GCSE AQA Chemistry: Worked Examples Explained | GCSE AQA 化学：典型例题详解

📚 GCSE AQA Chemistry: Worked Examples Explained | GCSE AQA 化学：典型例题详解

This article provides a collection of fully worked typical exam questions for the GCSE AQA Chemistry specification. Each section covers a core topic area, presenting a realistic question followed by a step-by-step, bilingual explanation of the method and key marking points. Use these examples to consolidate your understanding, spot common mistakes, and refine your exam technique. All content is aligned with the AQA 9–1 syllabus, including required practical skills and mathematical demands.

本文精选了 GCSE AQA 化学考纲中的典型例题，并提供了逐步详解。每个章节聚焦一个核心主题，先给出真实考试风格的题目，然后通过中英双语的解读，剖析解题思路和得分要点。读者可以借助这些范例巩固知识、避开常见错误，并提升应试技巧。所有内容均紧扣 AQA 9–1 课程大纲，涵盖必要实验技能和数学要求。

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1. Atomic Structure & Periodic Table | 原子结构与周期表

Question: An atom of element X has 15 protons and 16 neutrons. (a) State the mass number of X. (b) Identify element X using the modern periodic table. (c) Write the electronic configuration of element X. (d) Deduce the group and period of X.

题目： 元素 X 的一个原子含有 15 个质子和 16 个中子。(a) 说出 X 的质量数。(b) 利用现代周期表确定元素 X。(c) 写出元素 X 的电子排布。(d) 推断 X 所在的族和周期。

Worked solution: (a) Mass number = protons + neutrons = 15 + 16 = 31. (b) Atomic number = number of protons = 15, corresponding to phosphorus (P). (c) 15 electrons are arranged as 2,8,5. (d) The period is given by the number of occupied shells (3 shells occupied), so period 3. The group is determined by the number of outer‑shell electrons (5 electrons), so the element is in group 5 (or group 15). Common pitfall: confusing atomic number and mass number; remember the atomic number is the proton count, while mass number includes neutrons.

详解： (a) 质量数 = 质子数 + 中子数 = 15 + 16 = 31。(b) 原子序数 = 质子数 = 15，对应元素是磷（P）。(c) 15 个电子排布为 2, 8, 5。(d) 周期由已占据的电子层数决定（3 层），因此是第 3 周期。族由最外层电子数决定（5 个电子），因此属于第 5 主族（或第 15 族）。常见错误：混淆原子序数和质量数；记住原子序数等于质子数，而质量数是质子数与中子数之和。

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2. Bonding, Structure and Properties | 化学键、结构与性质

Question: Sodium chloride (NaCl) has a high melting point (801 °C), while iodine (I₂) sublimates at room temperature. Explain these differences in terms of structure and bonding. Your answer should refer to the type of particles, the forces overcome, and the energy required.

题目： 氯化钠（NaCl）的熔点很高（801 °C），而碘（I₂）在室温下升华。从结构和键合的角度解释这些差异。你的回答应提及微粒类型、需要克服的作用力以及能量需求。

Worked solution: Sodium chloride is an ionic compound consisting of a giant lattice of oppositely charged Na⁺ and Cl⁻ ions held together by strong electrostatic forces in all directions. Melting requires breaking many of these strong ionic bonds, which demands a large amount of energy, hence a high melting point. Iodine, on the other hand, is a simple molecular substance. It exists as discrete I₂ molecules. The atoms within a molecule are joined by strong covalent bonds, but the intermolecular forces (London dispersion forces) between molecules are weak. Sublimation involves overcoming only these weak intermolecular forces, requiring little energy, so iodine turns directly into a gas at room temperature. Key points to include: ‘giant ionic lattice’ vs. ‘simple molecular’, ‘strong electrostatic forces’ vs. ‘weak intermolecular forces’.

详解： 氯化钠是离子化合物，由带相反电荷的 Na⁺ 和 Cl⁻ 离子通过各方向的强静电引力构成巨型离子晶格。熔化需要破坏大量这些强离子键，这需要很大的能量，因此熔点很高。碘则是简单分子物质，以独立的 I₂ 分子存在。分子内原子由强共价键连接，但分子间的分子间作用力（伦敦色散力）很弱。升华只需克服这些弱分子间作用力，所需能量很少，因此碘在室温下直接变为气体。关键得分点：需要写出“巨型离子晶格”与“简单分子”、“强静电引力”与“弱分子间作用力”。

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3. Quantitative Chemistry: Moles and Mass | 定量化学：摩尔与质量

