Task number 1
Sodium was heated in a hydrogen atmosphere. When water was added to the resulting substance, gas evolution and the formation of a clear solution were observed. A brown gas was passed through this solution, which was obtained as a result of the interaction of copper with a concentrated solution of nitric acid. Write the equations for the four described reactions.
1) When sodium is heated in a hydrogen atmosphere (T = 250-400 o C), sodium hydride is formed):
2Na + H2 = 2NaH
2) When water is added to sodium hydride, alkali NaOH is formed, and hydrogen is released:
NaH + H 2 O \u003d NaOH + H 2
3) When copper interacts with a concentrated solution of nitric acid, brown gas is released - NO 2:
Cu + 4HNO 3 (conc.) = Cu(NO 3) 2 + 2NO 2 + 2H 2 O
4) When brown gas NO 2 is passed through an alkali solution, a disproportionation reaction proceeds - nitrogen N +4 is simultaneously oxidized and reduced to N +5 and N +3:
2NaOH + 2NO 2 \u003d NaNO 3 + NaNO 2 + H 2 O
(disproportionation reaction 2N +4 → N +5 + N +3).
Task number 2
Iron scale was dissolved in concentrated nitric acid. Sodium hydroxide solution was added to the resulting solution. The precipitate formed was separated and calcined. The resulting solid residue was fused with iron. Write the equations for the four described reactions.
The formula of iron oxide is Fe 3 O 4.
When iron oxide reacts with concentrated nitric acid, iron nitrate is formed and nitric oxide NO 2 is released:
Fe 3 O 4 + 10HNO 3 (conc.) → 3Fe (NO 3) 3 + NO 2 + 5H 2 O
When iron nitrate reacts with sodium hydroxide, a precipitate is released - iron (III) hydroxide:
Fe(NO 3) 3 + 3NaOH → Fe(OH) 3 ↓ + 3NaNO 3
Fe (OH) 3 - amphoteric hydroxide, insoluble in water, decomposes when heated into iron oxide (III) and water:
2Fe(OH) 3 → Fe 2 O 3 + 3H 2 O
When iron(III) oxide is fused with iron, iron(II) oxide is formed:
Fe 2 O 3 + Fe → 3FeO
Task number 3
The sodium was burned in the air. The resulting substance was treated with hydrogen chloride when heated. The resulting simple yellow-green substance reacted with chromium (III) oxide in the presence of potassium hydroxide when heated. When a solution of one of the formed salts was treated with barium chloride, a yellow precipitate formed. Write the equations for the four described reactions.
1) When sodium is burned in air, sodium peroxide is formed:
2Na + O 2 → Na 2 O 2
2) When sodium peroxide interacts with hydrogen chloride, Cl 2 gas is released when heated:
Na 2 O 2 + 4HCl → 2NaCl + Cl 2 + 2H 2 O
3) B alkaline environment chlorine reacts when heated with amphoteric chromium oxide to form chromate and potassium chloride:
Cr 2 O 3 + 3Cl 2 + 10KOH → 2K 2 CrO 4 + 6KCl + 5H 2 O
2Cr +3 -6e → 2Cr +6 | . 3 - oxidation
Cl 2 + 2e → 2Cl - | . 1 - recovery
4) Sediment yellow color(BaCrO 4) is formed by the interaction of potassium chromate and barium chloride:
K 2 CrO 4 + BaCl 2 → BaCrO 4 ↓ + 2KCl
Task number 4
Zinc was completely dissolved in a concentrated potassium hydroxide solution. The resulting clear solution was evaporated and then calcined. The solid residue was dissolved in the required amount of hydrochloric acid. Ammonium sulfide was added to the resulting clear solution and a white precipitate formed. Write the equations for the four described reactions.
1) Zinc reacts with potassium hydroxide to form potassium tetrahydroxozincate (Al and Be behave similarly):
2) Potassium tetrahydroxozincate, after calcination, loses water and turns into potassium zincate:
3) Potassium zincate when interacting with hydrochloric acid forms zinc chloride, potassium chloride and water:
4) Zinc chloride, as a result of interaction with ammonium sulfide, turns into insoluble zinc sulfide - a white precipitate:
Task number 5
The hydroiodic acid was neutralized with potassium bicarbonate. The resulting salt reacted with a solution containing potassium dichromate and sulfuric acid. When the resulting simple substance reacted with aluminum, a salt was obtained. This salt was dissolved in water and mixed with a potassium sulfide solution, resulting in a precipitate and gas evolution. Write the equations for the four described reactions.
1) Hydroiodic acid is neutralized with a weak acid salt carbonic acid, resulting in the release of carbon dioxide and the formation of NaCl:
HI + KHCO 3 → KI + CO 2 + H 2 O
2) Potassium iodide enters into a redox reaction with potassium dichromate in an acidic medium, while Cr +6 is reduced to Cr +3, I - is oxidized to molecular I 2, which precipitates:
6KI + K 2 Cr 2 O 7 + 7H 2 SO 4 → Cr 2 (SO 4) 3 + 4K 2 SO 4 + 3I 2 ↓ + 7H 2 O
2Cr +6 + 6e → 2Cr +3 │ 1
2I − -2e → I 2 │ 3
3) When molecular iodine interacts with aluminum, aluminum iodide is formed:
2Al + 3I 2 → 2AlI 3
4) When aluminum iodide interacts with a solution of potassium sulfide, Al (OH) 3 precipitates and H 2 S is released. The formation of Al 2 S 3 does not occur due to the complete hydrolysis of the salt in an aqueous solution:
2AlI 3 + 3K 2 S + 6H 2 O → 2Al(OH) 3 ↓ + 6KI + 3H 2 S
Task number 6
Aluminum carbide is completely dissolved in hydrobromic acid. Potassium sulfite solution was added to the resulting solution, whereby a white precipitate formed and a colorless gas evolved. The gas was absorbed with a solution of potassium dichromate in the presence of sulfuric acid. The resulting chromium salt was isolated and added to a solution of barium nitrate, and a precipitate was observed. Write the equations for the four described reactions.
1) When aluminum carbide is dissolved in hydrobromic acid, a salt is formed - aluminum bromide, and methane is released:
Al 4 C 3 + 12HBr → 4AlBr 3 + 3CH 4
2) When aluminum bromide interacts with a solution of potassium sulfite, Al (OH) 3 precipitates and sulfur dioxide is released - SO 2:
2AlBr 3 + 3K 2 SO 3 + 3H 2 O → 2Al(OH) 3 ↓ + 6KBr + 3SO 2
3) Passing sulfur dioxide through an acidified potassium dichromate solution, while Cr +6 is reduced to Cr +3, S +4 is oxidized to S +6:
3SO 2 + K 2 Cr 2 O 7 + H 2 SO 4 → Cr 2 (SO 4) 3 + K 2 SO 4 + H 2 O
2Cr +6 + 6e → 2Cr +3 │ 1
S +4 -2e → S +6 │ 3
4) When chromium (III) sulfate reacts with a solution of barium nitrate, chromium (III) nitrate is formed, and white barium sulfate precipitates:
Cr 2 (SO 4) 3 + 3Ba(NO 3) 2 → 3BaSO 4 ↓ + 2Cr(NO 3) 3
Task number 7
Aluminum powder was added to the sodium hydroxide solution. Excess was passed through the solution of the obtained substance carbon dioxide. The precipitate formed was separated and calcined. The resulting product was fused with sodium carbonate. Write the equations for the four described reactions.
1) Aluminum, as well as beryllium and zinc, is able to react both with aqueous solutions of alkalis and with anhydrous alkalis during fusion. When aluminum is treated with an aqueous solution of sodium hydroxide, sodium tetrahydroxoaluminate and hydrogen are formed:
2) When carbon dioxide is passed through water solution sodium tetrahydroxoaluminate precipitates crystalline aluminum hydroxide. Since, according to the condition, an excess of carbon dioxide is passed through the solution, not carbonate is formed, but sodium bicarbonate:
Na + CO 2 → Al(OH) 3 ↓ + NaHCO 3
3) Aluminum hydroxide is an insoluble metal hydroxide, therefore, when heated, it decomposes into the corresponding metal oxide and water:
4) Aluminum oxide, which is an amphoteric oxide, when fused with carbonates, displaces carbon dioxide from them to form aluminates (not to be confused with tetrahydroxoaluminates!):
Task number 8
Aluminum reacted with sodium hydroxide solution. The released gas was passed over the heated powder of copper oxide (II). The resulting simple substance was dissolved by heating in concentrated sulfuric acid. The resulting salt was isolated and added to a solution of potassium iodide. Write the equations for the four described reactions.
