The problems are on the reactivity series worksheet. Try them on your own first, then compare with the answers below.
Quick answer: The reactivity series ranks metals by how easily they lose electrons. These worked answers correct the 20 problems of the Chemistry 9-10 worksheet on reactivity series, from put them in order to which metal is M?: each solution shows the formula, the substitution and the final answer with its unit.
Each answer gives the chemical reasoning as well as the result. For equations, check the balancing last; for redox, always say which substance gains or loses oxygen or electrons.
Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: GCSE C4
Problem 1 Put them in order
1. Potassium, magnesium, zinc, iron, copper, gold.2. Potassium reacts with cold water, violently.
3. Copper and gold.
Problem 2 Metals and dilute acid
1. Magnesium, zinc, iron, and copper does not fizz at all.2. Magnesium + hydrochloric acid → magnesium chloride + hydrogen.
3. A lighted splint gives a squeaky pop.
Problem 3 Will it react?
1. Reaction: zinc sulfate and copper.2. No reaction: copper is less reactive than zinc.
3. Reaction: magnesium sulfate and iron.
4. No reaction: iron is below magnesium.
Problem 4 Vocabulary of redox
1. Oxidation is gain of oxygen.2. Reduction is gain of electrons.
3. Hydrogen gains oxygen, so H₂ is oxidised; copper oxide loses oxygen, so CuO is reduced.
4. Oxidation Is Loss, Reduction Is Gain (of electrons).
Problem 5 Choose the method
1. Electrolysis.2. Heating with carbon.
3. Found uncombined.
4. Electrolysis.
5. Heating with carbon.
Problem 6 Inside a blast furnace
1. Coke is carbon: it burns to give heat and forms carbon monoxide.2. Carbon monoxide.
3. Limestone removes sandy impurities as slag: CaCO₃ → CaO + CO₂, then CaO + SiO₂ → CaSiO₃.
Problem 7 Balance the equations
1. 2Na + 2H₂O → 2NaOH + H₂.2. 2Al + 6HCl → 2AlCl₃ + 3H₂.
3. Fe₂O₃ + 3CO → 2Fe + 3CO₂.
4. Already balanced: ZnO + C → Zn + CO.
Problem 8 Ionic equation and half-equations
1. Zn + Cu²⁺ → Zn²⁺ + Cu.2. Zn → Zn²⁺ + 2e⁻: oxidation.
3. Cu²⁺ + 2e⁻ → Cu (reduction). The spectator is the sulfate ion, SO₄²⁻.
Problem 9 Redox spotting
1. Mg is oxidised, O₂ is reduced; reducing agent Mg.2. CO is oxidised, Fe₂O₃ is reduced; reducing agent CO.
3. Fe is oxidised (Fe → Fe²⁺), Cu²⁺ is reduced; reducing agent Fe.
4. H₂ is oxidised, CuO is reduced; reducing agent H₂.
Problem 10 Deduce the order
1. W displaces X, Y and Z, and Y displaces X and Z, X displaces only Z. The order is W > Y > X > Z.2. Z, because it displaces nothing.
3. No reaction, because Y is less reactive than W.
Problem 11 Welding rails
1. Aluminium is more reactive than iron, so it displaces iron from its oxide.2. n = 5400 ÷ 27 = 200 mol.
3. n(Fe) = 200 mol, m = 200 × 56 = 11.2 kg. n(Fe₂O₃) = 100 mol, m = 100 × 160 = 16 kg.
Problem 12 Galvanised roofs
1. Area = 20 000 cm²; V = 20 000 × 0.0020 = 40 cm³.2. m = 40 × 7.1 = 284 g.
3. Zinc is more reactive than iron, so it loses electrons first and corrodes instead of the iron (sacrificial protection).
Problem 13 Recycling aluminium
1. 1000 × 15 g = 15 000 g = 15 kg.2. Extraction: 15 × 50 = 750 MJ; recycling: 15 × 2.5 = 37.5 MJ; saving = 712.5 MJ.
3. 712.5 ÷ 750 × 100 = 95%. Recycling only melts the metal, whereas extraction requires electrolysis of molten aluminium oxide at high temperature.
Problem 14 How much iron in the ore?
1. 112 ÷ 160 × 100 = 70.0%.2. Fe₂O₃ = 0.80 × 50 = 40 t; iron = 0.700 × 40 = 28 t.
3. 0.92 × 28 = 25.8 t (25.76 t).
Problem 15 Reactions with water
1. Sodium > calcium > magnesium.2. Sodium loses one outer electron very easily (Na → Na⁺ + e⁻), whereas magnesium must lose two electrons, which takes more energy.
3. 2Na + 2H₂O → 2NaOH + H₂.
Problem 16 Smelting copper oxide
1. M(CuO) = 79.5, so n = 159 ÷ 79.5 = 2.00 mol; n(C) = 6.0 ÷ 12 = 0.50 mol.2. 2.00 mol CuO needs 1.00 mol C, so carbon is limiting.
3. 0.50 mol C reacts with 1.00 mol CuO and forms 1.00 mol Cu: m = 63.5 g. CuO left = 1.00 mol × 79.5 = 79.5 g.
Problem 17 Zinc in copper sulfate
1. n(Zn) = 3.25 ÷ 65 = 0.0500 mol; n(CuSO₄) = 0.400 × 0.100 = 0.0400 mol.2. Ratio 1 : 1, so copper sulfate is limiting; n(Cu) = 0.0400 mol, m = 2.54 g.
3. Zinc left = 0.0100 mol × 65 = 0.65 g. c(ZnSO₄) = 0.0400 ÷ 0.100 = 0.400 mol/dm³.
Problem 18 Spot the errors
1. (a) Cost is irrelevant; carbon is less reactive than aluminium. (b) Use as coins depends on resisting corrosion, which shows copper is less reactive. (c) Sodium reacts with the water first.2. (a) Carbon cannot displace aluminium because aluminium is more reactive than carbon. (b) Copper is less reactive than iron. (c) Sodium reacts with water in the solution.
3. 2Na + 2H₂O → 2NaOH + H₂, and the NaOH then forms a blue precipitate of copper(II) hydroxide, not copper metal.
Problem 19 Aluminium electrolysis
1. 1000 ÷ 27 = 37.0 kmol (37.04).2. n(Al₂O₃) = 37.04 ÷ 2 = 18.52 kmol; M = 102; m = 1890 kg (1889 kg).
3. n(O₂) = 37.04 × 3/4 = 27.78 kmol; m = 889 kg (check: 1000 + 889 = 1889). The oxygen burns the carbon anodes away to form carbon dioxide.
Problem 20 Which metal is M?
1. M is below iron and above copper, and above hydrogen because it reacts with acid.2. M + Cu²⁺ → M²⁺ + Cu.
3. n(Cu) = 6.35 ÷ 63.5 = 0.100 mol = n(M); Ar = 5.90 ÷ 0.100 = 59.0, which is nickel.
4. M is below carbon, so its oxide can be reduced by heating with carbon.
