📚 GCSE CCEA Chemistry: Typical Worked Example Questions | GCSE CCEA 化学：典型例题详解
Mastering GCSE CCEA Chemistry requires not only understanding concepts but also practising exam-style questions. This article walks you through typical worked examples covering key topics such as quantitative chemistry, bonding, rates, organic chemistry and chemical analysis. Each question is presented with a clear step-by-step solution, helping you develop problem-solving skills and build confidence for the exam.
掌握 GCSE CCEA 化学不仅需要理解概念，还需要练习考试题型。本文将带你学习涵盖定量化学、化学键、反应速率、有机化学和化学分析等重要主题的典型例题。每道题都配有清晰的分步解答，帮助你培养解题技巧，提升考试信心。
1. Relative Formula Mass and Moles | 相对分子质量与摩尔计算
Question: Aluminium sulfate has the formula Al₂(SO₄)₃. The relative atomic masses are Al = 27, S = 32, O = 16. (a) Calculate the relative formula mass (Mᵣ) of aluminium sulfate. (b) Calculate the mass of 0.2 moles of aluminium sulfate.
题目：硫酸铝的化学式为 Al₂(SO₄)₃。相对原子质量为 Al = 27, S = 32, O = 16。(a) 计算硫酸铝的相对分子质量 (Mᵣ)。(b) 计算 0.2 摩尔硫酸铝的质量。
Solution (a): Identify the number of each type of atom: 2 Al, 3 S and 12 O atoms (since (SO₄)₃ means 3×1 S and 3×4 O). Mᵣ = (2 × 27) + (3 × 32) + (12 × 16) = 54 + 96 + 192 = 342.
解答 (a)：确定每种原子的个数：2 个 Al、3 个 S 和 12 个 O 原子（因为 (SO₄)₃ 表示 3×1 个 S 和 3×4 个 O）。Mᵣ = (2 × 27) + (3 × 32) + (12 × 16) = 54 + 96 + 192 = 342。
Solution (b): Use the formula mass = moles × Mᵣ. Mass = 0.2 mol × 342 g/mol = 68.4 g.
解答 (b)：使用公式 质量 = 摩尔数 × 相对分子质量。质量 = 0.2 mol × 342 g/mol = 68.4 g。
2. Balancing Chemical Equations | 配平化学方程式
Question: Balance the following equation: __Ca + __O₂ → __CaO
题目：配平下列方程式：__Ca + __O₂ → __CaO
Solution: On the right there is 1 Ca and 1 O, while on the left we have 2 oxygen atoms in O₂. Place a ‘2’ before CaO to give 2 oxygen atoms on the right: Ca + O₂ → 2CaO. Now the right has 2 Ca atoms, so put a ‘2’ before Ca on the left: 2Ca + O₂ → 2CaO. The equation is now balanced with 2 Ca and 2 O on each side.
解答：右边有 1 个 Ca 和 1 个 O，左边 O₂ 中有 2 个氧原子。在 CaO 前放上系数 ‘2’，使右边也有 2 个氧原子：Ca + O₂ → 2CaO。此时右边有 2 个 Ca 原子，所以在左边 Ca 前放上 ‘2’：2Ca + O₂ → 2CaO。现在方程式两边各有 2 个 Ca 和 2 个 O，已配平。
3. Reacting Mass Calculations | 反应质量计算
Question: 5.6 g of iron reacts with excess copper(II) sulfate solution according to the equation: Fe + CuSO₄ → FeSO₄ + Cu. (Aᵣ: Fe = 56, Cu = 63.5) What mass of copper is produced?
题目：5.6 g 铁与过量硫酸铜溶液反应，方程式为：Fe + CuSO₄ → FeSO₄ + Cu。(相对原子质量：Fe = 56, Cu = 63.5) 求生成铜的质量。
Solution: Calculate moles of iron: moles Fe = mass / Aᵣ = 5.6 / 56 = 0.10 mol. From the equation, the mole ratio of Fe : Cu is 1 : 1, so moles of Cu produced = 0.10 mol. Mass of Cu = moles × Aᵣ = 0.10 × 63.5 = 6.35 g.
