This, of course, is the minimum information that can be useful for solving tasks C2.
In the process of preparing students for solving tasks C2, you can offer them compose texts of assignments in accordance with the schemes of transformations . This task will allow students to master the terminology and remember the characteristic features of substances.
Example 1:
t o C t o C/H 2 HNO 3 (conc) NaOH, 0 o C
(CuOH) 2 CO 3 → CuO → Cu → NO 2 → X
Text: Malachite was calcined, the resulting black solid was heated in a stream of hydrogen. The resulting red substance was completely dissolved in concentrated nitric acid. The liberated brown gas was passed through a cold solution of sodium hydroxide.
Example 2:
O 2 H 2 S solution t o C/Al H 2 O
ZnS → SO 2 → S → Al 2 S 3 → X
Text: The zinc sulfide was calcined. The resulting gas with a pungent odor was passed through a solution of hydrogen sulfide until a yellow precipitate formed. The precipitate was filtered off, dried and fused with aluminum. The resulting compound was placed in water until the reaction terminated.
The next step is to ask students to draw up both schemes for the transformation of substances and texts of tasks. Of course, the "authors" of the tasks must submit and own solution . At the same time, students repeat all the properties of inorganic substances. And the teacher can form a bank of tasks C2.
After that you can go to the solution of tasks С2 . At the same time, students draw up a scheme of transformations according to the text, and then the corresponding reaction equations. To do this, reference points are highlighted in the text of the task: the names of substances, an indication of their classes, physical properties, conditions for conducting reactions, names of processes.
Let's give examples of some tasks.
Example 1 Manganese (II) nitrate was calcined, and concentrated hydrochloric acid was added to the resulting brown solid. The evolved gas was passed through hydrosulfide acid. The resulting solution forms a precipitate with barium chloride.
Solution:
Manganese(II) nitrate- Mn(NO 3) 2,
calcined- heated to decomposition,
solid brown matter- MnO 2,
Concentrated hydrochloric acid–HCl,
Hydrosulphuric acid - solution H 2 S,
Barium chloride - BaCl 2 forms a precipitate with the sulfate ion.
t o C HCl H 2 Sp-p BaCl 2
Mn(NO 3) 2 → MnO 2 → X → Y → ↓ (BaSO 4 ?)
1) Mn(NO 3) 2 → MnО 2 + 2NO 2
2) MnO 2 + 4 HCl → MnCl 2 + 2H 2 O + Cl 2 (gas X)
3) Cl 2 + H 2 S → 2HCl + S (not suitable, because there is no product that precipitates with barium chloride) or 4Cl 2 + H 2 S + 4H 2 O → 8HCl + H 2 SO 4
4) H 2 SO 4 + BaCl 2 → BaSO 4 + 2HCl
Example 2 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 resulting blue precipitate was filtered off, dried and calcined. The solid black substance thus obtained was placed in a glass tube, heated, and ammonia was passed over it.
Solution:
Selection of support moments:
Orange copper oxide– Cu 2 O,
concentrated sulfuric acid- H 2 SO 4,
blue solution- copper (II) salt, СuSO 4
Potassium hydroxide-KOH,
Blue precipitate - Cu(OH)2,
Calcined - heated to decomposition
Solid black matter CuO,
Ammonia- NH3.
· Drawing up a scheme of transformations:
H 2 SO 4 KOH t o C NH 3
Cu 2 O → СuSO 4 → Cu(OH) 2 ↓ → CuO → X
Drawing up reaction equations:
1) Cu 2 O + 3Н 2 SO 4 → 2СuSO 4 + SO 2 + 3H 2 O
2) СuSO 4 + 2KOH → Cu(OH) 2 + K 2 SO 4
3) Cu(OH) 2 → CuO + H 2 O
4) 3CuO + 2NH 3 → 3Cu + 3H 2 O + N 2
Fe(OH)3 write the dissociation equation.
Attention! Solutions are provided by ordinary people, so there may be errors or inaccuracies in the solutions. When using solutions, don't forget to double-check them!
Fe(OH)3 was calcined. What is it like?
Attention! Solutions are provided by ordinary people, so there may be errors or inaccuracies in the solutions. When using solutions, don't forget to double-check them!
Fe(OH)2 + HNO3 = ..; Fe(OH)3 + H2SO4 = ..; MgO + HCl = .. .
Attention! Solutions are provided by ordinary people, so there may be errors or inaccuracies in the solutions. When using solutions, don't forget to double-check them!
