- Write equations for the reactions of aluminium, its oxide and its hydroxide with acids and alkalis, and explain amphoterism.
- Calculate and graph how the mass of precipitate changes as alkali is added to AlCl₃ solution.
- Know which reagents turn iron into Fe²⁺ and which into Fe³⁺ compounds, and identify these ions.
- Calculate the mass change of a metal plate in a salt solution and the metal obtainable from an ore; explain the blast-furnace process.
Aircraft bodies, foil and power lines are made of aluminium; rails, bridges and car bodies of iron alloys. Aluminium is the most abundant metal in the Earth’s crust and iron the most widely used. In Azerbaijan, Dashkasan is known for its iron-ore (magnetite) deposits, and Zaylik in the same district for its alunite deposit: aluminium oxide is made from alunite in Ganja, and aluminium metal is produced by electrolysis in Sumgayit. In 2026 the group IV DİM situation task was built on iron reacting with dilute sulfuric acid and a magnesium plate.
Aluminium: production and properties
Aluminium (Al, Z = 13) is in group IIIA: 1s²2s²2p⁶3s²3p¹, +3 in compounds. It is the third element of the Earth’s crust after oxygen and silicon (≈ 8 %); its main minerals are bauxite Al₂O₃·nH₂O, corundum Al₂O₃ (ruby and sapphire are coloured varieties), kaolin Al₂O₃·2SiO₂·2H₂O and alunite K₂SO₄·Al₂(SO₄)₃·4Al(OH)₃. The metal is made by electrolysis of Al₂O₃ dissolved in molten cryolite (Na₃AlF₆) at ≈ 950 °C: 2Al₂O₃ → 4Al + 3O₂ — the cryolite lowers the melting point. Aluminium is a light (2.7 g/cm³), silvery, highly conductive and malleable metal that melts at 660 °C.
- Its surface is covered by a thin, tough Al₂O₃ film, so aluminium survives in air and water. As a powder it burns with a dazzling flame: 4Al + 3O₂ → 2Al₂O₃. If the film is removed (for example with mercury), it reacts with water: 2Al + 6H₂O → 2Al(OH)₃ + 3H₂↑.
- With non-metals: 2Al + 3Cl₂ → 2AlCl₃, 2Al + 3S → Al₂S₃ (on heating). With acids: 2Al + 6HCl → 2AlCl₃ + 3H₂↑; it is passivated by cold concentrated HNO₃ and H₂SO₄.
- With alkali solutions it also releases hydrogen — it is an amphoteric metal: 2Al + 2NaOH + 6H₂O → 2Na[Al(OH)₄] + 3H₂↑ (sodium tetrahydroxoaluminate).
- Aluminothermy — reducing metals from their oxides with aluminium, with a lot of heat: 2Al + Fe₂O₃ → Al₂O₃ + 2Fe (welding rails), 2Al + Cr₂O₃ → Al₂O₃ + 2Cr.
- n(Al)amount of aluminium reacting, mol
- n(H₂)amount of hydrogen released, mol
The same with an acid or an alkali: Al⁰ → Al⁺³ (3 electrons), 2H⁺¹ → H₂ (2 electrons). In a mixture only Al reacts with an alkali, whereas an acid attacks every metal before hydrogen.
One sample of an aluminium–iron mixture released 6.72 L of hydrogen with excess sodium hydroxide solution, and a second sample of the same mass released 8.96 L (STP) with excess hydrochloric acid. Calculate the mass of each metal in the mixture (g). Ar(Al) = 27, Ar(Fe) = 56
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With the acid both metals react: n(H₂) = 8.96 / 22.4 = 0.4 mol; aluminium again gives 0.3 mol and iron 0.1 mol (Fe + 2HCl → FeCl₂ + H₂↑).
n(Fe) = 0.1 mol → m(Fe) = 5.6 g.
Aluminium oxide and hydroxide. Amphoterism
Al₂O₃ is a very hard, refractory (2050 °C) white substance insoluble in water; it is an amphoteric oxide: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O, Al₂O₃ + 2NaOH + 3H₂O → 2Na[Al(OH)₄]; fused with an alkali it gives a metaaluminate: Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O. It does not react with water, so Al(OH)₃ is made from a salt: AlCl₃ + 3NaOH → Al(OH)₃↓ + 3NaCl. This white jelly-like precipitate is amphoteric too: Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O, Al(OH)₃ + NaOH → Na[Al(OH)₄]. That is why it is better made with ammonia solution, an excess of which does not dissolve it: AlCl₃ + 3NH₃·H₂O → Al(OH)₃↓ + 3NH₄Cl. CO₂ brings the hydroxide back from an aluminate: Na[Al(OH)₄] + CO₂ → Al(OH)₃↓ + NaHCO₃; on heating 2Al(OH)₃ → Al₂O₃ + 3H₂O. Test for Al³⁺: a little alkali gives a white jelly-like precipitate, which dissolves in excess alkali.
