📚 GCSE CIE Chemistry: Worked Examples Explained Step by Step | GCSE CIE 化学：典型例题详解
Mastering GCSE CIE Chemistry requires more than memorising facts; it demands the ability to apply concepts to unfamiliar problems. In this article, we walk through a carefully selected set of exam-style worked examples covering core topics: stoichiometry, electrolysis, energetics, rates and equilibrium, acid–base chemistry, and organic reactions. Each question is followed by a clear, stepwise solution that highlights exactly how marks are awarded, common pitfalls, and the reasoning you need to demonstrate. Use these examples to sharpen your problem-solving skills and build the confidence to tackle any question the exam board can present.
掌握 GCSE CIE 化学不仅需要记忆知识点，更要求将概念应用于陌生问题中。本文精选了一组贴近考试的典型例题，涵盖化学计量、电解、能量变化、速率与平衡、酸碱化学和有机反应等核心主题。每道题后都配有清晰的逐步解析，突出评分要点、常见错误以及你需要展示的推理过程。利用这些例题打磨解题技巧，积累应对任何考题的信心。
1. Moles and Mass Calculation | 摩尔与质量计算
A student heats 5.00 g of calcium carbonate, CaCO₃, until it decomposes completely according to the equation: CaCO₃(s) → CaO(s) + CO₂(g). Calculate the mass of calcium oxide produced. (Aᵣ values: Ca = 40, C = 12, O = 16)
某学生将 5.00 g 碳酸钙 CaCO₃ 加热至完全分解，反应方程式为：CaCO₃(s) → CaO(s) + CO₂(g)。计算生成的氧化钙的质量。（相对原子质量：Ca = 40，C = 12，O = 16）
Solution: First find the molar mass of CaCO₃: 40 + 12 + (3 × 16) = 100 g/mol. Moles of CaCO₃ = mass / molar mass = 5.00 / 100 = 0.0500 mol. From the equation, 1 mol CaCO₃ produces 1 mol CaO, so moles of CaO = 0.0500 mol. Molar mass of CaO = 40 + 16 = 56 g/mol. Mass of CaO = moles × molar mass = 0.0500 × 56 = 2.80 g.
解析：首先求 CaCO₃ 的摩尔质量：40 + 12 + (3 × 16) = 100 g/mol。CaCO₃ 物质的量 = 质量 / 摩尔质量 = 5.00 / 100 = 0.0500 mol。由方程式可知，1 mol CaCO₃ 生成 1 mol CaO，因此 CaO 物质的量 = 0.0500 mol。CaO 的摩尔质量 = 40 + 16 = 56 g/mol。CaO 质量 = 物质的量 × 摩尔质量 = 0.0500 × 56 = 2.80 g。
2. Reacting Masses with Limiting Reactant | 限量反应物质量计算
Magnesium reacts with oxygen to form magnesium oxide: 2Mg(s) + O₂(g) → 2MgO(s). A mixture of 4.86 g of magnesium and 3.20 g of oxygen is ignited. Identify the limiting reactant and calculate the mass of magnesium oxide formed. (Mg = 24; O = 16)
镁与氧气反应生成氧化镁：2Mg(s) + O₂(g) → 2MgO(s)。将 4.86 g 镁和 3.20 g 氧气混合并点燃。确定限量反应物并计算生成氧化镁的质量。（Mg = 24；O = 16）
Solution: Moles of Mg = 4.86 / 24 = 0.2025 mol. Moles of O₂ = 3.20 / (2 × 16) = 3.20 / 32 = 0.100 mol. According to the equation, 2 mol Mg react with 1 mol O₂. Required moles of O₂ for 0.2025 mol Mg = 0.2025 / 2 = 0.10125 mol. Available O₂ is 0.100 mol, which is slightly less, so O₂ is the limiting reactant. Using O₂, moles of MgO produced = 2 × moles of O₂ = 2 × 0.100 = 0.200 mol. Molar mass of MgO = 24 + 16 = 40 g/mol. Mass of MgO = 0.200 × 40 = 8.00 g.
