equation it is assumed that their equal amount. - The assumption that their molar ratio corresponds to the coefficients in the reaction equations.
For example:
Zn + 2HCl = ZnCl 2 + H 2
2Al + 6HCl = 2AlCl 3 + 3H 2
The amount of zinc is taken as x, and the amount of aluminum is taken as 2x (in accordance with the coefficient in the reaction equation). This is also wrong. These quantities can be any and they are not related to each other. - Attempts to find the "amount of substance in a mixture" by dividing its mass by the sum of the molar masses of the components.
This action does not make any sense at all. Each molar mass can only refer to a single substance.
The reaction of metals with acids is often used in such problems. To solve such problems, you need to know exactly which metals interact with which acids and which do not.
Necessary theoretical information.
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"] = m component / m mixture
- Mole fraction of a component in a mixture- the ratio of the number of moles (amount of substance) of the component to the total number of moles of all substances in the mixture. For example, if the mixture includes substances A, B and C, then:
χ ["chi"] component A = 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:
N component A: n component 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"] = V component / V mixture
Electrochemical series of metal voltages.
| Li Rb K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb H Sb Bi Cu Hg Ag Pd Pt Au |
Reactions of metals with acids.
- With mineral acids, which include all soluble acids ( except for nitrogen and concentrated sulfuric, the interaction of which with metals occurs in a special way), react only metals, in the electrochemical series of voltages are to (to the left of) hydrogen.
- In this case, metals that have several oxidation states (iron, chromium, manganese, cobalt) exhibit the lowest possible oxidation state - usually +2.
- 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 indication of a specific substance.
Nitric acid reduction products.
| The more active the metal and the lower the acid concentration, the further nitrogen is reduced. | ||||
| NO 2 | NO | N 2 O | N 2 | NH 4 NO 3 |
| Inactive metals (to the right of iron) + conc. acid Non-metals + conc. acid |
Inactive metals (to the right of iron) + dil. acid | Active metals (alkali, alkaline earth, zinc) + conc. acid | Active metals (alkali, alkaline earth, zinc) + acid of medium dilution | Active metals (alkali, alkaline earth, zinc) + very decomp. acid |
| Passivation: do not react with cold concentrated nitric acid: Al, Cr, Fe, Be, Co. |
||||
| Do not react with nitric acid at no concentration: Au, Pt, Pd. |
||||
Sulfuric acid reduction products.
Reactions of metals with water and with alkalis.
- They dissolve in water at room temperature only metals, which 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.
- 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.
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 example 1.
- Find the amount of hydrogen:
n = V / V m = 5.6 / 22.4 = 0.25 mol. - 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. - 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%
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 (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 using a very simple system of equations, taking for x - the number of moles of one of the metals, and for y - the amount of substance of the second.
Solution of example 2.
- Find the amount of hydrogen:
n = V / V m = 8.96 / 22.4 = 0.4 mol. - 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:
It is much more convenient to solve such systems by the subtraction method, multiplying the first equation by 18:
27x + 18y = 7.2
and subtracting the first equation from the second:(56 - 18) y = 11 - 7.2
y = 3.8 / 38 = 0.1 mol (Fe)
x = 0.2 mol (Al) - Then 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.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.
The next three problem examples (# 4, 5, 6) involve metal reactions 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.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.
With alkali reacts only aluminum- amphoteric metal (in addition to aluminum, zinc and tin also dissolve in alkalis, in hot concentrated alkali - you can still dissolve beryllium).
Solution example 4.
- Only copper reacts with concentrated sulfuric acid, the number of moles of gas:
n SO 2 = V / Vm = 5.6 / 22.4 = 0.25 mol - The number of moles of hydrogen:
n H 2 = 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 - The remainder is iron, weighing 3 g. You can find the mass of the mixture:
m mixture = 16 + 2.7 + 3 = 21.7 g. - 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 a nitric acid solution containing 20 wt. % HNO 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.
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.