Fe(OH)3 + HCl = ..; Fe(OH)3 + H2SO4 = .. .
Attention! Solutions are provided by ordinary people, so there may be errors or inaccuracies in the solutions. When using solutions, don't forget to double-check them!
Fe(OH)3 + NaOH = .. .
Attention! Solutions are provided by ordinary people, so there may be errors or inaccuracies in the solutions. When using solutions, don't forget to double-check them!
HNO3 + Zi2O -> ..; HNO3 + ZnCO3 -> ..; HNO3 + Fe(OH)3 -> .. .
Attention! Solutions are provided by ordinary people, so there may be errors or inaccuracies in the solutions. When using solutions, don't forget to double-check them!
Fe(OH)3 + acid oxide = .. .
Attention! Solutions are provided by ordinary people, so there may be errors or inaccuracies in the solutions. When using solutions, don't forget to double-check them!
Fe(OH)3 + H2SO4 = ..; write ions.
Attention! Solutions are provided by ordinary people, so there may be errors or inaccuracies in the solutions. When using solutions, don't forget to double-check them!
Fe(OH)3 + HNO3 -> ..; do an ion exchange reaction.
Attention! Solutions are provided by ordinary people, so there may be errors or inaccuracies in the solutions. When using solutions, don't forget to double-check them!
Fe -> FeCl3 -> Fe(OH)3; OVR reaction.
Attention! Solutions are provided by ordinary people, so there may be errors or inaccuracies in the solutions. When using solutions, don't forget to double-check them!
H2SO4 + Fe(OH)3; write an equation for the reaction.
Attention! Solutions are provided by ordinary people, so there may be errors or inaccuracies in the solutions. When using solutions, don't forget to double-check them!
The human body contains about 5 g of iron, most of it (70%) is part of the hemoglobin in the blood.
Physical Properties
In the free state, iron is a silvery-white metal with a grayish tint. Pure iron is ductile and has ferromagnetic properties. In practice, iron alloys are commonly used - cast irons and steels.
Fe is the most important and most common element of the nine d-metals of the secondary subgroup of group VIII. Together with cobalt and nickel, it forms the "iron family".
When forming compounds with other elements, it often uses 2 or 3 electrons (B \u003d II, III).
Iron, like almost all d-elements of group VIII, does not show a higher valency equal to the group number. Its maximum valency reaches VI and is extremely rare.
The most typical compounds are those in which the Fe atoms are in the +2 and +3 oxidation states.
Methods for obtaining iron
1. Commercial iron (in an alloy with carbon and other impurities) is obtained by carbothermal reduction of its natural compounds according to the scheme:
Recovery occurs gradually, in 3 stages:
1) 3Fe 2 O 3 + CO = 2Fe 3 O 4 + CO 2
2) Fe 3 O 4 + CO = 3FeO + CO 2
3) FeO + CO \u003d Fe + CO 2
The cast iron resulting from this process contains more than 2% carbon. In the future, steels are obtained from cast iron - iron alloys containing less than 1.5% carbon.
2. Very pure iron is obtained in one of the following ways:
a) decomposition of pentacarbonyl Fe
Fe(CO) 5 = Fe + 5CO
b) hydrogen reduction of pure FeO
FeO + H 2 \u003d Fe + H 2 O
c) electrolysis of aqueous solutions of Fe +2 salts
FeC 2 O 4 \u003d Fe + 2СO 2
iron(II) oxalate
Chemical properties
Fe - a metal of medium activity, exhibits general properties characteristic of metals.
A unique feature is the ability to "rust" in humid air:
In the absence of moisture with dry air, iron begins to noticeably react only at T > 150°C; when calcined, “iron scale” Fe 3 O 4 is formed:
3Fe + 2O 2 = Fe 3 O 4
Iron does not dissolve in water in the absence of oxygen. At very high temperatures, Fe reacts with water vapor, displacing hydrogen from water molecules:
3 Fe + 4H 2 O (g) \u003d 4H 2
The rusting process in its mechanism is electrochemical corrosion. The rust product is presented in a simplified form. In fact, a loose layer of a mixture of oxides and hydroxides of variable composition is formed. Unlike the Al 2 O 3 film, this layer does not protect the iron from further destruction.
Types of corrosion
Corrosion protection of iron
1. Interaction with halogens and sulfur at high temperature.
2Fe + 3Cl 2 = 2FeCl 3
2Fe + 3F 2 = 2FeF 3
Fe + I 2 \u003d FeI 2
Compounds are formed in which the ionic type of bond predominates.