解析：Mg 物质的量 = 4.86 / 24 = 0.2025 mol。O₂ 物质的量 = 3.20 / (2 × 16) = 3.20 / 32 = 0.100 mol。由方程式，2 mol Mg 与 1 mol O₂ 反应。0.2025 mol Mg 所需 O₂ 的量 = 0.2025 / 2 = 0.10125 mol。实际 O₂ 仅为 0.100 mol，稍少，因此 O₂ 是限量反应物。以 O₂ 为基准，生成的 MgO 物质的量 = 2 × O₂ 的物质的量 = 2 × 0.100 = 0.200 mol。MgO 摩尔质量 = 24 + 16 = 40 g/mol。MgO 质量 = 0.200 × 40 = 8.00 g。
3. Percentage Yield | 产率计算
In an experiment, 2.80 g of iron is heated with excess sulfur and 3.81 g of iron(II) sulfide, FeS, is obtained. The equation is Fe(s) + S(s) → FeS(s). Calculate the percentage yield. (Fe = 56, S = 32)
实验中，将 2.80 g 铁与过量的硫加热，得到 3.81 g 硫化亚铁 FeS。反应方程式为 Fe(s) + S(s) → FeS(s)。计算产率。（Fe = 56，S = 32）
Solution: Moles of Fe used = 2.80 / 56 = 0.0500 mol. Theoretical moles of FeS = 0.0500 mol (1:1 ratio). Molar mass of FeS = 56 + 32 = 88 g/mol. Theoretical mass of FeS = 0.0500 × 88 = 4.40 g. Actual mass obtained = 3.81 g. Percentage yield = (actual / theoretical) × 100 = (3.81 / 4.40) × 100 = 86.6%.
解析：所用 Fe 的物质的量 = 2.80 / 56 = 0.0500 mol。根据 1:1 化学计量比，理论生成 FeS 物质的量 = 0.0500 mol。FeS 摩尔质量 = 56 + 32 = 88 g/mol。理论 FeS 质量 = 0.0500 × 88 = 4.40 g。实际得到 3.81 g。产率 = (实际 / 理论) × 100 = (3.81 / 4.40) × 100 = 86.6%。
4. Gas Volume Calculation at RTP | 常温常压下气体体积计算
Calcium carbonate reacts with excess hydrochloric acid: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g). Calculate the volume of carbon dioxide produced at room temperature and pressure (RTP) when 2.50 g of CaCO₃ reacts completely. (Molar volume at RTP = 24 dm³/mol; CaCO₃ = 100)
碳酸钙与过量盐酸反应：CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)。计算 2.50 g CaCO₃ 完全反应后在常温常压 (RTP) 下产生的二氧化碳体积。（RTP 下气体摩尔体积 = 24 dm³/mol；CaCO₃ = 100）
Solution: Moles of CaCO₃ = 2.50 / 100 = 0.0250 mol. 1 mol CaCO₃ produces 1 mol CO₂, so moles of CO₂ = 0.0250 mol. Volume of CO₂ at RTP = moles × 24 = 0.0250 × 24 = 0.60 dm³ (or 600 cm³).
解析：CaCO₃ 物质的量 = 2.50 / 100 = 0.0250 mol。1 mol CaCO₃ 生成 1 mol CO₂，因此 CO₂ 物质的量 = 0.0250 mol。RTP 下 CO₂ 体积 = 物质的量 × 24 = 0.0250 × 24 = 0.60 dm³（或 600 cm³）。
5. Concentration and Titration | 浓度与滴定计算
25.0 cm³ of sodium hydroxide solution is neutralised by 23.5 cm³ of 0.100 mol/dm³ sulfuric acid: 2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l). Calculate the concentration of the sodium hydroxide solution in mol/dm³.
25.0 cm³ 氢氧化钠溶液被 23.5 cm³ 0.100 mol/dm³ 的硫酸中和：2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)。计算氢氧化钠溶液的浓度，单位为 mol/dm³。
Solution: Moles of H₂SO₄ = concentration × volume in dm³ = 0.100 × (23.5 / 1000) = 0.00235 mol. From the equation, 1 mol H₂SO₄ reacts with 2 mol NaOH, so moles of NaOH = 2 × 0.00235 = 0.00470 mol. Volume of NaOH = 25.0 / 1000 = 0.0250 dm³. Concentration of NaOH = moles / volume = 0.00470 / 0.0250 = 0.188 mol/dm³.
解析：H₂SO₄ 物质的量 = 浓度 × 体积 (dm³) = 0.100 × (23.5 / 1000) = 0.00235 mol。由方程式，1 mol H₂SO₄ 与 2 mol NaOH 反应，因此 NaOH 物质的量 = 2 × 0.00235 = 0.00470 mol。NaOH 溶液的体积 = 25.0 / 1000 = 0.0250 dm³。NaOH 浓度 = 物质的量 / 体积 = 0.00470 / 0.0250 = 0.188 mol/dm³。
6. Empirical Formula from Combustion Data | 由燃烧数据求实验式
A hydrocarbon contains 85.7% carbon by mass. Its molar mass is found to be 56 g/mol. Determine the empirical formula and the molecular formula. (C = 12, H = 1)
某碳氢化合物含碳 85.7%（质量分数），其摩尔质量为 56 g/mol。求其实验式和分子式。（C = 12，H = 1）
Solution: Assume 100 g of compound: mass of C = 85.7 g, mass of H = 100 – 85.7 = 14.3 g. Moles C = 85.7 / 12 = 7.14 mol, moles H = 14.3 / 1 = 14.3 mol. Divide by smallest (7.14): C = 1, H = 2.006 ≈ 2. Empirical formula = CH₂. Empirical formula mass = 12 + (2 × 1) = 14. Molecular formula multiplier = molar mass / empirical mass = 56 / 14 = 4. Molecular formula = C₄H₈.