- aamount of AlCl₃ in the solution, mol
- xamount of NaOH added, mol
Adding alkali drop by drop: at x = 3a the precipitate is at its maximum (a mol), at x = 4a it has fully dissolved. The same mass of precipitate occurs at two different values of x.
NaOH solution is added drop by drop, with stirring, to a solution containing 0.15 mol of AlCl₃.
(1) Find the maximum mass of precipitate (g) and the amount of NaOH added up to that moment (mol).
(2) How many grams of precipitate remain after 0.55 mol of NaOH has been added? At what other amount of NaOH is the same mass obtained?
(3) How many grams of NaOH in total are needed to dissolve the precipitate completely? Mr(Al(OH)₃) = 78, Mr(NaOH) = 40
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(2) After 0.45 mol, Al(OH)₃ + NaOH → Na[Al(OH)₄]: the precipitate left is 4a − x = 0.6 − 0.55 = 0.05 mol → 3.9 g.
On the rising part 0.05 mol of precipitate forms at x / 3 = 0.05 → x = 0.15 mol of NaOH.
(3) x = 4a = 0.6 mol → m(NaOH) = 0.6 · 40 = 24 g.
Iron and its compounds
Iron (Fe, Z = 26) is a group VIIIB element: [Ar]3d⁶4s², mostly +2 and +3 in compounds. It is the second most abundant metal in the Earth’s crust; its ores are magnetite Fe₃O₄, haematite Fe₂O₃, limonite 2Fe₂O₃·3H₂O and siderite FeCO₃. It is a silvery-grey metal attracted by a magnet and melts at 1539 °C. Which oxidation state forms? Strong oxidising agents (Cl₂, HNO₃, hot concentrated H₂SO₄) give Fe³⁺; weak ones (S, HCl, dilute H₂SO₄, salts of less active metals) give Fe²⁺:
- Fe³⁺: 2Fe + 3Cl₂ → 2FeCl₃; Fe + 4HNO₃(dil.) → Fe(NO₃)₃ + NO↑ + 2H₂O; 2Fe + 6H₂SO₄(conc., hot) → Fe₂(SO₄)₃ + 3SO₂↑ + 6H₂O. Cold concentrated HNO₃ and H₂SO₄ passivate iron.
- Fe²⁺: Fe + S → FeS; Fe + 2HCl → FeCl₂ + H₂↑; Fe + H₂SO₄(dil.) → FeSO₄ + H₂↑; Fe + CuSO₄ → FeSO₄ + Cu.
- Both — Fe₃O₄ (FeO·Fe₂O₃): burning in oxygen 3Fe + 2O₂ → Fe₃O₄, red-hot iron with steam 3Fe + 4H₂O → Fe₃O₄ + 4H₂; in acid it gives two salts: Fe₃O₄ + 8HCl → FeCl₂ + 2FeCl₃ + 4H₂O.
| Fe(II) | Fe(III) | |
|---|---|---|
| Oxide | FeO — black, basic | Fe₂O₃ — red-brown |
| Hydroxide | Fe(OH)₂ — white, turns greenish and then brown in air | Fe(OH)₃ — brown |
| Salt solutions | pale green (FeSO₄·7H₂O — green vitriol) | yellow-brown (FeCl₃) |
| Test | K₃[Fe(CN)₆] (red prussiate) — a blue precipitate (Turnbull’s blue) | K₄[Fe(CN)₆] (yellow prussiate) — a blue precipitate (Prussian blue); KSCN — blood-red colour |
| Redox | reducing agent: 2FeCl₂ + Cl₂ → 2FeCl₃; 4Fe(OH)₂ + O₂ + 2H₂O → 4Fe(OH)₃ | oxidising agent: 2FeCl₃ + Fe → 3FeCl₂; 2FeCl₃ + Cu → 2FeCl₂ + CuCl₂; 2FeCl₃ + 2KI → 2FeCl₂ + 2KCl + I₂ |
Which substances give an iron(II) compound when they react with iron?
I. chlorine (on heating)
II. dilute sulfuric acid
III. sulfur (on heating)
IV. copper(II) chloride solution
V. excess dilute nitric acid
A) I, II B) II, III, IV C) I, V D) III, IV, V E) II, V
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I: 2Fe + 3Cl₂ → 2FeCl₃ and V: Fe + 4HNO₃ → Fe(NO₃)₃ + NO + 2H₂O — strong oxidising agents give Fe³⁺.
