Task No. 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. 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 equations for the four reactions described.
1) When sodium is heated in a hydrogen atmosphere (T = 250-400 o C), sodium hydride is formed:
2Na + H 2 = 2NaH
2) When water is added to sodium hydride, an alkali NaOH is formed and hydrogen is released:
NaH + H 2 O = NaOH + H 2
3) When copper reacts 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 occurs - nitrogen N +4 is simultaneously oxidized and reduced to N +5 and N +3:
2NaOH + 2NO2 = NaNO3 + NaNO2 + H2O
(disproportionation reaction 2N +4 → N +5 + N +3).
Task No. 2
Iron scale was dissolved in concentrated nitric acid. A sodium hydroxide solution was added to the resulting solution. The resulting precipitate was separated and calcined. The resulting solid residue was fused with iron. Write equations for the four reactions described.
The formula of iron scale is Fe 3 O 4.
When iron scale interacts with concentrated nitric acid, iron nitrate is formed and nitrogen 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 is an amphoteric hydroxide, insoluble in water, decomposes when heated into iron (III) oxide and water:
2Fe(OH) 3 → Fe 2 O 3 + 3H 2 O
When iron(III) oxide fuses with iron, iron(II) oxide is formed:
Fe 2 O 3 + Fe → 3FeO
Task No. 3
The sodium was burned in air. The resulting substance was treated with hydrogen chloride when heated. The resulting simple yellow-green substance, when heated, reacted with chromium (III) oxide in the presence of potassium hydroxide. When a solution of one of the resulting salts was treated with barium chloride, a yellow precipitate formed. Write equations for the four reactions described.
1) When sodium is burned in air, sodium peroxide is formed:
2Na + O 2 → Na 2 O 2
2) When sodium peroxide reacts with hydrogen chloride when heated, Cl 2 gas is released:
Na 2 O 2 + 4HCl → 2NaCl + Cl 2 + 2H 2 O
3) In an 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) A yellow precipitate (BaCrO 4) is formed by the interaction of potassium chromate and barium chloride:
K 2 CrO 4 + BaCl 2 → BaCrO 4 ↓ + 2KCl
Task No. 4
Zinc is completely dissolved in a concentrated solution of potassium hydroxide. 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 the formation of a white precipitate was observed. Write equations for the four reactions described.
1) Zinc reacts with potassium hydroxide to form potassium tetrahydroxocinate (Al and Be behave similarly):
2) After calcination, potassium tetrahydroxozincate loses water and turns into potassium zincate:
3) Potassium zincate, when reacting 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 No. 5
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 solution of potassium sulfide, resulting in the formation of a precipitate and the release of gas. Write equations for the four reactions described.
1) Hydroiodic acid is neutralized by the acid salt of weak 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 environment, 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 reacts with aluminum, aluminum iodide is formed:
2Al + 3I 2 → 2AlI 3
4) When aluminum iodide reacts 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 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 No. 6
Aluminum carbide was completely dissolved in hydrobromic acid. A solution of potassium sulfite was added to the resulting solution, and the formation of a white precipitate and the release of a colorless gas were observed. The gas was absorbed by a solution of potassium dichromate in the presence of sulfuric acid. The resulting chromium salt was isolated and added to the barium nitrate solution, and the formation of a precipitate was observed. Write equations for the four reactions described.
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 reacts 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) By passing sulfur dioxide through an acidified solution of potassium dichromate, 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 No. 7
Aluminum powder was added to the sodium hydroxide solution. Excess carbon dioxide was passed through the solution of the resulting substance. The precipitate that formed was separated and calcined. The resulting product was fused with sodium carbonate. Write equations for the four reactions described.
1) Aluminum, as well as beryllium and zinc, is capable of reacting with both aqueous solutions of alkalis and anhydrous alkalis during fusion. When aluminum is treated with an aqueous solution of sodium hydroxide, sodium tetrahydroxyaluminate and hydrogen are formed:
2) When carbon dioxide is passed through an aqueous solution of sodium tetrahydroxoaluminate, crystalline aluminum hydroxide precipitates. Since, according to the condition, an excess of carbon dioxide is passed through the solution, it is not carbonate that 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 No. 8
Aluminum reacted with sodium hydroxide solution. The released gas was passed over heated copper (II) oxide powder. The resulting simple substance was dissolved by heating in concentrated sulfuric acid. The resulting salt was isolated and added to the potassium iodide solution. Write equations for the four reactions described.