1) Aluminum (also beryllium and zinc) reacts both with aqueous solutions of alkalis and with anhydrous alkalis during fusion. When aluminum is treated with an aqueous solution of sodium hydroxide, sodium tetrahydroxoaluminate and hydrogen are formed:
2NaOH + 2Al + 6H 2 O → 2Na + 3H 2
2) When hydrogen is passed over the heated copper (II) oxide powder, Cu +2 is reduced to Cu 0: the color of the powder changes from black (CuO) to red (Cu):
3) Copper dissolves in concentrated sulfuric acid to form copper (II) sulfate. In addition, sulfur dioxide is released:
4) When copper sulfate is added to a solution of potassium iodide, a redox reaction occurs: Cu +2 is reduced to Cu +1, I - is oxidized to I 2 (molecular iodine precipitates):
CuSO 4 + 4KI → 2CuI + 2K 2 SO 4 + I 2 ↓
Task number 9
Spent the electrolysis of a solution of sodium chloride. Iron(III) chloride was added to the resulting solution. The precipitate that formed was filtered off and calcined. The solid residue was dissolved in hydroiodic acid. Write the equations for the four described reactions.
1) Electrolysis of sodium chloride solution:
Cathode: 2H 2 O + 2e → H 2 + 2OH −
Anode: 2Cl − − 2e → Cl 2
Thus, as a result of its electrolysis, gaseous H 2 and Cl 2 are released from a sodium chloride solution, and Na + and OH ions remain in the solution. IN general view the equation is written as follows:
2H 2 O + 2NaCl → H 2 + 2NaOH + Cl 2
2) When iron (III) chloride is added to an alkali solution, an exchange reaction occurs, as a result of which Fe (OH) 3 precipitates:
3NaOH + FeCl 3 → Fe(OH) 3 ↓ + 3NaCl
3) When iron (III) hydroxide is calcined, iron oxide (III) and water are formed:
4) When iron oxide (III) is dissolved in hydroiodic acid, FeI 2 is formed, while I 2 precipitates:
Fe 2 O 3 + 6HI → 2FeI 2 + I 2 ↓ + 3H 2 O
2Fe +3 + 2e → 2Fe +2 │1
2I − − 2e → I 2 │1
Task number 10
Potassium chlorate was heated in the presence of a catalyst and a colorless gas evolved. By burning iron in an atmosphere of this gas, iron scale was obtained. It was dissolved in an excess of hydrochloric acid. To the solution thus obtained was added a solution containing sodium dichromate and hydrochloric acid.
1) When potassium chlorate is heated in the presence of a catalyst (MnO 2, Fe 2 O 3, CuO, etc.), potassium chloride is formed and oxygen is released:
2) When iron is burned in an oxygen atmosphere, iron scale is formed, the formula of which is Fe 3 O 4 (iron scale is a mixed oxide of Fe 2 O 3 and FeO):
3) When iron scale is dissolved in excess hydrochloric acid, a mixture of iron (II) and (III) chlorides is formed:
4) In the presence of a strong oxidizing agent - sodium dichromate, Fe +2 is oxidized to Fe +3:
6FeCl 2 + Na 2 Cr 2 O 7 + 14HCl → 6FeCl 3 + 2CrCl 3 + 2NaCl + 7H 2 O
Fe +2 – 1e → Fe +3 │6
2Cr +6 + 6e → 2Cr +3 │1
Task number 11
Ammonia was passed through hydrobromic acid. Silver nitrate solution was added to the resulting solution. The precipitate formed was separated and heated with zinc powder. The metal formed during the reaction was treated with a concentrated solution of sulfuric acid, and a gas with a pungent odor was released. Write the equations for the four described reactions.
1) When ammonia is passed through hydrobromic acid, ammonium bromide is formed (neutralization reaction):
NH 3 + HBr → NH 4 Br
2) When the solutions of ammonium bromide and silver nitrate are drained, an exchange reaction occurs between the two salts, as a result of which a light yellow precipitate forms - silver bromide:
NH 4 Br + AgNO 3 → AgBr↓ + NH 4 NO 3
3) When silver bromide is heated with zinc powder, a substitution reaction occurs - silver is released:
2AgBr + Zn → 2Ag + ZnBr 2
4) When concentrated sulfuric acid acts on the metal, silver sulfate is formed and a gas with an unpleasant odor is released - sulfur dioxide:
2Ag + 2H 2 SO 4 (conc.) → Ag 2 SO 4 + SO 2 + 2H 2 O
2Ag 0 – 2e → 2Ag + │1
S +6 + 2e → S +4 │1
Task number 12
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Chromium(VI) oxide reacted with potassium hydroxide. The resulting substance was treated with sulfuric acid, an orange salt was isolated from the resulting solution. This salt was treated with hydrobromic acid. The resulting simple substance reacted with hydrogen sulfide. Write the equations for the four described reactions.
1) Chromium (VI) oxide CrO 3 is an acidic oxide, therefore, it interacts with alkali to form a salt - potassium chromate:
CrO 3 + 2KOH → K 2 CrO 4 + H 2 O
2) Potassium chromate in an acidic medium turns without changing the oxidation state of chromium into dichromate K 2 Cr 2 O 7 - an orange salt:
2K 2 CrO 4 + H 2 SO 4 → K 2 Cr 2 O 7 + K 2 SO 4 + H 2 O
3) When potassium dichromate is treated with hydrobromic acid, Cr +6 is reduced to Cr +3, while molecular bromine is released:
K 2 Cr 2 O 7 + 14HBr → 2CrBr 3 + 2KBr + 3Br 2 + 7H 2 O
2Cr +6 + 6e → 2Cr +3 │1
2Br − − 2e → Br 2 │3
4) Bromine, as a stronger oxidizing agent, displaces sulfur from its hydrogen compound:
Br 2 + H 2 S → 2HBr + S↓
Task number 13
The magnesium powder was heated under a nitrogen atmosphere. When the resulting substance interacts with water, a gas is released. The gas was passed through an aqueous solution of chromium(III) sulfate, resulting in a gray precipitate. The precipitate was separated and treated with heating with a solution containing hydrogen peroxide and potassium hydroxide. Write the equations for the four described reactions.
1) When magnesium powder is heated in a nitrogen atmosphere, magnesium nitride is formed:
2) Magnesium nitride is completely hydrolyzed to form magnesium hydroxide and ammonia:
Mg 3 N 2 + 6H 2 O → 3Mg (OH) 2 ↓ + 2NH 3
3) Ammonia has basic properties due to the presence of a lone electron pair at the nitrogen atom and, as a base, enters into an exchange reaction with chromium (III) sulfate, as a result of which a gray precipitate is released - Cr (OH) 3:
6NH3. H 2 O + Cr 2 (SO 4) 3 → 2Cr (OH) 3 ↓ + 3 (NH 4) 2 SO 4
4) Hydrogen peroxide in an alkaline medium oxidizes Cr +3 to Cr +6, resulting in the formation of potassium chromate:
2Cr(OH) 3 + 3H 2 O 2 + 4KOH → 2K 2 CrO 4 + 8H 2 O
Cr +3 -3e → Cr +6 │2
2O - + 2e → 2O -2 │3
Task number 14
When aluminum oxide reacted with nitric acid, a salt was formed. The salt was dried and calcined. The solid residue formed during calcination was subjected to electrolysis in molten cryolite. The metal obtained by electrolysis was heated with a concentrated solution containing potassium nitrate and potassium hydroxide, and a gas with a pungent odor was released. Write the equations for the four described reactions.