Answer: B.
A metal plate in a salt solution. Cast iron and steel
- Δmchange in the mass of the plate, g (increase “+”, decrease “−”)
- namount of metal reacted, mol (1 : 1 for ions of equal charge)
- M₁, M₂molar masses of the dissolving metal (the plate) and the deposited metal, g/mol
An active metal displaces a less active one from its salt: M₁ “leaves” the plate and M₂ “arrives”. If the ion charges differ, multiply by the equation coefficients: for Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag, Δm = n(Cu) · (2 · 108 − 64).
1) An iron plate was placed in CuSO₄ solution, and after a while its mass had increased by 1.6 g. How many grams of copper were deposited, and how many grams of FeSO₄ formed in the solution?
2) An iron plate was placed in 122.5 g of dilute sulfuric acid. When all the acid had reacted (t₁), the plate was removed and a zinc plate was put in. At t₂ the zinc plate had lost 0.9 g and the amount of iron salt in the solution had fallen by 40%. Calculate the mass fraction of the acid (%), the volume of hydrogen released (L, STP) and the mass of ZnSO₄ formed (g). Ar(Fe) = 56, Ar(Cu) = 64, Ar(Zn) = 65, Mr(H₂SO₄) = 98, Mr(FeSO₄) = 152, Mr(ZnSO₄) = 161
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m(Cu) = 0.2 · 64 = 12.8 g; m(FeSO₄) = 0.2 · 152 = 30.4 g.
2) Zn + FeSO₄ → ZnSO₄ + Fe: Δm = n · (56 − 65) = −9n = −0.9 → n = 0.1 mol of FeSO₄ reacted — that is 40%, so at t₁ n(FeSO₄) = 0.1 / 0.4 = 0.25 mol.
Fe + H₂SO₄ → FeSO₄ + H₂↑: n(H₂SO₄) = 0.25 mol → 24.5 g → ω = 24.5 / 122.5 · 100% = 20%; V(H₂) = 0.25 · 22.4 = 5.6 L.
m(ZnSO₄) = 0.1 · 161 = 16.1 g. The amount of iron salt goes 0 → 0.25 mol (t₁) → 0.15 mol (t₂).
- 1Charging the blast furnace
Ore (Fe₂O₃, Fe₃O₄), coke and flux (CaCO₃) go in at the top; hot air is blown in at the bottom.
- 2Making the reducing agent
C + O₂ → CO₂ (gives heat), CO₂ + C → 2CO.
- 3Stepwise reduction of iron
3Fe₂O₃ + CO → 2Fe₃O₄ + CO₂; Fe₃O₄ + CO → 3FeO + CO₂; FeO + CO → Fe + CO₂.
- 4Forming slag
CaCO₃ → CaO + CO₂; with the SiO₂ of the waste rock: CaO + SiO₂ → CaSiO₃ (slag). Molten cast iron collects under the slag.
Cast iron contains more than 2 % carbon (usually 2–4 %, plus Si, Mn, S, P): it is hard and brittle and is cast in moulds. Steel contains less than 2 % carbon; it is strong and springy and can be forged. Steel is made from cast iron by oxidising the excess carbon and impurities (converter, open-hearth and electric furnaces): 2C + O₂ → 2CO, FeO + C → Fe + CO. Alloy steel with chromium and nickel does not rust. The rusting of iron is covered in “Redox reactions”.
Key points
- Aluminium is made by electrolysis of Al₂O₃ in molten cryolite; an oxide film protects it; it releases H₂ with acids and alkalis (n(H₂) = 1.5 · n(Al)) and is passivated by cold concentrated HNO₃ and H₂SO₄.
- Al₂O₃ and Al(OH)₃ are amphoteric; adding alkali to AlCl₃ gives the most precipitate at x = 3a and none at x = 4a; Al(OH)₃ is made with ammonia or by passing CO₂ into an aluminate.
- Aluminothermy: 2Al + Fe₂O₃ → Al₂O₃ + 2Fe — obtaining metals from their oxides and welding rails.
- Iron gives Fe³⁺ with strong oxidising agents (Cl₂, HNO₃) and Fe²⁺ with weak ones (HCl, dilute H₂SO₄, S, CuSO₄); Fe²⁺ is detected with red prussiate, Fe³⁺ with yellow prussiate and KSCN.
- Plate in a salt solution: Δm = n · (M₂ − M₁); in the blast furnace CO reduces iron and CaO turns waste rock into slag; cast iron has C > 2 %, steel C < 2 %.
Check yourself
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