解析：假设化合物为 100 g：C 质量 = 85.7 g，H 质量 = 100 – 85.7 = 14.3 g。C 物质的量 = 85.7 / 12 = 7.14 mol，H 物质的量 = 14.3 / 1 = 14.3 mol。除以最小值 (7.14)：C = 1，H = 2.006 ≈ 2。实验式为 CH₂。实验式质量 = 12 + (2 × 1) = 14。分子式倍率 = 摩尔质量 / 实验式质量 = 56 / 14 = 4。分子式为 C₄H₈。
7. Energy Change Calculation | 能量变化计算
When 0.50 g of ethanol (C₂H₅OH) is burned completely, the heat released raises the temperature of 200 cm³ of water by 13.2 °C. Given that the specific heat capacity of water is 4.2 J/g°C and the density of water is 1 g/cm³, calculate the enthalpy of combustion of ethanol in kJ/mol. (C = 12, H = 1, O = 16)
当 0.50 g 乙醇 (C₂H₅OH) 完全燃烧时，释放的热量使 200 cm³ 水的温度升高 13.2 °C。已知水的比热容为 4.2 J/g°C，密度为 1 g/cm³，计算乙醇的燃烧焓，单位 kJ/mol。（C = 12，H = 1，O = 16）
Solution: Mass of water = 200 cm³ × 1 g/cm³ = 200 g. Heat absorbed by water, q = m × c × ΔT = 200 × 4.2 × 13.2 = 11088 J = 11.088 kJ. This heat came from 0.50 g ethanol. Molar mass of C₂H₅OH = (2 × 12) + (6 × 1) + 16 = 46 g/mol. Moles of ethanol = 0.50 / 46 = 0.01087 mol. ΔH = –q / moles = –11.088 / 0.01087 ≈ –1020 kJ/mol (negative because exothermic). The experimental value is usually less negative due to heat loss; reporting as –1020 kJ/mol (to 3 significant figures: –1.02 × 10³ kJ/mol) is acceptable.
解析：水的质量 = 200 cm³ × 1 g/cm³ = 200 g。水吸收的热量 q = m × c × ΔT = 200 × 4.2 × 13.2 = 11088 J = 11.088 kJ。此热量来自 0.50 g 乙醇。乙醇摩尔质量 = (2 × 12) + (6 × 1) + 16 = 46 g/mol。乙醇物质的量 = 0.50 / 46 = 0.01087 mol。ΔH = –q / 物质的量 = –11.088 / 0.01087 ≈ –1020 kJ/mol（负值表示放热）。因热损失实验值通常负得少；以 –1020 kJ/mol（3 位有效数字：–1.02 × 10³ kJ/mol）报告即可。
8. Electrolysis of Aqueous Sodium Chloride | 氯化钠水溶液的电解
Concentrated aqueous sodium chloride is electrolysed using inert electrodes. Describe the products formed at each electrode and write the relevant half-equations. Explain why the products are different from those obtained from molten NaCl.
以惰性电极电解浓氯化钠水溶液。描述各电极生成的产物，并写出相关半方程式。解释为什么产物与电解熔融 NaCl 时不同。
Solution: At the cathode (negative electrode): hydrogen gas, H₂, is produced because water is reduced in preference to Na⁺ ions in aqueous solution. Half-equation: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq). At the anode (positive electrode): chlorine gas, Cl₂, is produced from oxidation of Cl⁻ ions. Half-equation: 2Cl⁻(aq) → Cl₂(g) + 2e⁻. The solution around the cathode becomes alkaline due to OH⁻ ions. In molten NaCl, only Na⁺ and Cl⁻ ions are present, so sodium metal forms at the cathode and chlorine at the anode. In aqueous solution, water molecules also ionise, introducing H⁺ and OH⁻ ions, and since H⁺ is easier to reduce than Na⁺, water reduction occurs at the cathode.
解析：在阴极（负极）：生成氢气 H₂，因为在溶液中水优先于 Na⁺ 离子被还原。半方程式：2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq)。在阳极（正极）：Cl⁻ 离子被氧化生成氯气 Cl₂。半方程式：2Cl⁻(aq) → Cl₂(g) + 2e⁻。阴极附近溶液因 OH⁻ 离子而呈碱性。在熔融 NaCl 中，仅有 Na⁺ 和 Cl⁻ 离子，因此阴极生成金属钠，阳极生成氯气。在水溶液中，水分子也会电离产生 H⁺ 和 OH⁻ 离子，而 H⁺ 比 Na⁺ 更易被还原，因此阴极发生水的还原。
9. Reversible Reactions and Equilibrium Position | 可逆反应与平衡位置
Consider the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ/mol. Predict the effect of increasing pressure and increasing temperature on the yield of ammonia. Explain your answers in terms of Le Chatelier’s principle.