- Determine the amount of gas substance:
n N 2 = V / Vm = 2.912 / 22.4 = 0.13 mol. - We 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
m HNO 3 = ω m solution = 0.2 630.3 = 126.06 g
n HNO 3 = m / M = 126.06 / 63 = 2 molPlease 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 was there an excess of acid, and how much of it remained after the reaction in the resulting solution.
- We compose the reaction equations ( do not forget about electronic balance) and, for the convenience of calculations, we take for 5x - the amount of zinc, and for 10y - the amount of aluminum. Then, in accordance with the coefficients in the equations, nitrogen in the first reaction will be x mol, and in the second - 3y mol:
5x x 5Zn + 12HNO 3 = 5Zn (NO 3) 2 + N 2 + 6H 2 O 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 molLet's check the mass of the mixture:
0.2 65 + 0.3 27 = 21.1 g. - 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:
0,2 0,48 0,2 0,03 5Zn + 12HNO 3 = 5Zn (NO 3) 2 + N 2 + 6H 2 O 0,3 1,08 0,3 0,09 10Al + 36HNO 3 = 10Al (NO 3) 3 + 3N 2 + 18H 2 O - 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 HNO 3 = 0.48 + 1.08 = 1.56 mol,
those. acid was in excess and you can calculate its remainder in solution:
n HNO 3 rest. = 2 - 1.56 = 0.44 mol. - 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 the amount of 0.44 mol:
m HNO 3 rest. = n M = 0.44 63 = 27.72 g - What is the mass of the final solution?
Recall that the mass of the final solution consists of those components that we mixed (solutions and substances) minus those reaction products that left the solution (precipitates and gases):Then for our task:
M new. solution = 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 - Now you can calculate the mass fractions of substances in the resulting solution:
ωZn (NO 3) 2 = m in-va / m p-ra = 37.8 / 648.04 = 0.0583
ωAl (NO 3) 3 = m in-va / m p-ra = 63.9 / 648.04 = 0.0986
ω HNO 3 rest. = m in-va / m p-ra = 27.72 / 648.04 = 0.0428
Answer: 5.83% zinc nitrate, 9.86% aluminum nitrate, 4.28% nitric acid.
Example 6.When treating 17.4 g of a mixture of copper, iron and aluminum 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 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 1.11 g of a mixture of iron and aluminum sawdust was dissolved 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 - were formed. Determine the quantitative composition of the initial mixture.
2-4. The interaction of iron weighing 28 g with chlorine formed a mixture of iron (II) and (III) chlorides weighing 77.7 g. Calculate the 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. A mixture of lithium and sodium with a total mass of 7.6 g was oxidized with an excess of oxygen; a total of 3.92 L (NU) 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. 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 sample 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) 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 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 components 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 weight of 7.1 g was treated with dilute hydrochloric acid, the weight 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);
1-3. 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, 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)
Task number 1
When a mixture of copper and copper (II) oxide was dissolved in concentrated nitric acid, 18.4 g of brown gas were 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 dissolve, 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 gas NO 2 is released only when copper interacts with concentrated nitric acid. Let's find its quantity by 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 the 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.
The 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 equal to:
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:
m total (Cu (NO 3) 2) = w (Cu (NO 3) 2). m (p − ra) / 100% = 20%. 470g / 100% = 94g.
The mass of copper nitrate formed by the interaction 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 interaction 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 (mixture) = m (CuO) + m (Cu) = 24 g + 12.8 g = 36.8 g
Mass fraction of copper oxide in the mixture is equal to:
w (CuO)% = m (CuO) / m (mixture). 100% = 24g / 36.8g. 100% = 65.22%
Task number 2
Determine the mass fraction of sodium carbonate in a solution obtained by boiling 150 g of an 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? Evaporation of water is negligible.
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 solute is calculated by the formula:
w (in-in)% = m (in-va) / m (r-ra). 100%,
where m (in-va) is the mass of the solute, m (p − ra) is the mass of the solution).