1) When amphoteric Al 2 O 3 interacts with nitric acid, a salt is formed - aluminum nitrate (exchange reaction):
Al 2 O 3 + 6HNO 3 → 2Al (NO 3) 3 + 3H 2 O
2) When aluminum nitrate is calcined, aluminum oxide is formed, and nitrogen dioxide and oxygen are also released (aluminum belongs to the group of metals (in the activity series from alkaline earth to Cu inclusive), whose nitrates decompose to metal oxides, NO 2 and O 2):
3) Metallic aluminum is formed by electrolysis of Al 2 O 3 in molten cryolite Na 2 AlF 6 at 960-970 o C.
Al 2 O 3 electrolysis scheme:
The dissociation of aluminum oxide proceeds in the melt:
Al 2 O 3 → Al 3+ + AlO 3 3-
K(-): Al 3+ + 3e → Al 0
A(+): 4AlO 3 3- − 12e → 2Al 2 O 3 + 3O 2
The overall equation of the process:
Liquid aluminum is collected at the bottom of the cell.
4) When aluminum is treated with a concentrated alkaline solution containing potassium nitrate, ammonia is released, and potassium tetrahydroxoaluminate (alkaline medium) is also formed:
8Al + 5KOH + 3KNO 3 + 18H 2 O → 3NH 3 + 8K
Al 0 – 3e → Al +3 │8
N +5 + 8e → N -3 │3
Task number 15
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A certain amount of iron (II) sulfide was divided into two parts. One of them was treated with hydrochloric acid, and the other was fired in air. During the interaction of the evolved gases, a simple yellow substance was formed. The resulting substance was heated with concentrated nitric acid, and a brown gas was released. Write the equations for the four described reactions.
1) When iron (II) sulfide is treated with hydrochloric acid, iron (II) chloride is formed and hydrogen sulfide is released (exchange reaction):
FeS + 2HCl → FeCl 2 + H 2 S
2) During the firing of iron (II) sulfide, iron is oxidized to an oxidation state of +3 (Fe 2 O 3 is formed) and sulfur dioxide is released:
3) When two sulfur-containing compounds SO 2 and H 2 S interact, a redox reaction (co-proportionation) occurs, as a result of which sulfur is released:
2H 2 S + SO 2 → 3S↓ + 2H 2 O
S -2 - 2e → S 0 │2
S +4 + 4e → S 0 │1
4) When sulfur is heated with concentrated nitric acid, sulfuric acid and nitrogen dioxide (redox reaction):
S + 6HNO 3 (conc.) → H 2 SO 4 + 6NO 2 + 2H 2 O
S 0 - 6e → S +6 │1
N +5 + e → N +4 │6
Task number 16
The gas obtained by treating calcium nitride with water was passed over hot powder of copper (II) oxide. Received at the same time solid dissolved in concentrated nitric acid, the solution was evaporated, and the resulting solid residue was calcined. Write equations for the four reactions described.
1) Calcium nitride reacts with water, forming alkali and ammonia:
Ca 3 N 2 + 6H 2 O → 3Ca (OH) 2 + 2NH 3
2) By passing ammonia over the hot powder of copper (II) oxide, the copper in the oxide is reduced to metallic, while nitrogen is released (hydrogen, coal, carbon monoxide, etc. are also used as reducing agents):
Cu +2 + 2e → Cu 0 │3
2N -3 – 6e → N 2 0 │1
3) Copper, located in a series of metal activities after hydrogen, interacts with concentrated nitric acid to form copper nitrate and nitrogen dioxide:
Cu + 4HNO 3 (conc.) → Cu(NO 3) 2 + 2NO 2 + 2H 2 O
Cu 0 - 2e → Cu +2 │1
N +5 +e → N +4 │2
4) When copper nitrate is calcined, copper oxide is formed, and nitrogen dioxide and oxygen are also released (copper belongs to the group of metals (in the activity series from alkaline earth to Cu inclusive), whose nitrates decompose to metal oxides, NO 2 and O 2):
Task number 17
Silicon was burned in an atmosphere of chlorine. The resulting chloride was treated with water. The precipitate thus formed was calcined. Then alloyed with calcium phosphate and coal. Write equations for the four reactions described.
1) The reaction of interaction of silicon and chlorine proceeds at a temperature of 340-420 o C in a stream of argon with the formation of silicon (IV) chloride:
2) Silicon (IV) chloride is completely hydrolyzed, with the formation of hydrochloric acid, and silicic acid precipitates:
SiCl 4 + 3H 2 O → H 2 SiO 3 ↓ + 4HCl
3) When calcined, silicic acid decomposes to silicon oxide (IV) and water:
4) When silicon dioxide is fused with coal and calcium phosphate, a redox reaction occurs, as a result of which calcium silicate, phosphorus are formed, and carbon monoxide is also released:
C 0 − 2e → C +2 │10
4P +5 +20e → P 4 0 │1
Task number 18
Note! This format of assignments is outdated, but nevertheless, assignments of this type deserve attention, since in fact they require writing the same equations that are found in KIMah USE new format.
Substances are given: iron, iron scale, dilute hydrochloric acid and concentrated nitric acid. Write the equations for the four possible reactions between all the proposed substances, without repeating the pairs of reactants.
1) Hydrochloric acid reacts with iron, oxidizing it to an oxidation state of +2, while hydrogen is released (substitution reaction):
Fe + 2HCl → FeCl 2 + H 2
2) Concentrated nitric acid passivates iron (i.e. a strong protective oxide film forms on its surface), however, under the influence of high temperature, iron is oxidized by concentrated nitric acid to an oxidation state of +3:
3) The formula of iron scale is Fe 3 O 4 (a mixture of iron oxides FeO and Fe 2 O 3). Fe 3 O 4 enters into an exchange reaction with hydrochloric acid, and a mixture of two iron (II) and (III) chlorides is formed:
Fe 3 O 4 + 8HCl → 2FeCl 3 + FeCl 2 + 4H 2 O
4) In addition, iron scale enters into a redox reaction with concentrated nitric acid, while the Fe +2 contained in it is oxidized to Fe +3:
Fe 3 O 4 + 10HNO 3 (conc.) → 3Fe(NO 3) 3 + NO 2 + 5H 2 O
5) Iron scale and iron, during their sintering, enter into a co-proportionation reaction (the oxidizing and reducing agents are the same chemical element):
Task #19
Substances are given: phosphorus, chlorine, aqueous solutions of sulfuric acid and potassium hydroxide. Write the equations for the four possible reactions between all the proposed substances, without repeating the pairs of reactants.
1) Chlorine is a highly reactive poisonous gas that reacts particularly vigorously with red phosphorus. In an atmosphere of chlorine, phosphorus ignites spontaneously and burns with a weak greenish flame. Depending on the ratio of the reactants, phosphorus (III) chloride or phosphorus (V) chloride can be obtained:
2P (red) + 3Cl 2 → 2PCl 3
2P (red) + 5Cl 2 → 2PCl 5
Cl 2 + 2KOH → KCl + KClO + H 2 O
If chlorine is passed through a hot concentrated alkali solution, molecular chlorine disproportionates into Cl +5 and Cl -1, resulting in the formation of chlorate and chloride, respectively:
3) As a result of the interaction of aqueous solutions of alkali and sulfuric acid, an acidic or medium salt of sulfuric acid is formed (depending on the concentration of the reagents):
KOH + H 2 SO 4 → KHSO 4 + H 2 O
2KOH + H 2 SO 4 → K 2 SO 4 + 2H 2 O (neutralization reaction)
4) Strong oxidizing agents such as sulfuric acid convert phosphorus to phosphoric acid:
2P + 5H 2 SO 4 → 2H 3 PO 4 + 5SO 2 + 2H 2 O
Task number 20
Substances are given: nitric oxide (IV), copper, potassium hydroxide solution and concentrated sulfuric acid. Write the equations for the four possible reactions between all the proposed substances, without repeating the pairs of reactants.