1) Aluminum (also beryllium and zinc) reacts with both aqueous solutions of alkalis and anhydrous alkalis when fused. When aluminum is treated with an aqueous solution of sodium hydroxide, sodium tetrahydroxyaluminate and hydrogen are formed:
2NaOH + 2Al + 6H 2 O → 2Na + 3H 2
2) When hydrogen is passed over 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, an oxidation-reduction reaction occurs: Cu +2 is reduced to Cu +1, I - is oxidized to I2 (molecular iodine precipitates):
CuSO 4 + 4KI → 2CuI + 2K 2 SO 4 + I 2 ↓
Task No. 9
We carried out electrolysis of a sodium chloride solution. Iron (III) chloride was added to the resulting solution. The precipitate that formed was filtered and calcined. The solid residue was dissolved in hydroiodic acid. Write equations for the four reactions described.
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, 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 (III) oxide and water are formed:
4) When iron (III) oxide 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 No. 10
Potassium chlorate was heated in the presence of a catalyst, releasing a colorless gas. By burning iron in an atmosphere of this gas, iron oxide was obtained. It was dissolved in excess hydrochloric acid. To the resulting solution 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 No. 11
Ammonia was passed through hydrobromic acid. A solution of silver nitrate was added to the resulting solution. The precipitate that formed was separated and heated with zinc powder. The metal formed during the reaction was exposed to a concentrated solution of sulfuric acid, which released a gas with a pungent odor. Write equations for the four reactions described.
1) When ammonia is passed through hydrobromic acid, ammonium bromide is formed (neutralization reaction):
NH 3 + HBr → NH 4 Br
2) When solutions of ammonium bromide and silver nitrate are combined, an exchange reaction occurs between the two salts, resulting in the formation of a light yellow precipitate - 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 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 No. 12
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Chromium(VI) oxide reacted with potassium hydroxide. The resulting substance was treated with sulfuric acid, and 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 equations for the four reactions described.
1) Chromium (VI) oxide CrO 3 is an acidic oxide, therefore, it reacts with alkali to form a salt - potassium chromate:
CrO 3 + 2KOH → K 2 CrO 4 + H 2 O
2) Potassium chromate in an acidic environment is converted 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 treating potassium dichromate with hydrobromic acid, Cr +6 is reduced to Cr +3, and 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 No. 13
Magnesium powder was heated in a nitrogen atmosphere. When the resulting substance interacted with water, a gas was released. The gas was passed through an aqueous solution of chromium(III) sulfate, resulting in the formation of a gray precipitate. The precipitate was separated and treated by heating with a solution containing hydrogen peroxide and potassium hydroxide. Write equations for the four reactions described.
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 on 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 environment 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 No. 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 equations for the four reactions described.
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), the nitrates of which decompose to metal oxides, NO 2 and O 2):
3) Metallic aluminum is formed during the electrolysis of Al 2 O 3 in molten cryolite Na 2 AlF 6 at 960-970 o C.
Al 2 O 3 electrolysis scheme:
Dissociation of aluminum oxide occurs 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
Overall process equation:
Liquid aluminum collects at the bottom of the electrolyser.
4) When aluminum is treated with a concentrated alkaline solution containing potassium nitrate, ammonia is released and potassium tetrahydroxyaluminate is also formed (alkaline medium):
8Al + 5KOH + 3KNO 3 + 18H 2 O → 3NH 3 + 8K
Al 0 – 3e → Al +3 │8
N +5 + 8e → N -3 │3
Task No. 15
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Some iron(II) sulfide was divided into two parts. One of them was treated with hydrochloric acid, and the other was fired in air. When the released gases interacted, a simple yellow substance was formed. The resulting substance was heated with concentrated nitric acid, and a brown gas was released. Write equations for the four reactions described.
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) When burning iron (II) sulfide, iron is oxidized to the oxidation state +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, an oxidation-reduction reaction (coproportionation) 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 are formed (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 No. 16
The gas obtained by treating calcium nitride with water was passed over hot copper (II) oxide powder. The resulting solid was dissolved in concentrated nitric acid, the solution was evaporated, and the resulting solid residue was calcined. Write down 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 a hot powder of copper (II) oxide, the copper in the oxide is reduced to metallic, and 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 the series of metal activities after hydrogen, reacts 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), the nitrates of which decompose to metal oxides, NO 2 and O 2):
Task No. 17
Silicon was burned in a chlorine atmosphere. The resulting chloride was treated with water. The precipitate released was calcined. Then fused with calcium phosphate and coal. Write down equations for the four reactions described.