2. Interaction with phosphorus, carbon, silicon (iron does not directly combine with N 2 and H 2, but dissolves them).
Fe + P = Fe x P y
Fe + C = Fe x C y
Fe + Si = FexSiy
Substances of variable composition are formed, since berthollides (the covalent nature of the bond prevails in the compounds)
3. Interaction with "non-oxidizing" acids (HCl, H 2 SO 4 dil.)
Fe 0 + 2H + → Fe 2+ + H 2
Since Fe is located in the activity series to the left of hydrogen (E ° Fe / Fe 2+ \u003d -0.44V), it is able to displace H 2 from ordinary acids.
Fe + 2HCl \u003d FeCl 2 + H 2
Fe + H 2 SO 4 \u003d FeSO 4 + H 2
4. Interaction with "oxidizing" acids (HNO 3 , H 2 SO 4 conc.)
Fe 0 - 3e - → Fe 3+
Concentrated HNO 3 and H 2 SO 4 "passivate" iron, so at ordinary temperatures the metal does not dissolve in them. With strong heating, slow dissolution occurs (without release of H 2).
In razb. HNO 3 iron dissolves, goes into solution in the form of Fe 3+ cations, and the acid anion is reduced to NO *:
Fe + 4HNO 3 \u003d Fe (NO 3) 3 + NO + 2H 2 O
It dissolves very well in a mixture of HCl and HNO 3
5. Attitude to alkalis
Fe does not dissolve in aqueous solutions of alkalis. It reacts with molten alkalis only at very high temperatures.
6. Interaction with salts of less active metals
Fe + CuSO 4 \u003d FeSO 4 + Cu
Fe 0 + Cu 2+ = Fe 2+ + Cu 0
7. Interaction with gaseous carbon monoxide (t = 200°C, P)
Fe (powder) + 5CO (g) \u003d Fe 0 (CO) 5 iron pentacarbonyl
Fe(III) compounds
Fe 2 O 3 - iron oxide (III).
Red-brown powder, n. R. in H 2 O. In nature - "red iron ore".
Ways to get:
1) decomposition of iron hydroxide (III)
2Fe(OH) 3 = Fe 2 O 3 + 3H 2 O
2) pyrite roasting
4FeS 2 + 11O 2 \u003d 8SO 2 + 2Fe 2 O 3
3) decomposition of nitrate
Chemical properties
Fe 2 O 3 is a basic oxide with signs of amphoterism.
I. The main properties are manifested in the ability to react with acids:
Fe 2 O 3 + 6H + = 2Fe 3+ + ZH 2 O
Fe 2 O 3 + 6HCI \u003d 2FeCI 3 + 3H 2 O
Fe 2 O 3 + 6HNO 3 \u003d 2Fe (NO 3) 3 + 3H 2 O
II. Weak acid properties. Fe 2 O 3 does not dissolve in aqueous solutions of alkalis, but when fused with solid oxides, alkalis and carbonates, ferrites are formed:
Fe 2 O 3 + CaO \u003d Ca (FeO 2) 2
Fe 2 O 3 + 2NaOH \u003d 2NaFeO 2 + H 2 O
Fe 2 O 3 + MgCO 3 \u003d Mg (FeO 2) 2 + CO 2
III. Fe 2 O 3 - feedstock for iron production in metallurgy:
Fe 2 O 3 + ZS \u003d 2Fe + ZSO or Fe 2 O 3 + ZSO \u003d 2Fe + ZSO 2
Fe (OH) 3 - iron (III) hydroxide
Ways to get:
Obtained by the action of alkalis on soluble salts Fe 3+:
FeCl 3 + 3NaOH \u003d Fe (OH) 3 + 3NaCl
At the time of receipt of Fe(OH) 3 - red-brown mucosamorphous precipitate.
Fe (III) hydroxide is also formed during the oxidation of Fe and Fe (OH) 2 in humid air:
4Fe + 6H 2 O + 3O 2 \u003d 4Fe (OH) 3
4Fe(OH) 2 + 2Н 2 O + O 2 = 4Fe(OH) 3
Fe(III) hydroxide is the end product of hydrolysis of Fe 3+ salts.