Let's calculate the mass of dissolved sodium bicarbonate (NaHCO 3):
M (NaHCO 3) = m (solution). 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 of carbon dioxide from the mass of the initial solution:
m (p-ra) = m (original r-ra) - m (CO 2) = 150 g - 3.3 g = 146.7 g
w (Na 2 CO 3)% = m (Na 2 CO 3) / m (solution). 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 the barium chloride solution is:
m (solution BaCl 2) = m (BaCl 2) / w (BaCl 2). 100% = 15.6g / 15.6%. 100% = 100 g
The volume of the solution is calculated by 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 number 3
In what mass ratios should 10% sodium hydroxide and sulfuric acid solutions be mixed to obtain a neutral sodium sulfate solution? What is the mass fraction of salt in such a solution?
Answer: m (solution of H 2 SO 4) / m (solution of NaOH) = 1.225; w (Na 2 SO 4) = 8%Explanation:
The interaction of sodium hydroxide and sulfuric acid solutions with the formation of a medium salt (neutral solution) proceeds according to the 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 of 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 (solution NaOH) / 40
m (solution H 2 SO 4) / m (solution NaOH) = 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 (solution H 2 SO 4) / 98 g / mol
m (Na 2 SO 4) = 0.1m (solution H 2 SO 4) / 98 g / mol. 142 g / mol = 0.145 m (solution H 2 SO 4)
m (solution H 2 SO 4) = 1.225. m (solution NaOH)
The mass of the resulting sodium sulfate solution is:
m (solution of Na 2 SO 4) = m (solution of H 2 SO 4) + m (solution of NaOH) = 2.225 m (solution of NaOH) = 2.225 (solution of H 2 SO 4) / 1,225
w (Na 2 SO 4)% = 0.145 m (solution H 2 SO 4). 1.225 / 2.225 m (solution H 2 SO 4). 100% = 8.0%
Task number 4
How many liters of chlorine (n.u.) 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 proceeds:
MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2 O
We 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
The 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 by 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 during normal operation. 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 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:
m (NaOH) = M (NaOH). ν (NaOH) = 40 g / mol. 0.6 mol = 24 g
Task number 5
A mixture of copper and copper (II) oxide can react with 219 g of 10% hydrochloric acid solution or 61.25 g of 80% sulfuric acid solution. Determine the mass fraction of copper in the mixture.
Answer: 21%
Explanation:
Since copper is among the activities of metals 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:
Consequently, the mass fraction of copper in the initial mixture
Task number 6
When 5.6 g of potassium hydroxide were 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 the 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 amount of reacting KOH and NH 4 Cl:
According to the reaction equation, KOH and NH 4 Cl react in equal amounts, and according to the condition 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
Therefore, an equal amount of nitric acid reacts with ammonia, i.e.
therefore, the mass of the reacted nitric acid is equal to:
The mass of the nitric acid solution is:
The volume of a 10% nitric acid solution is:
Task number 7
Phosphorus (V) chloride, 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 (solution KOH) = 84 ml
Explanation:
Phosphorus (V) chloride is completely hydrolyzed to form orthophosphoric acid and hydrogen chloride:
The formed orthophosphoric and hydrochloric acids are neutralized with a solution of potassium hydroxide, while medium salts are formed:
Let's calculate the amount of phosphorus (V) chloride substance that reacts with water:
Consequently, (according to the reaction equation)
And (5 times more than ν (PCl 5)).
To neutralize phosphoric acid, 0.06 mol of KOH is needed, and 0.1 mol of KOH is needed to neutralize hydrochloric acid.
Therefore, the total amount of alkali required to neutralize the acid solution is.
The mass of alkali is equal to:
The mass of the alkali solution is:
The volume of a 10% alkali solution is:
Task number 8
When draining 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 is formed - barium chromate:
Ba (NO 3) 2 + K 2 CrO 4 → BaCrO 4 ↓ + 2KNO 3
We calculate the masses and quantities of the reacting barium nitrate and potassium chromate:
According to the reaction equation, the salts Ba (NO 3) 2 and K 2 CrO 4 react 1 to 1, according to the condition of the problem ν (Ba (NO 3) 2)> ν (K 2 CrO 4), therefore, potassium chromate is deficient and fully reacts ...