解答：计算铁的摩尔数：摩尔数 = 质量 / 相对原子质量 = 5.6 / 56 = 0.10 mol。由方程式可知，Fe 与 Cu 的计量比为 1 : 1，因此生成的铜的摩尔数为 0.10 mol。铜的质量 = 摩尔数 × 相对原子质量 = 0.10 × 63.5 = 6.35 g。
4. Titration Calculations | 滴定计算
Question: 25.0 cm³ of sodium hydroxide solution is neutralised exactly by 20.0 cm³ of 0.100 mol/dm³ hydrochloric acid. The equation is: NaOH + HCl → NaCl + H₂O. Calculate the concentration of the sodium hydroxide solution in mol/dm³.
题目：25.0 cm³ 氢氧化钠溶液恰好被 20.0 cm³ 0.100 mol/dm³ 盐酸中和。方程式：NaOH + HCl → NaCl + H₂O。计算氢氧化钠溶液的浓度（mol/dm³）。
Solution: First find moles of HCl used: moles = (volume in dm³) × concentration = (20.0 / 1000) × 0.100 = 0.00200 mol. The reaction shows a 1 : 1 mole ratio, so moles of NaOH = 0.00200 mol. Concentration of NaOH = moles / volume in dm³ = 0.00200 / (25.0 / 1000) = 0.0800 mol/dm³.
解答：首先计算所用 HCl 的物质的量：物质的量 = (体积 dm³) × 浓度 = (20.0 / 1000) × 0.100 = 0.00200 mol。反应显示计量比为 1 : 1，因此 NaOH 的物质的量也为 0.00200 mol。NaOH 浓度 = 物质的量 / 体积 dm³ = 0.00200 / (25.0 / 1000) = 0.0800 mol/dm³。
5. Percentage Yield | 百分产率
Question: When 5.00 g of calcium carbonate is heated, it decomposes to calcium oxide and carbon dioxide. The theoretical yield of calcium oxide is 2.80 g. In an experiment, only 2.24 g of calcium oxide is collected. Calculate the percentage yield.
题目：加热 5.00 g 碳酸钙，分解生成氧化钙和二氧化碳。氧化钙的理论产量为 2.80 g。在一次实验中只收集到 2.24 g 氧化钙。计算百分产率。
Solution: Percentage yield = (actual yield / theoretical yield) × 100. Here, actual yield = 2.24 g, theoretical yield = 2.80 g. Percentage yield = (2.24 / 2.80) × 100 = 80.0%.
解答：百分产率 = (实际产量 / 理论产量) × 100。此处实际产量 = 2.24 g，理论产量 = 2.80 g。百分产率 = (2.24 / 2.80) × 100 = 80.0%。
6. Electrolysis Half-Equations | 电解半反应方程式
Question: Write the half-equations for the reactions occurring at the cathode and the anode during the electrolysis of molten lead(II) bromide, PbBr₂. Include state symbols.
题目：写出电解熔融溴化铅 (PbBr₂) 时阴极和阳极发生的半反应方程式，并标出状态符号。
Solution: In molten lead(II) bromide, the ions are Pb²⁺ and Br⁻. At the cathode (reduction): Pb²⁺(l) + 2e⁻ → Pb(l). At the anode (oxidation): 2Br⁻(l) → Br₂(g) + 2e⁻.
解答：在熔融溴化铅中，离子为 Pb²⁺ 和 Br⁻。在阴极（还原反应）：Pb²⁺(l) + 2e⁻ → Pb(l)。在阳极（氧化反应）：2Br⁻(l) → Br₂(g) + 2e⁻。
7. Energy Changes: Exothermic and Endothermic | 能量变化：放热与吸热
Question: A student mixes 50 cm³ of 1.0 mol/dm³ hydrochloric acid with 50 cm³ of 1.0 mol/dm³ sodium hydroxide solution. The temperature rises from 21.0 °C to 27.5 °C. The density of the solution is 1.0 g/cm³ and the specific heat capacity is 4.2 J/g/°C. (a) Calculate the heat energy released. (b) State whether the reaction is exothermic or endothermic.
题目：某学生将 50 cm³ 1.0 mol/dm³ 盐酸与 50 cm³ 1.0 mol/dm³ 氢氧化钠溶液混合。温度从 21.0 °C 上升到 27.5 °C。溶液密度为 1.0 g/cm³，比热容为 4.2 J/g/°C。(a) 计算放出的热量。(b) 指出该反应是放热还是吸热反应。
Solution (a): Total volume = 50 + 50 = 100 cm³, so mass of solution = 100 g (since density is 1.0 g/cm³). Temperature change ΔT = 27.5 – 21.0 = 6.5 °C. Heat energy released, Q = mcΔT = 100 g × 4.2 J/g/°C × 6.5 °C = 2730 J.