1) Copper, located in the series of metal activities to the right of hydrogen, can be oxidized by strong oxidizing acids (H 2 SO 4 (conc.), HNO 3, etc.):
Cu + 2H 2 SO 4 (conc.) → CuSO 4 + SO 2 + 2H 2 O
2) As a result of the interaction of a KOH solution with concentrated sulfuric acid, an acid salt is formed - potassium hydrogen sulfate:
KOH + H 2 SO 4 (conc.) → KHSO 4 + H 2 O
3) When passing brown gas, NO 2 N +4 disproportionates to N +5 and N +3, resulting in the formation of potassium nitrate and nitrite, respectively:
2NO 2 + 2KOH → KNO 3 + KNO 2 + H 2 O
4) When brown gas is passed through a concentrated solution of sulfuric acid, N +4 is oxidized to N +5 and sulfur dioxide is released:
2NO 2 + H 2 SO 4 (conc.) → 2HNO 3 + SO 2
Task number 21
Substances are given: chlorine, sodium hydrosulfide, potassium hydroxide (solution), iron. Write the equations for the four possible reactions between all the proposed substances, without repeating the pairs of reactants.
1) Chlorine, being a strong oxidizing agent, reacts with iron, oxidizing it to Fe +3:
2Fe + 3Cl 2 → 2FeCl 3
2) When chlorine is passed through a cold concentrated alkali solution, chloride and hypochlorite are formed (molecular chlorine disproportionates into Cl +1 and Cl -1):
2KOH + Cl 2 → KCl + KClO + H 2 O
If chlorine is passed through a hot concentrated alkali solution, molecular chlorine disproportionates into Cl +5 and Cl -1, resulting in the formation of chlorate and chloride, respectively:
3Cl 2 + 6KOH → 5KCl + KClO 3 + 3H 2 O
3) Chlorine, which has stronger oxidizing properties, is able to oxidize the sulfur that is part of the acid salt:
Cl 2 + NaHS → NaCl + HCl + S↓
4) Acid salt - sodium hydrosulfide in an alkaline environment turns into sulfide:
2NaHS + 2KOH → K 2 S + Na 2 S + 2H 2 O
CuCl 2 + 4NH 3 \u003d Cl 2
Na 2 + 4HCl \u003d 2NaCl + CuCl 2 + 4H 2 O
2Cl + K 2 S \u003d Cu 2 S + 2KCl + 4NH 3
When solutions are mixed, hydrolysis occurs both at the cation of a weak base and at the anion of a weak acid:
2CuSO 4 + Na 2 SO 3 + 2H 2 O \u003d Cu 2 O + Na 2 SO 4 + 2H 2 SO 4
2CuSO 4 + 2Na 2 CO 3 + H 2 O \u003d (CuOH) 2 CO 3 ↓ + 2Na 2 SO 4 + CO 2
Copper and copper compounds.
1) Through a solution of copper chloride (II) with the help of graphite electrodes, a constant electricity. The electrolysis product released at the cathode was dissolved in concentrated nitric acid. The resulting gas was collected and passed through a sodium hydroxide solution. The gaseous product of electrolysis released at the anode was passed through a hot solution of sodium hydroxide. Write the equations of the described reactions.
2) The substance obtained at the cathode during the electrolysis of a copper (II) chloride melt reacts with sulfur. The resulting product was treated with concentrated nitric acid, and the evolved gas was passed through a barium hydroxide solution. Write the equations of the described reactions.
3) The unknown salt is colorless and turns the flame yellow. When this salt is slightly heated with concentrated sulfuric acid, a liquid is distilled off in which copper is dissolved; the last transformation is accompanied by the evolution of brown gas and the formation of a copper salt. During the thermal decomposition of both salts, one of the decomposition products is oxygen. Write the equations of the described reactions.
4) When a solution of salt A reacted with alkali, a gelatinous substance insoluble in water was obtained blue color, which was dissolved in colorless liquid B to form a blue solution. The solid product remaining after careful evaporation of the solution was calcined; in this case, two gases were released, one of which is brown, and the second is part of the atmospheric air, and a black solid remains, which dissolves in liquid B with the formation of substance A. Write the equations of the described reactions.
5) Copper shavings were dissolved in dilute nitric acid and the solution was neutralized with caustic potash. The liberated blue substance was separated, calcined (the color of the substance changed to black), mixed with coke, and calcined again. Write the equations of the described reactions.
6) Copper shavings were added to a solution of mercury (II) nitrate. After completion of the reaction, the solution was filtered, and the filtrate was added dropwise to a solution containing sodium hydroxide and ammonium hydroxide. At the same time, a short-term formation of a precipitate was observed, which dissolved with the formation of a bright blue solution. When an excess of sulfuric acid solution was added to the resulting solution, a color change occurred. Write the equations of the described reactions.
7) Copper (I) oxide was treated with concentrated nitric acid, the solution was carefully evaporated and the solid residue was calcined. The gaseous reaction products were passed through a large amount of water and magnesium shavings were added to the resulting solution, as a result, a gas used in medicine was released. Write the equations of the described reactions.
8) The solid substance formed when malachite is heated was heated in a hydrogen atmosphere. The reaction product was treated with concentrated sulfuric acid, added to a sodium chloride solution containing copper filings, and a precipitate formed as a result. Write the equations of the described reactions.
9) The salt obtained by dissolving copper in dilute nitric acid was subjected to electrolysis using graphite electrodes. The substance released at the anode was introduced into interaction with sodium, and the resulting reaction product was placed in a vessel with carbon dioxide. Write the equations of the described reactions.
10) The solid product of the thermal decomposition of malachite was dissolved by heating in concentrated nitric acid. The solution was carefully evaporated and the solid residue was calcined to give a black substance which was heated in excess ammonia (gas). Write the equations of the described reactions.
11) A solution of dilute sulfuric acid was added to a black powdery substance and heated. A solution of caustic soda was added to the resulting blue solution until the precipitation ceased. The precipitate was filtered off and heated. The reaction product was heated in an atmosphere of hydrogen, resulting in a red substance. Write the equations of the described reactions.
12) An unknown red substance was heated in chlorine, and the reaction product was dissolved in water. Alkali was added to the resulting solution, the blue precipitate that formed was filtered off and calcined. When the calcination product, which is black, was heated with coke, a red starting material was obtained. Write the equations of the described reactions.
13) The solution obtained by the interaction of copper with concentrated nitric acid was evaporated and the precipitate was calcined. Gaseous products are completely absorbed by water, and hydrogen is passed over the solid residue. Write the equations of the described reactions.
14) Black powder, which was formed during the combustion of red metal in excess air, was dissolved in 10% sulfuric acid. Alkali was added to the resulting solution, and the resulting blue precipitate was separated and dissolved in an excess of ammonia solution. Write the equations of the described reactions.
15) A black substance was obtained by calcining the precipitate, which is formed by the interaction of sodium hydroxide and copper (II) sulfate. When this substance is heated with coal, a red metal is obtained, which dissolves in concentrated sulfuric acid. Write the equations of the described reactions.
16) Metallic copper was treated by heating with iodine. The resulting product was dissolved in concentrated sulfuric acid with heating. The resulting solution was treated with potassium hydroxide solution. The precipitate that formed was calcined. Write the equations of the described reactions.
17) An excess of soda solution was added to the copper (II) chloride solution. The precipitate formed was calcined, and the resulting product was heated in a hydrogen atmosphere. The resulting powder was dissolved in dilute nitric acid. Write the equations of the described reactions.
18) Copper was dissolved in dilute nitric acid. An excess of ammonia solution was added to the resulting solution, observing first the formation of a precipitate, and then its complete dissolution with the formation of a dark blue solution. The resulting solution was treated with sulfuric acid until the characteristic blue color of copper salts appeared. Write the equations of the described reactions.
19) Copper was dissolved in concentrated nitric acid. An excess of ammonia solution was added to the resulting solution, observing first the formation of a precipitate, and then its complete dissolution with the formation of a dark blue solution. The resulting solution was treated with an excess of hydrochloric acid. Write the equations of the described reactions.
20) The gas obtained by the interaction of iron filings with a solution of hydrochloric acid was passed over heated copper (II) oxide until the metal was completely reduced. the resulting metal was dissolved in concentrated nitric acid. The resulting solution was subjected to electrolysis with inert electrodes. Write the equations of the described reactions.