1) The reaction between silicon and chlorine occurs at a temperature of 340-420 o C in a flow of argon with the formation of silicon (IV) chloride:
2) Silicon (IV) chloride is completely hydrolyzed, resulting in 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 (IV) oxide and water:
4) When silicon dioxide is fused with coal and calcium phosphate, an oxidation-reduction reaction occurs, resulting in the formation of calcium silicate, phosphorus, and the release of carbon monoxide:
C 0 − 2e → C +2 │10
4P +5 +20e → P 4 0 │1
Task No. 18
Note! This format of tasks is outdated, but nevertheless tasks of this type deserve attention, since in fact they require writing down the same equations that are found in the Unified State Exam KIMs of the new format.
The following substances are given: iron, iron scale, dilute hydrochloric and concentrated nitric acid. Write equations for four possible reactions between all the proposed substances, without repeating pairs of reactants.
1) Hydrochloric acid reacts with iron, oxidizing it to the oxidation state +2, and hydrogen is released (substitution reaction):
Fe + 2HCl → FeCl 2 + H 2
2) Concentrated nitric acid passivates iron (i.e., a strong protective oxide film is formed on its surface), however, under the influence of high temperature, iron is oxidized by concentrated nitric acid to oxidation state +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, resulting in a mixture of two iron (II) and (III) chlorides:
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, and 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, when sintered, enter into a comporportionation reaction (the same chemical element acts as an oxidizing agent and a reducing agent):
Task No. 19
The following substances are given: phosphorus, chlorine, aqueous solutions of sulfuric acid and potassium hydroxide. Write equations for four possible reactions between all the proposed substances, without repeating pairs of reactants.
1) Chlorine is a poisonous gas with high chemical activity and reacts especially vigorously with red phosphorus. In an atmosphere of chlorine, phosphorus spontaneously ignites 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, the molecular chlorine is disproportionated 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 average 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 into phosphoric acid:
2P + 5H 2 SO 4 → 2H 3 PO 4 + 5SO 2 + 2H 2 O
Task No. 20
The substances given are: nitric oxide (IV), copper, potassium hydroxide solution and concentrated sulfuric acid. Write equations for four possible reactions between all the proposed substances, without repeating pairs of reactants.
1) Copper, located in the series of metal activities to the right of hydrogen, is capable of oxidation 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 is disproportioned into 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 No. 21
The following substances are given: chlorine, sodium hydrosulfide, potassium hydroxide (solution), iron. Write equations for four possible reactions between all the proposed substances, without repeating 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 is disproportionate to Cl +1 and Cl -1):
2KOH + Cl 2 → KCl + KClO + H 2 O
If chlorine is passed through a hot concentrated alkali solution, the 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 capable of oxidizing the sulfur contained in 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
1 . Sodium was burned in excess 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 phosphorus was burned in its atmosphere. 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 = 2Na 2 CO 3 + O 2
3) 4P + 5O 2 = 2P 2 O 5
4) P 2 O 5 + 6 NaOH = 2Na 3 PO 4 + 3H 2 O
2. Aluminum carbide was treated with hydrochloric acid. The released gas was burned, the combustion products were passed through lime water until a white precipitate was 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 = CO 2 + 2H 2 O
3) CO 2 + Ca(OH) 2 = CaCO 3 + H 2 O
4) CaCO 3 + H 2 O + CO 2 = Ca(HCO 3) 2
3. The pyrite was fired, and the resulting gas with a pungent odor was passed through hydrogen sulfide acid. The resulting yellowish precipitate was filtered, dried, mixed with concentrated nitric acid and heated. The resulting solution gives a precipitate containing barium nitrate.
1) 4FeS 2 + 11O 2 → 2Fe 2 O 3 + 8SO 2
2) SO 2 + 2H 2 S = 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 = 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 = 2OH
5 . Iron filings were dissolved in dilute sulfuric acid, and the resulting solution was treated with an excess of sodium hydroxide solution. The resulting precipitate was filtered and left in air until it acquired a brown color. The brown substance was calcined to constant mass.
1) Fe + H 2 SO 4 = FeSO 4 + H 2
2) FeSO 4 + 2NaOH = 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 . 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 = 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 when zinc reacted with hydrochloric acid was mixed with chlorine and exploded. The resulting gaseous product was dissolved in water and acted on 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 = 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, and 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 when heated. The solid decomposition product was dissolved in sulfuric acid. A solution of sodium hydroxide was added to the resulting solution until a precipitate formed. Upon 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 = 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 a chlorine atmosphere. The product of this reaction was dissolved in excess 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 = 2Al(OH) 3 + 3H 2 S
3) Al(OH) 3 + 3HCl= AlCl 3 + 3H 2 O
4) Al(OH) 3 + NaOH = Na
12 . Silicon was placed in a solution of potassium hydroxide, and after the reaction was completed, excess hydrochloric acid was added to the resulting solution. The precipitate that formed was filtered, 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 = SiO 2 + H 2 O
4) SiO 2 + 4HF = SiF 4 + 2H 2 O
Tasks for independent solution.