解答 (a)：总体积 = 50 + 50 = 100 cm³，因此溶液质量 = 100 g（密度为 1.0 g/cm³）。温度变化 ΔT = 27.5 – 21.0 = 6.5 °C。放出的热量 Q = mcΔT = 100 g × 4.2 J/g/°C × 6.5 °C = 2730 J。
Solution (b): Since the temperature increases, heat is released to the surroundings. Therefore, the reaction is exothermic.
解答 (b)：由于温度升高，热量释放到周围环境中，因此该反应为放热反应。
8. Organic Chemistry: Alkenes and Addition Reactions | 有机化学：烯烃与加成反应
Question: Ethene, C₂H₄, is bubbled through orange-brown bromine water. (a) Name the product formed and write its structural formula. (b) Describe the colour change and explain why it occurs.
题目：将乙烯 (C₂H₄) 通入橙棕色的溴水中。(a) 写出生成物的名称和结构式。(b) 描述颜色变化并解释原因。
Solution (a): Ethene undergoes an addition reaction with bromine. The product is 1,2-dibromoethane, with the formula C₂H₄Br₂. Its displayed formula can be drawn as H–CBr–CBr–H with two hydrogen atoms on each carbon, or written as CH₂BrCH₂Br.
解答 (a)：乙烯与溴发生加成反应。产物是 1,2-二溴乙烷，化学式 C₂H₄Br₂。其结构式可表示为每个碳原子上连有两个氢原子和两个溴原子互相连接，即 CH₂BrCH₂Br。
Solution (b): The orange-brown colour of bromine water disappears, turning colourless. This is because bromine molecules add across the carbon‑carbon double bond, forming a saturated compound. The consumption of Br₂ removes the colour.
解答 (b)：溴水的橙棕色消失，变为无色。这是因为溴分子加成到碳碳双键上，形成饱和化合物。Br₂ 被消耗，颜色褪去。
9. Chemical Tests for Ions | 离子的化学检验
Question: Describe how you could distinguish between a solution containing chloride ions (Cl⁻) and a solution containing sulfate ions (SO₄²⁻). Include reagents and expected observations.
题目：描述如何区分含有氯离子 (Cl⁻) 的溶液和含有硫酸根离子 (SO₄²⁻) 的溶液，需写出所用试剂和预期现象。
Solution: For chloride ions: add a few drops of dilute nitric acid followed by silver nitrate solution. A white precipitate of silver chloride (AgCl) forms. For sulfate ions: add a few drops of dilute hydrochloric acid followed by barium chloride solution. A white precipitate of barium sulfate (BaSO₄) forms. Both precipitates are white, but the tests use different reagents and reactions.
解答：检验氯离子：加入几滴稀硝酸，再滴加硝酸银溶液。产生氯化银 (AgCl) 白色沉淀。检验硫酸根离子：加入几滴稀盐酸，再滴加氯化钡溶液。产生硫酸钡 (BaSO₄) 白色沉淀。两种沉淀均为白色，但通过不同的试剂和反应可以区分。
10. Water Hardness and Soap | 水的硬度与肥皂
Question: Hard water contains dissolved calcium ions, Ca²⁺. Explain why hard water requires more soap to form a lather and write a balanced ionic equation to illustrate the reaction between calcium ions and soap ions (represented as RCOO⁻).
题目：硬水中含有溶解的钙离子 (Ca²⁺)。解释为什么硬水需要更多肥皂才能产生泡沫，并写出钙离子与肥皂离子（表示为 RCOO⁻）反应的离子方程式。
Solution: Soap contains stearate or similar ions (RCOO⁻) that help form lather. In hard water, Ca²⁺ ions react with RCOO⁻ to form an insoluble precipitate called scum (calcium stearate). This consumes soap before it can produce lather, so more soap is needed. The ionic equation is: Ca²⁺(aq) + 2RCOO⁻(aq) → (RCOO)₂Ca(s).
解答：肥皂含有硬脂酸根或类似离子 (RCOO⁻)，可产生泡沫。在硬水中，Ca²⁺ 离子与 RCOO⁻ 反应，生成不溶性沉淀（钙皂垢）。这消耗了肥皂，使其无法产生泡沫，因此需要更多的肥皂。离子方程式为：Ca²⁺(aq) + 2RCOO⁻(aq) → (RCOO)₂Ca(s)。
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