21) Iodine was placed in a test tube with concentrated hot nitric acid. The evolved gas was passed through water in the presence of oxygen. Copper (II) hydroxide was added to the resulting solution. The resulting solution was evaporated and the dry solid residue was calcined. Write the equations of the described reactions.
22) Orange copper oxide was placed in concentrated sulfuric acid and heated. An excess of potassium hydroxide solution was added to the resulting blue solution. the precipitated blue precipitate was filtered off, dried and calcined. The resulting black solid was heated into a glass tube, and ammonia was passed over it. Write the equations of the described reactions.
23) Copper (II) oxide was treated with a solution of sulfuric acid. During the electrolysis of the resulting solution on an inert anode, gas is released. The gas was mixed with nitric oxide (IV) and absorbed with water. Magnesium was added to a dilute solution of the obtained acid, as a result of which two salts were formed in the solution, and no evolution of the gaseous product occurred. Write the equations of the described reactions.
24) Copper (II) oxide was heated in a stream of carbon monoxide. The resulting substance was burned in an atmosphere of chlorine. The reaction product was dissolved in water. The resulting solution was divided into two parts. A solution of potassium iodide was added to one part, a solution of silver nitrate was added to the second. In both cases, the formation of a precipitate was observed. Write the equations of the described reactions.
25) Copper (II) nitrate was calcined, the resulting solid was dissolved in dilute sulfuric acid. The resulting salt solution was subjected to electrolysis. The substance released at the cathode was dissolved in concentrated nitric acid. The dissolution proceeds with the release of brown gas. Write the equations of the described reactions.
26) Oxalic acid was heated with a small amount of concentrated sulfuric acid. The evolved gas was passed through a solution of calcium hydroxide. in which the precipitate fell. Part of the gas was not absorbed, it was passed over a black solid obtained by calcining copper (II) nitrate. As a result, a dark red solid was formed. Write the equations of the described reactions.
27) Concentrated sulfuric acid reacted with copper. The gas evolved was completely absorbed by an excess of potassium hydroxide solution. The copper oxidation product was mixed with the calculated amount of sodium hydroxide until precipitation ceased. The latter was dissolved in an excess of hydrochloric acid. Write the equations of the described reactions.
Copper. copper compounds.
1. CuCl 2 Cu + Cl 2
at the cathode at the anode
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
6NaOH (gor.) + 3Cl 2 = NaClO 3 + 5NaCl + 3H 2 O
2. CuCl 2 Cu + Cl 2
at the cathode at the anode
CuS + 8HNO 3 (conc. horizon) = CuSO 4 + 8NO 2 + 4H 2 O
or CuS + 10HNO 3 (conc.) = Cu(NO 3) 2 + H 2 SO 4 + 8NO 2 + 4H 2 O
4NO 2 + 2Ba(OH) 2 = Ba(NO 3) 2 + Ba(NO 2) 2 + 2H 2 O
3. NaNO 3 (solid) + H 2 SO 4 (conc.) = HNO 3 + NaHSO 4
Cu + 4HNO 3 (conc.) = Cu(NO 3) 2 + 2NO 2 + 2H 2 O
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
2NaNO 3 2NaNO 2 + O 2
4. Cu(NO 3) 2 + 2NaOH = Cu(OH) 2 ↓ + 2NaNO 3
Cu(OH) 2 + 2HNO 3 = Cu(NO 3) 2 + 2H 2 O
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
CuO + 2HNO 3 \u003d Cu (NO 3) 2 + H 2 O
5. 3Cu + 8HNO 3(razb.) = 3Cu(NO 3) 2 + 2NO + 4H 2 O
Cu (NO 3) 2 + 2KOH \u003d Cu (OH) 2 ↓ + 2KNO 3
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
CuO + C Cu + CO
6. Hg (NO 3) 2 + Cu \u003d Cu (NO 3) 2 + Hg
Cu(NO 3) 2 + 2NaOH = Cu(OH) 2 ↓ + 2NaNO 3
(OH) 2 + 5H 2 SO 4 \u003d CuSO 4 + 4NH 4 HSO 4 + 2H 2 O
7. Cu 2 O + 6HNO 3 (conc.) = 2Cu (NO 3) 2 + 2NO 2 + 3H 2 O
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
4NO 2 + O 2 + 2H 2 O \u003d 4HNO 3
10HNO 3 + 4Mg \u003d 4Mg (NO 3) 2 + N 2 O + 5H 2 O
8. (CuOH) 2 CO 3 2CuO + CO 2 + H 2 O
CuO + H 2 Cu + H 2 O
CuSO 4 + Cu + 2NaCl \u003d 2CuCl ↓ + Na 2 SO 4
9. 3Cu + 8HNO 3(razb.) = 3Cu(NO 3) 2 + 2NO + 4H 2 O
at the cathode at the anode
2Na + O 2 \u003d Na 2 O 2
2Na 2 O 2 + CO 2 \u003d 2Na 2 CO 3 + O 2
10. (CuOH) 2 CO 3 2CuO + CO 2 + H 2 O
CuO + 2HNO 3 Cu(NO 3) 2 + H 2 O
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
11. CuO + H 2 SO 4 CuSO 4 + H 2 O
CuSO 4 + 2NaOH \u003d Cu (OH) 2 + Na 2 SO 4
Cu(OH) 2 CuO + H 2 O
CuO + H 2 Cu + H 2 O
12. Cu + Cl 2 CuCl 2
CuCl 2 + 2NaOH = Cu(OH) 2 ↓ + 2NaCl
Cu(OH) 2 CuO + H 2 O
CuO + C Cu + CO
13. Cu + 4HNO 3 (conc.) = Cu (NO 3) 2 + 2NO 2 + 2H 2 O
4NO 2 + O 2 + 2H 2 O \u003d 4HNO 3
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
CuO + H 2 Cu + H 2 O
14. 2Cu + O 2 \u003d 2CuO
CuSO 4 + NaOH \u003d Cu (OH) 2 ↓ + Na 2 SO 4
Сu (OH) 2 + 4 (NH 3 H 2 O) \u003d (OH) 2 + 4H 2 O
15. СuSO 4 + 2NaOH \u003d Cu (OH) 2 + Na 2 SO 4
Cu(OH) 2 CuO + H 2 O
CuO + C Cu + CO
Cu + 2H 2 SO 4 (conc.) = CuSO 4 + SO 2 + 2H 2 O
16) 2Cu + I 2 = 2CuI
2CuI + 4H 2 SO 4 2CuSO 4 + I 2 + 2SO 2 + 4H 2 O
Cu(OH) 2 CuO + H 2 O
17) 2CuCl 2 + 2Na 2 CO 3 + H 2 O = (CuOH) 2 CO 3 + CO 2 + 4NaCl
(CuOH) 2 CO 3 2CuO + CO 2 + H 2 O
CuO + H 2 Cu + H 2 O
3Cu + 8HNO 3 (diff.) \u003d 3Cu (NO 3) 2 + 2NO + 4H 2 O
18) 3Cu + 8HNO 3 (razb.) \u003d 3Cu (NO 3) 2 + 2NO + 4H 2 O
(OH) 2 + 3H 2 SO 4 \u003d CuSO 4 + 2 (NH 4) 2 SO 4 + 2H 2 O
19) Cu + 4HNO 3 (conc.) = Cu (NO 3) 2 + 2NO + 2H 2 O
Сu (NO 3) 2 + 2NH 3 H 2 O \u003d Cu (OH) 2 ↓ + 2NH 4 NO 3
Cu(OH) 2 + 4NH 3 H 2 O = (OH) 2 + 4H 2 O
(OH) 2 + 6HCl \u003d CuCl 2 + 4NH 4 Cl + 2H 2 O
20) Fe + 2HCl = FeCl 2 + H 2
CuO + H 2 \u003d Cu + H 2 O
Cu + 4HNO 3 (conc.) = Cu(NO 3) 2 + 2NO 2 + 2H 2 O
2Cu(NO 3) 2 + 2H 2 O 2Cu + O 2 + 4HNO 3
21) I 2 + 10HNO 3 \u003d 2HIO 3 + 10NO 2 + 4H 2 O
4NO 2 + 2H 2 O + O 2 \u003d 4HNO 3
Cu(OH) 2 + 2HNO 3 Cu(NO 3) 2 + 2H 2 O
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
22) Cu 2 O + 3H 2 SO 4 = 2CuSO 4 + SO 2 + 3H 2 O
СuSO 4 + 2KOH \u003d Cu (OH) 2 + K 2 SO 4
Cu(OH) 2 CuO + H 2 O
3CuO + 2NH 3 3Cu + N 2 + 3H 2 O
23) CuO + H 2 SO 4 = CuSO 4 + H 2 O
4NO 2 + O 2 + 2H 2 O \u003d 4HNO 3
10HNO 3 + 4Mg \u003d 4Mg (NO 3) 2 + NH 4 NO 3 + 3H 2 O
24) CuO + CO Cu + CO 2
Cu + Cl 2 = CuCl 2
2CuCl 2 + 2KI = 2CuCl↓ + I 2 + 2KCl
CuCl 2 + 2AgNO 3 \u003d 2AgCl ↓ + Cu (NO 3) 2
25) 2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
CuO + H 2 SO 4 \u003d CuSO 4 + H 2 O
2CuSO 4 + 2H 2 O 2Cu + O 2 + 2H 2 SO 4
Cu + 4HNO 3 (conc.) = Cu(NO 3) 2 + 2NO 2 + 2H 2 O
26) H 2 C 2 O 4 CO + CO 2 + H 2 O
CO 2 + Ca (OH) 2 \u003d CaCO 3 + H 2 O
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
CuO + CO Cu + CO 2
27) Cu + 2H 2 SO 4 (conc.) = CuSO 4 + SO 2 + 2H 2 O
SO 2 + 2KOH \u003d K 2 SO 3 + H 2 O
СuSO 4 + 2NaOH \u003d Cu (OH) 2 + Na 2 SO 4
Cu(OH) 2 + 2HCl CuCl 2 + 2H 2 O
Manganese. manganese compounds.