Chemical properties
Fe(OH) 3 is a very weak base (much weaker than Fe(OH) 2). Shows noticeable acidic properties. Thus, Fe (OH) 3 has an amphoteric character:
1) reactions with acids proceed easily:
2) a fresh precipitate of Fe(OH) 3 is dissolved in hot conc. solutions of KOH or NaOH with the formation of hydroxo complexes:
Fe (OH) 3 + 3KOH \u003d K 3
In an alkaline solution, Fe (OH) 3 can be oxidized to ferrates (salts of iron acid H 2 FeO 4 not isolated in the free state):
2Fe(OH) 3 + 10KOH + 3Br 2 = 2K 2 FeO 4 + 6KBr + 8H 2 O
Fe 3+ salts
The most practically important are: Fe 2 (SO 4) 3, FeCl 3, Fe (NO 3) 3, Fe (SCN) 3, K 3 4 - yellow blood salt \u003d Fe 4 3 Prussian blue (dark blue precipitate)
b) Fe 3+ + 3SCN - \u003d Fe (SCN) 3 Fe (III) thiocyanate (blood red solution)
In the gravimetric determination of iron in solutions, it is first oxidized to Fe3+, and then the hydrolysis of the iron salt is completed by the action of NH4OH:
Fe3+ + H20 FeOH2+ + H+
Fe(OH)++ H2O -> Fe(OH)3I+ H+
When calcined, Fe (OH) 3 (or rather, hydrous iron oxide Fe2O3-ZiH2O) loses water and turns into an anhydrous oxide:
2Fe(OH)3 -* Fe2O3+ 3H2Of
which is weighed. The solubility of iron hydroxide is very low (PR = 3.2 10~38), so it precipitates quantitatively even from slightly acidic solutions. The low solubility leads to the fact that the relative supersaturation of the solution during precipitation is very high, so the precipitate is amorphous and has a very large surface. For better coagulation of the precipitate, the precipitation is carried out by heating in the presence of an electrolyte (ammonium salts). The precipitate is not easily peptized and can be washed with hot water, but if the washing continues for a long time, it is better to use a 1% NH4Cl solution instead of water. It is necessary to ignite the sediment on the burner with air access, especially during the burning of the paper filter; prolonged calcination should be avoided so that there is no partial reduction of iron oxide by carbon to Fe3O4 (and even to metallic iron).
Definition progress. To a solution of Mohr's salt (7-10 ml, containing no more than 0.1 g of iron) is added 10 ml of H2O, 3 g x. h. NH4Cl1 solution is heated almost to boiling (but not boiled), 1-2 ml of concentrated HNO3 is added dropwise with stirring and heating is continued for another 3-5 minutes. Then 100-150 ml of hot water and NH4OH (1:I) are added to the solution with stirring until a clear smell of ammonia ** appears; the solution with the precipitate is left for 5 minutes and filtering is started.
* This definition is educational in nature and serves as a good example of the deposition of amorphous precipitates. In practice, titrimetric methods for determining iron are usually used as more accurate and faster.
** When adding ammonia, you need to make sure that the solution smells like it; iron hydroxide precipitate is not amphoteric, so a slight excess of NH4OH is not by Credig's definition.
It should be filtered through a medium-density filter (white tape) with a diameter of 9 cm. After draining the liquid from the sediment onto the filter, the filter is washed several times by decantation with hot water. After that, the sediment is transferred to the filter, the sediment particles remaining on the glass and stick are removed with pieces of an ashless filter.
Washing of the precipitate on the filter is continued until Cb is completely removed, i.e., until a portion of the washing water, acidified with HNO3, ceases to produce turbidity with AgNO3. sediment on the filter is impossible, it dries up, channels form in it, and in the future the washing liquid will not extract pollutants from the sediment.
The washed precipitate is dried and, still slightly damp, together with the filter, is transferred into a crucible calcined to constant weight. Next, the filter is carefully dried and charred on a small burner flame so that it does not catch fire. Then it is ashed and, gradually increasing the heating, the crucible with the precipitate is calcined to constant weight. It is better to calcinate the precipitate in a muffle furnace at 800-900 °C.
Calculation. Having found the mass of the sediment, calculate how much iron it contains, using the conversion factor.
Similarly, iron is determined in various objects containing it. For example, when analyzing an iron wire, a sample * of it (about 0.1 g) is dissolved when heated in 10-15 ml of 2N. HNO3. The Fe(NO3J3) solution is analyzed as described above. Having found the amount of iron in the Fe2O3 precipitate, the percentage of iron in the sample of wire is calculated.