考虑反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ/mol。预测增大压力和升高温度对氨产率的影响。用勒夏特列原理解释你的答案。
Solution: Increasing pressure: The forward reaction produces fewer gas molecules (4 mol on left → 2 mol on right), so an increase in pressure shifts the equilibrium to the right, increasing the yield of ammonia. Increasing temperature: Since the forward reaction is exothermic (ΔH negative), heat is a product. An increase in temperature adds heat, shifting the equilibrium to the left (endothermic direction) to absorb the added heat, thus decreasing the yield of ammonia.
解析：增大压力：正向反应生成较少的气体分子（左侧 4 mol → 右侧 2 mol），因此增大压力使平衡向右移动，提高氨的产率。升高温度：正向反应是放热的（ΔH 为负），热量相当于产物。升高温度增加了热量，平衡向左（吸热方向）移动以吸收增加的热量，从而降低氨的产率。
10. Rate of Reaction – Interpreting Graphs | 反应速率 – 图像解读
A student measures the volume of hydrogen gas produced every 30 seconds when magnesium ribbon is added to excess dilute hydrochloric acid. Explain why the rate of reaction decreases over time, even though the acid is in excess. Sketch the volume–time graph you would expect.
某学生将镁条加入过量稀盐酸中，每隔 30 秒测量产生氢气的体积。解释为什么即使酸是过量的，反应速率仍随时间减慢。并勾勒预期的体积–时间曲线。
Solution: The rate decreases because the concentration of H⁺ ions falls as the reaction proceeds. Even though the acid is in excess, the concentration of the reacting particles drops, leading to a lower frequency of successful collisions between Mg and H⁺ ions. Additionally, the surface area of the magnesium ribbon decreases as it is consumed. The expected graph is a curve that rises steeply at first, then levels off to a plateau when the reaction completes. The gradient (steepness) decreases with time, reflecting the decreasing rate.
解析：速率减慢是因为随着反应进行，H⁺ 离子浓度下降。即使酸过量，反应物粒子浓度降低，导致镁与 H⁺ 离子之间的成功碰撞频率下降。此外，镁条在消耗过程中表面积也会减小。预期图像为一条起初陡峭上升、随后趋于平缓的曲线，当反应完成时达到平台。曲线的斜率（陡度）随时间减小，反映出速率下降。
11. Organic Chemistry – Cracking and Alkenes | 有机化学 – 裂化与烯烃
Decane, C₁₀H₂₂, is cracked to produce ethene and one other hydrocarbon. Write a balanced equation for the reaction. Explain how you would use bromine water to distinguish between decane and the products.
癸烷 C₁₀H₂₂ 裂化生成乙烯和另一种烃。写出反应的化学方程式。说明如何用溴水区分癸烷和产物。
Solution: Cracking decane can produce ethene (C₂H₄) and octane: C₁₀H₂₂ → C₂H₄ + C₈H₁₈. The equation is already balanced. To distinguish, add orange bromine water to samples. Decane, being an alkane, does not react and the bromine water remains orange. The product mixture contains an alkene (ethene), which decolourises bromine water immediately by addition reaction, turning it colourless. Similarly, if the other product is an alkane, it will not decolourise bromine water.
解析：癸烷裂化可生成乙烯 (C₂H₄) 和辛烷：C₁₀H₂₂ → C₂H₄ + C₈H₁₈。方程式已配平。区分方法：向样品中加入橙色的溴水。癸烷作为烷烃不反应，溴水保持橙色。产物混合物中含有烯烃（乙烯），会通过加成反应立即使溴水褪色，变为无色。同样，若另一产物是烷烃，则不会使溴水褪色。
12. Building a Logical Approach to CIE Exam Questions | 建立应对 CIE 考试题的逻辑方法
Across all the examples above, a consistent pattern emerges: read the question carefully to identify what is given and what is asked, write the relevant chemical equation, convert all quantities to moles if necessary, and use the mole ratio to link substances. For non-calculation questions, apply fundamental principles (collision theory, Le Chatelier, reactivity series, bonding). Always show your working step by step, and state units. With practice, this structured approach becomes second nature and helps you score full marks on the most challenging questions.
从以上所有例题中可以总结出一个一致的模式：仔细读题以明确已知量和所问内容，写出相关化学方程式，必要时将所有量转换为物质的量，并利用物质的量比建立物质间的联系。对于非计算题，则运用基本原理（碰撞理论、勒夏特列原理、活动性顺序、化学键等）。始终逐步展示推理过程，并注明单位。通过练习，这种结构化的方法会变成你的第二天性，帮助你在最困难的题目上获得满分。
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