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 for the reactions described.
2. A solution of hydrochloric acid was added to the solution obtained by reacting sodium peroxide with water when heated until the reaction was completed. The solution of the resulting salt 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 for the reactions described.
3. The precipitate formed as a result of the interaction of a solution of iron(II) sulfate and sodium hydroxide was filtered and calcined. The solid residue was completely dissolved in concentrated nitric acid. Copper shavings were added to the resulting solution. Write the equations for the reactions described.
4. The gas produced 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 for the reactions described.
5. The gas obtained by reacting 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 for the reactions described.
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 oxygen in the air. The resulting brown gas was passed through barite water. Write the equations for the reactions described.
7. Iodine was placed in a test tube with concentrated hot nitric acid. The released 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 for the reactions described.
8. When a solution of aluminum sulfate interacted with a solution of potassium sulfide, a gas was released, which was passed through a solution of potassium hexahydroxyaluminate. The resulting precipitate was filtered, washed, dried and heated. The solid residue was fused with caustic soda. Write the equations for the reactions described.
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 resulting precipitate was separated and treated with hydrochloric acid. The released gas was passed through a cold solution of potassium hydroxide. Write the equations for the reactions described.
10. A mixture of silicon(IV) oxide and metallic magnesium was calcined. The simple substance obtained as a result of the reaction was treated with a concentrated solution of sodium hydroxide. The released gas was passed over heated sodium. The resulting substance was placed in water. Write the equations for the reactions described.
Topic 7. Chemical properties and production of organic substances in tasks C3. Reactions that cause the greatest difficulties in schoolchildren that go beyond the scope of the school course.
To solve C3 tasks, schoolchildren need to know the entire organic chemistry course at a specialized level.
The chemical properties of most elements are based on their ability to dissolve in aqueous media and acids. The study of the characteristics of copper is associated with a low-active effect 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. It has special chemical properties that allow the compound to be used in various industries.
Item Description
Copper is considered the oldest metal, which people learned to mine even before our era. This substance is obtained from natural sources in the form of ore. Copper is an element of the chemical table with the Latin name cuprum, the serial number of which is 29. In the periodic table, it is located in the fourth period and belongs to the first group.
The naturally occurring substance is a pink-red heavy metal with a soft and malleable structure. Its boiling and melting point is more than 1000 °C. Considered a good guide.
Chemical structure and properties
If you study the electronic formula of a copper atom, you will find that it has 4 levels. There is only one electron in the 4s valence 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 divalent derivatives are most stable.
In chemical reactions it acts as a low-reactive metal. Under normal conditions, copper has no solubility in water. Corrosion is not observed in dry air, but when heated, the metal surface becomes covered with a black coating of divalent oxide. The chemical stability of copper is manifested under the action of anhydrous gases, carbon, a number of organic compounds, phenolic resins and alcohols. It is characterized by complex formation reactions with the release of colored compounds. Copper has slight similarities with alkali group metals due to the formation of monovalent derivatives.
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 fractions or weight fractions. The solubility of copper in water, like other solid compounds, is subject only to changes in temperature conditions. This dependence is expressed using curves. If the indicator is very small, then the substance is considered insoluble.
Solubility of copper in aqueous media
The metal exhibits corrosion resistance when exposed to sea water. This proves its inertness under normal conditions. The solubility of copper in water (fresh) is practically not observed. But in a humid environment and under the influence of carbon dioxide, a green film forms on the metal surface, 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 salts, then their insignificant dissolution is observed. Such substances are subject to rapid oxidation. The result is divalent copper compounds. These salts have good solubility in aqueous media. Their complete dissociation into ions occurs.
Solubility in acids
The usual conditions for reactions of copper with weak or dilute acids do not favor their interaction. The chemical process of the metal with alkalis is not observed. Copper solubility in acids is possible if they are strong oxidizing agents. Only in this case does interaction take place.
Solubility of copper in nitric acid
This reaction is possible due to the fact that the process occurs with a strong reagent. Nitric acid in diluted and concentrated form exhibits oxidizing properties with the dissolution of copper.