I. Manganese.
In air, manganese is covered with an oxide film, which protects it even when heated from further oxidation, but in a finely divided state (powder), it oxidizes quite easily. Manganese interacts with sulfur, halogens, nitrogen, phosphorus, carbon, silicon, boron, forming compounds with a degree of +2:
3Mn + 2P = Mn 3 P 2
3Mn + N 2 \u003d Mn 3 N 2
Mn + Cl 2 \u003d MnCl 2
2Mn + Si = Mn 2 Si
When interacting with oxygen, manganese forms manganese (IV) oxide:
Mn + O 2 \u003d MnO 2
4Mn + 3O 2 = 2Mn 2 O 3
2Mn + O 2 \u003d 2MnO
When heated, manganese interacts with water:
Mn+ 2H 2 O (steam) Mn(OH) 2 + H 2
In the electrochemical series of voltages, manganese is located before hydrogen, therefore it easily dissolves in acids, forming manganese (II) salts:
Mn + H 2 SO 4 \u003d MnSO 4 + H 2
Mn + 2HCl \u003d MnCl 2 + H 2
Manganese reacts with concentrated sulfuric acid when heated:
Mn + 2H 2 SO 4 (conc.) MnSO 4 + SO 2 + 2H 2 O
With nitric acid at normal conditions:
Mn + 4HNO 3 (conc.) = Mn(NO 3) 2 + 2NO 2 + 2H 2 O
3Mn + 8HNO 3 (diff..) = 3Mn(NO 3) 2 + 2NO + 4H 2 O
Alkali solutions practically do not affect manganese, but it reacts with alkaline melts of oxidizing agents, forming manganates (VI)
Mn + KClO 3 + 2KOH K 2 MnO 4 + KCl + H 2 O
Manganese can reduce oxides of many metals.
3Mn + Fe 2 O 3 \u003d 3MnO + 2Fe
5Mn + Nb 2 O 5 \u003d 5MnO + 2Nb
II. Manganese compounds (II, IV, VII)
1) Oxides.
Manganese forms a number of oxides, the acid-base properties of which depend on the oxidation state of manganese.
Mn +2 O Mn +4 O 2 Mn 2 +7 O 7
basic amphoteric acid
Manganese(II) oxide
Manganese (II) oxide is obtained by reduction of other manganese oxides with hydrogen or carbon monoxide (II):
MnO 2 + H 2 MnO + H 2 O
MnO2 + CO MnO + CO2
The main properties of manganese (II) oxide are manifested in their interaction with acids and acid oxides:
MnO + 2HCl \u003d MnCl 2 + H 2 O
MnO + SiO 2 = MnSiO 3
MnO + N 2 O 5 \u003d Mn (NO 3) 2
MnO + H 2 \u003d Mn + H 2 O
3MnO + 2Al = 2Mn + Al 2 O 3
2MnO + O 2 = 2MnO 2
3MnO + 2KClO 3 + 6KOH = 3K 2 MnO 4 + 2KCl + 3H 2 O
1 . Sodium was burned in an excess of oxygen, the resulting crystalline substance was placed in a glass tube and carbon dioxide was passed through it. The gas coming out of the tube was collected and burned in its atmosphere of phosphorus. The resulting substance was neutralized with an excess of sodium hydroxide solution.
1) 2Na + O 2 = Na 2 O 2
2) 2Na 2 O 2 + 2CO 2 \u003d 2Na 2 CO 3 + O 2
3) 4P + 5O 2 \u003d 2P 2 O 5
4) P 2 O 5 + 6 NaOH = 2Na 3 PO 4 + 3H 2 O
2. Aluminum carbide treated with hydrochloric acid. The released gas was burned, the combustion products were passed through lime water until a white precipitate formed, further passing the combustion products into the resulting suspension led to the dissolution of the precipitate.
1) Al 4 C 3 + 12HCl = 3CH 4 + 4AlCl 3
2) CH 4 + 2O 2 \u003d CO 2 + 2H 2 O
3) CO 2 + Ca (OH) 2 \u003d CaCO 3 + H 2 O
4) CaCO 3 + H 2 O + CO 2 \u003d Ca (HCO 3) 2
3. Pyrite was roasted, the resulting gas with a pungent odor was passed through hydrosulfide acid. The resulting yellowish precipitate was filtered off, dried, mixed with concentrated nitric acid, and heated. The resulting solution gives a precipitate with barium nitrate.
1) 4FeS 2 + 11O 2 → 2Fe 2 O 3 + 8SO 2
2) SO 2 + 2H 2 S \u003d 3S + 2H 2 O
3) S+ 6HNO 3 = H 2 SO 4 + 6NO 2 + 2H 2 O
4) H 2 SO 4 + Ba(NO 3) 2 = BaSO 4 ↓ + 2 HNO 3
4 . Copper was placed in concentrated nitric acid, the resulting salt was isolated from the solution, dried and calcined. The solid reaction product was mixed with copper shavings and calcined in an inert gas atmosphere. The resulting substance was dissolved in ammonia water.
1) Cu + 4HNO 3 \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O
2) 2Cu(NO 3) 2 = 2CuO + 4NO 2 + O 2
3) Cu + CuO = Cu 2 O
4) Cu 2 O + 4NH 3 + H 2 O \u003d 2OH
5 . Iron filings were dissolved in dilute sulfuric acid, the resulting solution was treated with an excess of sodium hydroxide solution. The precipitate formed was filtered and left in air until it turned brown. The brown substance was calcined to constant weight.
1) Fe + H 2 SO 4 \u003d FeSO 4 + H 2
2) FeSO 4 + 2NaOH \u003d Fe (OH) 2 + Na 2 SO 4
3) 4Fe(OH) 2 + 2H 2 O + O 2 = 4Fe(OH) 3
4) 2Fe(OH) 3 = Fe 2 O 3 + 3H 2 O
6 . The zinc sulfide was calcined. The resulting solid reacted completely with the potassium hydroxide solution. Carbon dioxide was passed through the resulting solution until a precipitate formed. The precipitate was dissolved in hydrochloric acid.