We calculate the amount and mass of barium nitrate remaining after the reaction:
Since barium chromate precipitates, the mass of the solution obtained by merging solutions of barium nitrate and potassium chromate is:
Calculate the mass of precipitated barium chromate:
Then the mass of the solution is equal to:
We find the mass fraction of barium nitrate in the resulting solution:
Task number 9
If a mixture of potassium and calcium chlorides is added to a sodium carbonate solution, 10 g of precipitate is formed. If the same mixture is added to the silver nitrate solution, 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, as a result of which calcium carbonate precipitates:
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 10 g. Let's 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:
We calculate the mass fraction of potassium chloride in the mixture:
Task number 10
A mixture of sodium and sodium oxide was dissolved in water. At the same time, 4.48 liters (NU) of gas were released and 240 g of a solution with a mass fraction of sodium hydroxide of 10% were formed. Determine the mass fraction of sodium in the original mixture.
Answer: ω (KCl) = 59.74%
Explanation:
When sodium oxide interacts with water, an alkali is formed, and when sodium interacts 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 substance released during the interaction of sodium with water:
According to the reaction equation (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:
m (Na) = 23 g / mol 0.4 mol = 9.2 g and m I (NaOH) = 40 g / mol 0.4 mol = 16 g
Let's calculate the total mass of alkali in solution:
The mass and amount of the 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
We 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 number 11
A mixture of sodium carbonate and sodium bicarbonate can be reacted 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 original mixture.
Answer: ω (Na 2 CO 3) = 38.7%
Explanation:
The interaction of sodium carbonate and sodium 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 us calculate the mass and amount of the NaOH substance reacting according to reaction (III):
m (NaOH) = 80 g * 0.1 = 8 g
Therefore, according to reaction (III) 0.2 mol NaHCO 3 reacts, since ν III (NaHCO 3) = ν (NaOH).
Let us calculate the total mass and amount of the 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 with sodium carbonate according to reaction (I) interacts
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.
We calculate the masses of sodium 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 equal to:
m (mixture) = m (NaHCO 3) + m (Na 2 CO 3) = 16.8 g + 10.6 g = 27.4 g
Mass fraction of sodium carbonate is equal to:
Task number 12
A mixture of aluminum sulfide and aluminum was treated with water, while 6.72 L (NL) of gas was evolved. If the same mixture is dissolved in an excess of sodium hydroxide solution, 3.36 L (NU) gas will be released. Determine the mass fraction of aluminum in the original mixture.
Answer: ω (Al) = 15.25%
Explanation:
When processing a mixture of aluminum and aluminum sulfide, only aluminum sulfide reacts with water:
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 the substance of which is:
n (H 2 S) = V (H 2 S) / V m = 6.72 / 22.4 = 0.3 mol,
The amount of released hydrogen sulfide depends only on the mass of aluminum sulfide (it does not depend on the amount of metallic aluminum). 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 before Al 2 S 3, and 3 is the coefficient before 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 interacts in an excess of sodium hydroxide solution, gas is released only when sodium hydroxide interacts with aluminum:
Al 2 S 3 + 8NaOH → 2Na + 3Na 2 S (II)
Let us 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) = n III (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 original mixture:
m (mixture) = 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 number 13
For the complete combustion of a mixture of carbon and silicon dioxide, oxygen with a mass of 22.4 g was consumed. What volume of a 20% potassium hydroxide solution (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 formed:
C + O 2 → CO 2
Let's calculate the amount of carbon substance 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 the burned carbon:
m (C) = M (C) ν (C) = 12 g / mol 0.7 mol = 8.4 g
Let us calculate the mass of the initial mixture of carbon and silicon dioxide:
We calculate the mass and amount of the silicon dioxide substance:
m (SiO2) = m (mixture) - m (C) = 12 g - 8.4 g = 3.6 g
Only silicon dioxide interacts with alkali:
2KOH + SiO 2 → K 2 SiO 3 + H 2 O
As a result of this reaction, alkali is consumed in an amount:
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 number 14
A mixture of bicarbonate and potassium carbonate with a mass fraction of carbonate in it 73.4% can react with 40 g of a 14% potassium hydroxide solution. The initial mixture was treated with an excess of sulfuric acid solution. What volume (n.o.) of gas is released in this case?