In the first option, the reaction produces copper nitrate and nitrogen divalent oxide 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 obtained, divalent and tetravalent, the ratio of which is 1 to 1. This process involves 1 mole of metal and 3 moles of concentrated nitric acid. When copper dissolves, the solution heats up strongly, resulting in thermal decomposition of the oxidizing agent and the release of an additional volume of nitrogen oxides:
4HNO 3 + Cu → Cu(NO 3) 2 + NO 2 + NO 2 + 2H 2 O.
The reaction is used in small-scale production associated with recycling scrap or removing coatings from waste. However, this method of dissolving copper has a number of disadvantages associated with the release of large amounts of nitrogen oxides. Special equipment is required to capture or neutralize them. These processes are very expensive.
The dissolution of copper is considered complete when the production of volatile nitrogen oxides completely ceases. The reaction temperature ranges from 60 to 70 °C. The next step is to drain the solution from the solution. Small pieces of metal that have not reacted remain at its bottom. Water is added to the resulting liquid and filtered.
Solubility in sulfuric acid
Under normal conditions, this 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 concentrated copper 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 sulfuric acid, it is designated as: CuSO 4. It is a substance without a characteristic odor and does not exhibit volatility. In its anhydrous form, salt is colorless, opaque, and highly hygroscopic. Copper (sulfate) has good solubility. Water molecules, when added to salt, can form crystalline hydrate compounds. An example is which is the blue pentahydrate. Its formula: CuSO 4 5H 2 O.
Crystalline hydrates have a transparent structure with a bluish tint and exhibit a bitter, metallic taste. Their molecules are capable of losing bound water over time. They are found in nature in the form of minerals, which include chalcanthite and butite.
Susceptible to copper sulfate. Solubility is an exothermic reaction. The process of salt hydration generates a significant amount of heat.
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 of 1099.85 °C. The degree of solubility of copper in the solid form of iron is 8.5%. These are small numbers. The dissolution of metallic iron in the solid form of copper is about 4.2%.
Reducing the temperature to room values makes the mutual processes insignificant. When metallic copper is melted, it is able to well wet iron in solid form. When producing Fe and Cu pseudo-alloys, special blanks are used. They are created by pressing or baking iron powder in pure or alloyed form. Such workpieces are impregnated with liquid copper, forming pseudo-alloys.
Dissolution in ammonia
The process often occurs by passing NH 3 in gaseous form over a 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 ammonia solution. The addition of such a reagent to copper chloride leads to the formation of a precipitate in the form of hydroxide:
CuCl 2 + NH 3 + NH 3 + 2H 2 O → 2NH 4 Cl + Cu(OH) 2 ↓.
Excess ammonia promotes the formation of a complex type compound that is dark blue in color:
Cu(OH) 2 ↓+ 4NH 3 → (OH) 2.
This process is used to determine cupric ions.
Solubility in cast iron
In the structure of malleable pearlitic cast iron, in addition to the main components, there is an additional element in the form of ordinary copper. It is this that increases the graphitization of carbon atoms and helps to increase the 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 alloy the original 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 strength increases to 50%. This significantly changes the characteristics of the alloy. Increasing the amount of metal alloying to 2% leads to a change in strength to 65%, and the fluidity rate becomes 70%. With a higher copper content in cast iron, spheroidal graphite is more difficult to form. The introduction of an alloying element into the structure does not change the technology for forming a tough and soft alloy. The time allotted for annealing coincides with the duration of such a reaction without copper admixture. It is about 10 hours.
The use of copper for the production of 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 disappears only during heating. The metal must first be crushed. When solid copper is wetted with liquid mercury, mutual penetration of one substance into another or a process of diffusion occurs. The solubility value is expressed as a percentage and is 7.4 * 10 -3. The reaction produces a hard, simple amalgam similar to cement. If you heat it up a little, it softens. As a result, this mixture is used to repair porcelain products. There are also complex amalgams with an optimal content of metals. For example, dental alloy contains elements of copper and zinc. Their percentage ratio is 65: 27: 6:2. Amalgam with this composition is called silver. Each component of the alloy performs a specific function, which allows you to obtain a high-quality filling.
Another example is an amalgam alloy, which has a high copper content. It is also called copper alloy. 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 a filling leads to cheaper prices. To prepare amalgam, it is advisable to use an inert atmosphere or a protective liquid that forms a film. The metals that make up the alloy can be quickly oxidized by air. The process of heating cuprum amalgam in the presence of hydrogen causes the mercury to be distilled off, allowing the elemental copper to be separated. As you can see, this topic is not difficult to learn. Now you know how copper interacts not only with water, but also with acids and other elements.