METHODOLOGY OF PREPARING STUDENTS FOR THE DECISION
ASSIGNMENTS C 2 (thought experiment) USE IN CHEMISTRY
In 2012, task C2 of the Unified State Examination in chemistry provides for a change. Students will be offered a description of a chemical experiment, according to which they will need to write 4 reaction equations.
We can judge the content and level of complexity of this task by the demo version of the 2012 USE version. The task is formulated as follows: The salt obtained by dissolving iron in hot concentrated sulfuric acid was treated with an excess of sodium hydroxide solution. The brown precipitate formed was filtered off and dried. The resulting substance was fused with iron. Write the equations of the described reactions.
An analysis of the content of the assignment shows that the first two substances that enter into the reaction are indicated in clear form. For all other reactions, the reagent and conditions are indicated. Tips can be considered as indications of the class of the obtained substance, its state of aggregation, characteristic features (color, smell). Note that two reaction equations characterize the special properties of substances (1 - the oxidizing properties of concentrated sulfuric acid; 4 - the oxidizing properties of iron oxide (III)), two equations characterize the typical properties of the most important classes of inorganic substances (2 - the ion exchange reaction between solutions of salt and alkali , 3 – thermal decomposition of the insoluble base).
t o C NaOH (ex.) t o C + Fe/t o C
Fe + H 2 SO 4 (c) → salt → brown precipitate → X → Y
Highlight clues, key points, for example: a brown precipitate - iron (III) hydroxide, indicates that the salt is formed by an iron ion (3+).
2Fe + 6H 2 SO 4 (c) → Fe 2 (SO 4) 3+ 3SO 2 + 6H 2 O
Fe 2 (SO 4) 3+ 6NaOH(c) → 2 Fe(OH)3+ 3Na2SO4
2Fe(OH)3→ Fe2O3+ 3H2O
Fe2O3+ Fe → 3 FeO
What difficulties can such tasks cause for students?
- Description of actions with substances (filtration, evaporation, roasting, calcination, sintering, fusion). Students need to understand where a physical phenomenon occurs with a substance, and where a chemical reaction occurs. The most commonly used actions with substances are described below.
Filtration- a method for separating heterogeneous mixtures using filters - porous materials that pass liquid or gas, but retain solids. When separating mixtures containing a liquid phase, a solid remains on the filter, filtrate .
Evaporation - the process of concentrating solutions by evaporating the solvent. Sometimes evaporation is carried out until saturated solutions are obtained, with the aim of further crystallization of a solid substance in the form of a crystalline hydrate, or until the solvent is completely evaporated in order to obtain a pure solute.
Ignition - heating a substance to change its chemical composition.
The calcination can be carried out in air and in an inert gas atmosphere.
When calcined in air, crystalline hydrates lose water of crystallization:
CuSO 4 ∙5H 2 O → CuSO 4 + 5H 2 O
Thermally unstable substances decompose (insoluble bases, some salts, acids, oxides): Cu(OH) 2 →CuO + H 2 O; CaCO 3 → CaO + CO 2
Substances that are unstable to the action of air components oxidize when calcined, react with air components: 2Сu + O 2 → 2CuO;
4Fe(OH) 2 + O 2 →2Fe 2 O 3 + 4H 2 O
In order to prevent oxidation during calcination, the process is carried out in an inert atmosphere: Fe (OH) 2 → FeO + H 2 O
Sintering, fusion - This is the heating of two or more solid reactants, leading to their interaction. If the reagents are resistant to the action of oxidizing agents, then sintering can be carried out in air:
Al 2 O 3 + Na 2 CO 3 → 2NaAlO 2 + CO 2
If one of the reactants or the reaction product can be oxidized by air components, the process is carried out with an inert atmosphere, for example: Сu + CuO → Cu 2 O
Burning- a heat treatment process leading to the combustion of a substance (in the narrow sense. In a broader sense, roasting is a variety of thermal effects on substances in chemical production and metallurgy). It is mainly used in relation to sulfide ores. For example, firing pyrite:
4FeS 2 + 11O 2 → 2Fe 2 O 3 + 8SO 2
2. Description of the characteristic features of substances (color, smell, state of aggregation).
An indication of the characteristic features of substances should serve as a hint for students or a check on the correctness of the actions performed. However, if students are not familiar with the physical properties of substances, such information cannot provide an auxiliary function when performing a thought experiment. Below are the most characteristic features of gases, solutions, solids.
GASES:
Painted: Cl2- yellow-green; NO 2- brown; O 3- blue (all have smells). All are poisonous, dissolve in water, Cl2 and NO 2 react with her.