Answer: V (CO 2) = 6.72 L
Explanation:
Only acidic salt - potassium bicarbonate can react with alkali:
KHCO 3 + KOH → K 2 CO 3 + H 2 O
We calculate the mass and amount of the potassium hydroxide substance:
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:
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%
We calculate the mass of the mixture of salts:
Let's calculate the mass and amount of sodium carbonate substance:
When the salts of carbonate and potassium bicarbonate react with an excess of sulfuric acid, an acidic 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 emitted carbon dioxide are equal:
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 number 15
A mixture of magnesium and zinc sawdust was treated with an excess of dilute sulfuric acid, while 22.4 liters (NU) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 13.44 liters (NU) of hydrogen will be released. Calculate the mass fraction of magnesium in the original mixture.
Answer: 19.75%
Explanation:
When processing a mixture of magnesium and zinc sawdust with dilute sulfuric acid, hydrogen is released:
Zn + H 2 SO 4 (dil.) → ZnSO 4 + H 2 (I)
Mg + H 2 SO 4 (dil.) → MgSO 4 + H 2 (II)
Only Zn (amphoteric metal) reacts with an excess of NaOH solution:
Zn + 2NaOH + 2H 2 O → Na 2 + H 2 (III)
Let us calculate the amount of hydrogen released by reaction (III):
According to the reaction equation (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 the reaction equations (I) and (II) ν I (H 2) = ν (Zn) and ν II (H 2) = ν (Mg), the total amount of released hydrogen 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 (mixture) = m (Mg) + m (Zn) = 9.6 g + 39 g = 48.6 g
Mass fraction of magnesium in the initial mixture is equal to:
Task number 16
In excess of oxygen, 8 g of sulfur were burned. The resulting gas was passed through 200 g of 8% sodium hydroxide solution. Determine the mass fraction of salts in the resulting solution.
Answer: ω (NaHSO 3) ≈ 4.81%; ω (Na 2 SO 3) ≈ 8.75%
Explanation:
When sulfur burns in an excess of oxygen, sulfur dioxide is formed:
S + O 2 → SO 2 (I)
Let's calculate the amount of the burnt sulfur substance:
According to the reaction equation ν (S) = ν (SO 2), therefore, ν (SO 2) = 0.25 mol
The mass of the 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 substance:
The reaction between alkali and NaOH proceeds stepwise: first, an acidic salt is formed, then it turns into a medium 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.
During the interaction of sodium hydrogen sulfate and alkali (reaction (III)), sodium sulfite is formed with the amount of substance ν (Na 2 SO 3) = 0.15 mol, and ν (NaHSO 3) = 0.25 mol - 0.15 mol = 0, 1 mole.
We calculate the masses of sodium 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:
m (p-pa) = m (p-pa NaOH) + m (SO 2) = 200 g + 16 g = 216 g
Mass fractions of salts in the resulting solution are equal to:
Task number 17
A mixture of aluminum and iron filings was treated with an excess of dilute hydrochloric acid, while 8.96 L (NU) hydrogen was released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 6.72 liters of hydrogen will be released. Calculate the mass fraction of iron in the original mixture.
Answer: ω (Fe) ≈ 50.91%
Explanation:
The reactions of aluminum and iron filings in an excess of dilute hydrochloric acid proceed with the evolution of hydrogen and the formation of salts:
Fe + 2HCl → FeCl 2 + H 2 (I)
2Al + 6HCl → 2AlCl 3 + 3H 2 (II)
Only aluminum sawdust reacts 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 substance released by reaction (III):
Let us calculate the amount of hydrogen substance 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 equal to:
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 substance 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 equal to:
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 to.