1) 2ZnS + 3O 2 = 2ZnO + 2SO 2
2) ZnO + 2NaOH + H 2 O = Na 2
3 Na 2 + CO 2 \u003d Na 2 CO 3 + H 2 O + Zn (OH) 2
4) Zn(OH) 2 + 2 HCl = ZnCl 2 + 2H 2 O
7. The gas released during the interaction of zinc with hydrochloric acid was mixed with chlorine and exploded. The resulting gaseous product was dissolved in water and treated with manganese dioxide. The resulting gas was passed through a hot solution of potassium hydroxide.
1) Zn+ 2HCl = ZnCl 2 + H 2
2) Cl 2 + H 2 \u003d 2HCl
3) 4HCl + MnO 2 = MnCl 2 + 2H 2 O + Cl 2
4) 3Cl 2 + 6KOH = 5KCl + KClO 3 + 3H 2 O
8. Calcium phosphide was treated with hydrochloric acid. The released gas was burned in a closed vessel, the combustion product was completely neutralized with a solution of potassium hydroxide. A solution of silver nitrate was added to the resulting solution.
1) Ca 3 P 2 + 6HCl = 3CaCl 2 + 2PH 3
2) PH 3 + 2O 2 = H 3 PO 4
3) H 3 PO 4 + 3KOH = K 3 PO 4 + 3H 2 O
4) K 3 PO 4 + 3AgNO 3 = 3KNO 3 + Ag 3 PO 4
9 . Ammonium dichromate decomposed on heating. The solid decomposition product was dissolved in sulfuric acid. Sodium hydroxide solution was added to the resulting solution until a precipitate formed. On further addition of sodium hydroxide to the precipitate, it dissolved.
1) (NH 4) 2 Cr 2 O 7 = Cr 2 O 3 + N 2 + 4H 2 O
2) Cr 2 O 3 + 3H 2 SO 4 = Cr 2 (SO 4) 3 + 3H 2 O
3) Cr 2 (SO 4) 3 + 6NaOH \u003d 3Na 2 SO 4 + 2Cr (OH) 3
4) 2Cr(OH) 3 + 3NaOH = Na 3
10 . Calcium orthophosphate was calcined with coal and river sand. The resulting white glow-in-the-dark substance was burned in an atmosphere of chlorine. The product of this reaction was dissolved in an excess of potassium hydroxide. A solution of barium hydroxide was added to the resulting mixture.
1) Ca 3 (PO 4) 2 + 5C + 3SiO 2 = 3CaSiO 3 + 5CO + 2P
2) 2P + 5Cl 2 = 2PCl 5
3) PCl 5 + 8KOH = K 3 PO 4 + 5KCl + 4H 2 O
4) 2K 3 PO 4 + 3Ba(OH) 2 = Ba 3 (PO 4) 2 + 6KOH
11. Aluminum powder was mixed with sulfur and heated. The resulting substance was placed in water. The resulting precipitate was divided into two parts. Hydrochloric acid was added to one part, and sodium hydroxide solution was added to the other until the precipitate was completely dissolved.
1) 2Al + 3S = Al 2 S 3
2) Al 2 S 3 + 6H 2 O \u003d 2Al (OH) 3 + 3H 2 S
3) Al(OH) 3 + 3HCl= AlCl 3 + 3H 2 O
4) Al (OH) 3 + NaOH \u003d Na
12 . Silicon was placed in a solution of potassium hydroxide, after the completion of the reaction, an excess of hydrochloric acid was added to the resulting solution. The precipitate formed was filtered off, dried and calcined. The solid calcination product reacts with hydrogen fluoride.
1) Si + 2KOH + H 2 O = K 2 SiO 3 + 2H 2
2) K 2 SiO 3 + 2HCl = 2KCl + H 2 SiO 3
3) H 2 SiO 3 \u003d SiO 2 + H 2 O
4) SiO 2 + 4HF = SiF 4 + 2H 2 O
Tasks for independent decision.
1. As a result of the thermal decomposition of ammonium dichromate, a gas was obtained, which was passed over heated magnesium. The resulting substance was placed in water. The resulting gas was passed through freshly precipitated copper(II) hydroxide. Write the equations of the described reactions.
2. To the solution obtained as a result of the interaction of sodium peroxide with water during heating, a solution of hydrochloric acid was added until the end of the reaction. The resulting salt solution was subjected to electrolysis with inert electrodes. The gas formed as a result of electrolysis at the anode was passed through a suspension of calcium hydroxide. Write the equations of the described reactions.
3. The precipitate formed as a result of the interaction of iron(II) sulfate solution and sodium hydroxide was filtered off and calcined. The solid residue was completely dissolved in concentrated nitric acid. Copper shavings were added to the resulting solution. Write the equations of the described reactions.
4. The gas obtained by roasting pyrite reacted with hydrogen sulfide. The yellow substance obtained as a result of the reaction was treated with concentrated nitric acid while heating. A solution of barium chloride was added to the resulting solution. Write the equations of the described reactions.
5. The gas obtained by the interaction of iron filings with a solution of hydrochloric acid was passed over heated copper (II) oxide until the metal was completely reduced. The resulting metal was dissolved in concentrated nitric acid. The resulting solution was subjected to electrolysis with inert electrodes. Write the equations of the described reactions.
6. The gas released at the anode during the electrolysis of mercury(II) nitrate was used for the catalytic oxidation of ammonia. The resulting colorless gas instantly reacted with atmospheric oxygen. The resulting brown gas was passed through barite water. Write the equations of the described reactions.
7. Iodine was placed in a test tube with concentrated hot nitric acid. The evolved gas was passed through water in the presence of oxygen. Copper (II) hydroxide was added to the resulting solution. The resulting solution was evaporated and the dry solid residue was calcined. Write the equations of the described reactions.
8. When a solution of aluminum sulfate reacted with a solution of potassium sulfide, a gas was released, which was passed through a solution of potassium hexahydroxoaluminate. The precipitate formed was filtered off, washed, dried and heated. The solid residue was fused with caustic soda. Write the equations of the described reactions.
9. Sulfur dioxide was passed through a solution of sodium hydroxide until a medium salt was formed. An aqueous solution of potassium permanganate was added to the resulting solution. The formed precipitate was separated and treated with hydrochloric acid. The evolved gas was passed through a cold solution of potassium hydroxide. Write the equations of the described reactions.
10. A mixture of silicon(IV) oxide and magnesium metal was calcined. The simple substance obtained as a result of the reaction was treated with a concentrated solution of sodium hydroxide. The evolved gas was passed over heated sodium. The resulting substance was placed in water. Write the equations of the described reactions.
Topic 7. Chemical properties and production organic matter in tasks C3. Reactions that cause the greatest difficulties for schoolchildren, which go beyond the scope of the school course.
To solve C3 tasks, students need to know the entire course organic chemistry at the profile level.
The chemical properties of most elements are based on their ability to dissolve in an aqueous medium and acids. The study of the characteristics of copper is associated with low activity under normal conditions. A feature of its chemical processes is the formation of compounds with ammonia, mercury, nitrogen and low solubility of copper in water is not capable of causing corrosion processes. She has special Chemical properties, allowing the connection to be used in various industries.
Item description
Copper is considered the oldest of the metals that people learned to extract even before our era. This substance is obtained from natural sources in the form of ore. Copper is called an element chemical table with the Latin name cuprum, the serial number of which is 29. In periodic system it is located in the fourth period and belongs to the first group.
The natural substance is rose red heavy metal with soft and malleable structure. Its boiling and melting point is over 1000 °C. Considered a good conductor.
Chemical structure and properties
If you study electronic formula copper atom, it can be found that it has 4 levels. There is only one electron in the valence 4s orbital. During chemical reactions, from 1 to 3 negatively charged particles can be split off from an atom, then copper compounds with an oxidation state of +3, +2, +1 are obtained. Its bivalent derivatives are the most stable.
IN chemical reactions it acts as an inactive metal. Under normal conditions, the solubility of copper in water is absent. In dry air, corrosion is not observed, but when heated, the metal surface is covered with a black coating of divalent oxide. The chemical stability of copper is manifested under the action of anhydrous gases, carbon, a series organic compounds, phenolic resins and alcohols. It is characterized by complex formation reactions with the release of colored compounds. Copper has a slight resemblance to the alkali group metals associated with the formation of derivatives of the monovalent series.