Colorless, odorless: H 2 , N 2 , O 2 , CO 2 , CO (poison), NO (poison), inert gases. All are poorly soluble in water.
Colorless with an odor: HF, HCl, HBr, HI, SO 2 (pungent odors), NH 3 (ammonia) - highly soluble in water and poisonous,
PH 3 (garlic), H 2 S (rotten eggs) - slightly soluble in water, poisonous.
COLORED SOLUTIONS:
PAINTED DRAINAGE,
PRODUCED IN THE INTERACTION OF SOLUTIONS
OTHER COLORED SUBSTANCES
This, of course, is the minimum information that can be useful for solving tasks C2.
In the process of preparing students for solving tasks C2, you can offer them compose texts of assignments in accordance with the schemes of transformations . This task will allow students to master the terminology and remember the characteristic features of substances.
Example 1:
t o C t o C/H 2 HNO 3 (conc) NaOH, 0 o C
(CuOH) 2 CO 3 → CuO → Cu → NO 2 → X
Text: Malachite was calcined, the resulting black solid was heated in a stream of hydrogen. The resulting red substance was completely dissolved in concentrated nitric acid. The liberated brown gas was passed through a cold solution of sodium hydroxide.
Example 2:
O 2 H 2 S solution t o C/Al H 2 O
ZnS → SO 2 → S → Al 2 S 3 → X
Text: The zinc sulfide was calcined. The resulting gas with a pungent odor was passed through a solution of hydrogen sulfide until a yellow precipitate formed. The precipitate was filtered off, dried and fused with aluminum. The resulting compound was placed in water until the reaction terminated.
The next step is to ask students to draw up both schemes for the transformation of substances and texts of tasks. Of course, the "authors" of the tasks must submit and own solution . At the same time, students repeat all the properties of inorganic substances. And the teacher can form a bank of tasks C2. After that you can go to the solution of tasks С2 . At the same time, students draw up a scheme of transformations according to the text, and then the corresponding reaction equations. To do this, reference points are highlighted in the text of the task: the names of substances, an indication of their classes, physical properties, conditions for conducting reactions, names of processes.
Let's give examples of some tasks.
Example 1 Manganese (II) nitrate was calcined, and concentrated hydrochloric acid was added to the resulting brown solid. The evolved gas was passed through hydrosulfide acid. The resulting solution forms a precipitate with barium chloride.
Solution:
· Isolation of support moments:
Manganese(II) nitrate- Mn(NO 3) 2,
calcined- heated to decomposition,
solid brown matter- MnO 2,
Concentrated hydrochloric acid–HCl,
Hydrosulphuric acid - solution H 2 S,
Barium chloride - BaCl 2 forms a precipitate with the sulfate ion.
t o C HCl H 2 Sp-p BaCl 2
Mn(NO 3) 2 → MnO 2 → X → Y → ↓ (BaSO 4 ?)
1) Mn(NO 3) 2 → MnО 2 + 2NO 2
2) MnO 2 + 4 HCl → MnCl 2 + 2H 2 O + Cl 2 (gas X)
3) Cl 2 + H 2 S → 2HCl + S (not suitable, because there is no product that precipitates with barium chloride) or 4Cl 2 + H 2 S + 4H 2 O → 8HCl + H 2 SO 4
4) H 2 SO 4 + BaCl 2 → BaSO 4 + 2HCl
Example 2 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 resulting blue precipitate was filtered off, dried and calcined. The solid black substance thus obtained was placed in a glass tube, heated, and ammonia was passed over it.
Solution:
· Isolation of support moments:
Orange copper oxide– Cu 2 O,
concentrated sulfuric acid- H 2 SO 4,
blue solution- copper (II) salt, СuSO 4
Potassium hydroxide-KOH,
Blue precipitate - Cu(OH)2,
Calcined - heated to decomposition
Solid black matter CuO,
Ammonia- NH3.
· Drawing up a scheme of transformations:
H 2 SO 4 KOH t o C NH 3
Cu 2 O → СuSO 4 → Cu(OH) 2 ↓ → CuO → X
Drawing up reaction equations:
1) Cu 2 O + 3Н 2 SO 4 → 2СuSO 4 + SO 2 + 3H 2 O
2) СuSO 4 + 2KOH → Cu(OH) 2 + K 2 SO 4
3) Cu(OH) 2 → CuO + H 2 O
4) 3CuO + 2NH 3 → 3Cu + 3H 2 O + N 2