What is solubility?
This is the process of formation of homogeneous systems in the form of solutions when one compound interacts with other substances. Their components are individual molecules, atoms, ions and other particles. The degree of solubility is determined by the concentration of the substance that was dissolved when obtaining a saturated solution.
The unit of measurement is most often percentages, volume or weight fractions. The solubility of copper in water, like other solid compounds, is subject only to changes temperature conditions. This dependence is expressed using curves. If the indicator is very small, then the substance is considered insoluble.
Solubility of copper in the aquatic environment
The metal exhibits corrosion resistance under the action of sea water. This proves its inertia under normal conditions. The solubility of copper in water (fresh water) is practically not observed. But in a humid environment and under the action of carbon dioxide, a film is formed on the metal surface Green colour, which is the main carbonate:
Cu + Cu + O 2 + H 2 O + CO 2 → Cu (OH) 2 CuCO 2.
If we consider its monovalent compounds in the form of a salt, then their slight dissolution is observed. Such substances are subject to rapid oxidation. As a result, divalent copper compounds are obtained. These salts have good solubility in aqueous media. Their complete dissociation into ions occurs.
Solubility in acids
The usual conditions for the reactions of copper with weak or dilute acids are not conducive to their interaction. Not visible chemical process metal with alkalis. The solubility of copper in acids is possible if they are strong oxidizing agents. Only in this case the interaction takes place.
Solubility of copper in nitric acid
Such a reaction is possible due to the fact that the process takes place with a strong reagent. Nitric acid in dilute and concentrated form exhibits oxidizing properties with the dissolution of copper.
In the first variant, during the reaction, copper nitrate and nitrogen divalent oxide are obtained in a ratio of 75% to 25%. The process with dilute nitric acid can be described by the following equation:
8HNO 3 + 3Cu → 3Cu(NO 3) 2 + NO + NO + 4H 2 O.
In the second case, copper nitrate and nitrogen oxides are divalent and tetravalent, the ratio of which is 1 to 1. This process involves 1 mol of metal and 3 mol of concentrated nitric acid. When copper is dissolved, a strong heating of the solution occurs, as a result of which thermal decomposition of the oxidizer and the release of an additional volume of nitric oxides are observed:
4HNO 3 + Cu → Cu(NO 3) 2 + NO 2 + NO 2 + 2H 2 O.
The reaction is used in small-scale production associated with the processing of scrap or the removal of coatings from waste. However, this method of dissolving copper has a number of disadvantages associated with the release of a large amount of nitrogen oxides. To capture or neutralize them, special equipment is required. These processes are very costly.
The dissolution of copper is considered complete when there is a complete cessation of the production of volatile nitrogenous oxides. The reaction temperature ranges from 60 to 70 °C. The next step is to drain the solution from. Small pieces of metal remain at its bottom that have not reacted. Water is added to the resulting liquid and filtered.
Solubility in sulfuric acid
In the normal state, such a reaction does not occur. The factor determining the dissolution of copper in sulfuric acid is its strong concentration. A dilute medium cannot oxidize the metal. The dissolution of copper in concentrated proceeds with the release of sulfate.
The process is expressed by the following equation:
Cu + H 2 SO 4 + H 2 SO 4 → CuSO 4 + 2H 2 O + SO 2.
Properties of copper sulfate
Dibasic salt is also called sulfate, it is designated as follows: CuSO 4. It is a substance without a characteristic odor, not showing volatility. In its anhydrous form, salt is colorless, opaque, and highly hygroscopic. Copper (sulfate) has good solubility. Water molecules, joining the salt, can form crystal hydrate compounds. An example is which is a blue pentahydrate. Its formula: CuSO 4 5H 2 O.
Crystalline hydrates have a transparent structure of a bluish tint, they exhibit a bitter, metallic taste. Their molecules are capable of losing bound water over time. In nature, they are found in the form of minerals, which include chalcanthite and butite.
Affected by copper sulfate. Solubility is an exothermic reaction. In the process of salt hydration, a significant amount of heat is released.
Solubility of copper in iron
As a result of this process, pseudo-alloys of Fe and Cu are formed. For metallic iron and copper, limited mutual solubility is possible. Its maximum values are observed at a temperature index of 1099.85 °C. The degree of solubility of copper in the solid form of iron is 8.5%. These are small figures. The dissolution of metallic iron in the solid form of copper is about 4.2%.
Reducing the temperature to room values makes mutual processes insignificant. When metallic copper is melted, it is able to wet iron well in solid form. When obtaining Fe and Cu pseudo-alloys, special workpieces are used. They are created by pressing or baking iron powder, which is in pure or alloyed form. Such blanks are impregnated with liquid copper, forming pseudo-alloys.
Dissolution in ammonia
The process often proceeds by passing NH 3 in gaseous form over hot metal. The result is the dissolution of copper in ammonia, the release of Cu 3 N. This compound is called monovalent nitride.
Its salts are exposed to an ammonia solution. The addition of such a reagent to copper chloride leads to precipitation in the form of hydroxide:
CuCl 2 + NH 3 + NH 3 + 2H 2 O → 2NH 4 Cl + Cu(OH) 2 ↓.
An excess of ammonia contributes to the formation of a complex type compound, which has a dark blue color:
Cu(OH) 2 ↓+ 4NH 3 → (OH) 2.
This process is used to determine cuprous ions.
Solubility in cast iron
In the structure of malleable pearlitic iron, in addition to the main components, there is an additional element in the form of ordinary copper. It is she who increases the graphitization of carbon atoms, contributes to an increase in fluidity, strength and hardness of alloys. The metal has a positive effect on the level of perlite in the final product. The solubility of copper in cast iron is used to carry out the alloying of the initial composition. The main purpose of this process is to obtain a malleable alloy. It will have increased mechanical and corrosion properties, but reduced embrittlement.
If the copper content in cast iron is about 1%, then the tensile strength is equal to 40%, and the yield increases to 50%. This significantly changes the characteristics of the alloy. An increase in the amount of alloying metal to 2% leads to a change in strength to a value of 65%, and the yield index becomes 70%. With a higher copper content in the composition of cast iron, nodular graphite is more difficult to form. The introduction of an alloying element into the structure does not change the technology of forming a tough and soft alloy. The time allotted for annealing coincides with the duration of such a reaction at no copper impurity. It is about 10 hours.
The use of copper to make cast iron with a high concentration of silicon is not able to completely eliminate the so-called ferruginization of the mixture during annealing. The result is a product with low elasticity.
Solubility in mercury
When mercury is mixed with metals of other elements, amalgams are obtained. This process can take place at room temperature, because under such conditions Pb is a liquid. The solubility of copper in mercury passes only during heating. The metal must first be crushed. When wetting solid copper with liquid mercury, one substance interpenetrates another or diffuses. The solubility value is expressed as a percentage and is 7.4*10 -3 . The reaction produces a solid simple amalgam, similar to cement. If you heat it up a bit, it will soften. As a result, this mixture is used to repair porcelain items. There are also complex amalgams with an optimal metal content. For example, in a dental alloy there are elements of copper and zinc. Their number in percentage refers as 65:27:6:2. Amalgam with this composition is called silver. Each component of the alloy performs a specific function, which allows you to get a high quality seal.
Another example is the amalgam alloy, which has a high copper content. It is also called copper alloy. The composition of the amalgam contains from 10 to 30% Cu. The high copper content prevents the interaction of tin with mercury, which prevents the formation of a very weak and corrosive phase of the alloy. In addition, reducing the amount of silver in the filling leads to a reduction in price. For the preparation of amalgam, it is desirable to use an inert atmosphere or a protective liquid that forms a film. The metals that make up the alloy are able to quickly oxidize with air. The process of heating cuprum amalgam in the presence of hydrogen leads to the distillation of mercury, which allows the separation of elemental copper. As you can see, this topic is easy to learn. Now you know how copper interacts not only with water, but also with acids and other elements.