Question: Magnesium reacts with hydrochloric acid according to the equation: Mg + 2HCl → MgCl₂ + H₂. In an experiment, 2.40 g of magnesium is fully reacted with excess acid. Calculate the mass of magnesium chloride produced. (Aᵣ: Mg = 24, Cl = 35.5)

题目： 镁与盐酸反应的方程式为：Mg + 2HCl → MgCl₂ + H₂。某实验中，2.40 g 镁与过量酸完全反应。计算生成氯化镁的质量。（相对原子质量：Mg = 24, Cl = 35.5）

Worked solution: Step 1: Calculate moles of Mg used. Moles = mass / molar mass = 2.40 g / 24 g mol⁻¹ = 0.100 mol. Step 2: From the balanced equation, 1 mol Mg produces 1 mol MgCl₂. Therefore, moles of MgCl₂ = 0.100 mol. Step 3: Molar mass of MgCl₂ = 24 + (2 × 35.5) = 24 + 71 = 95 g mol⁻¹. Step 4: Mass of MgCl₂ = moles × molar mass = 0.100 × 95 = 9.50 g. Common error: forgetting to multiply the chlorine atomic mass by 2 when calculating molar mass. Also ensure you use the correct mole ratio from the equation.

详解： 第一步：计算所用镁的物质的量。物质的量 = 质量 / 摩尔质量 = 2.40 g / 24 g mol⁻¹ = 0.100 mol。第二步：根据配平方程式，1 mol Mg 生成 1 mol MgCl₂，因此 MgCl₂ 的物质的量 = 0.100 mol。第三步：MgCl₂ 的摩尔质量 = 24 + (2 × 35.5) = 24 + 71 = 95 g mol⁻¹。第四步：MgCl₂ 的质量 = 物质的量 × 摩尔质量 = 0.100 × 95 = 9.50 g。常见错误：计算摩尔质量时忘了将氯的相对原子质量乘以 2。另外需确保使用正确的方程式摩尔比。

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4. Chemical Changes: Electrolysis | 化学变化：电解

Question: Describe what is observed at each electrode during the electrolysis of molten lead(II) bromide (PbBr₂) using inert electrodes. Include the relevant half‑equations and explain why lead is formed at the cathode rather than at the anode.

题目： 描述使用惰性电极电解熔融溴化铅（PbBr₂）时，在每一电极上观察到的现象。写出相关半方程式，并解释为什么在阴极生成铅，而非在阳极。

Worked solution: At the cathode (negative electrode), a grey solid (molten lead, which may appear silvery as it sinks) is produced. Pb²⁺ ions migrate to the cathode, where they gain electrons: Pb²⁺ + 2e⁻ → Pb (reduction). At the anode (positive electrode), a brown/orange gas (bromine vapour) is evolved. Br⁻ ions migrate to the anode, lose electrons: 2Br⁻ → Br₂ + 2e⁻ (oxidation). Lead is produced at the cathode because positive lead ions are attracted to the negative electrode, where reduction occurs. The anode attracts negative bromide ions, which are oxidised. Common misconception: thinking metals form at the anode; remember that positive metal ions always move towards the cathode to gain electrons.

详解： 在阴极（负极）产生灰色固体（熔融铅，可能呈银白色沉入底部）。Pb²⁺ 离子移向阴极，获取电子：Pb²⁺ + 2e⁻ → Pb（还原）。在阳极（正极）产生棕/橙色气体（溴蒸气）。Br⁻ 离子移向阳极，失去电子：2Br⁻ → Br₂ + 2e⁻（氧化）。铅在阴极生成是因为正电荷的铅离子被吸引到负极，发生还原反应。阳极吸引负电荷的溴离子，发生氧化。常见误解：认为金属在阳极生成；记住正电荷的金属离子总是移向阴极以获得电子。

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5. Energy Changes in Reactions | 反应中的能量变化

Question: A student investigates the temperature change when zinc powder reacts with copper(II) sulfate solution. They add 2.5 g of zinc to 50 cm³ of 0.20 mol dm⁻³ CuSO₄ solution and record a temperature rise of 25.5 °C. The specific heat capacity of the solution is 4.18 J g⁻¹ °C⁻¹, and the density of the solution is 1.00 g cm⁻³. Calculate the energy released (in kJ) and the molar enthalpy change for the reaction, assuming zinc is the limiting reactant. (Aᵣ: Zn = 65.4)

题目： 某学生研究了锌粉与硫酸铜溶液反应的温度变化。他们将 2.5 g 锌加入 50 cm³、0.20 mol dm⁻³ 的 CuSO₄ 溶液中，记录温度升高 25.5 °C。溶液的比热容为 4.18 J g⁻¹ °C⁻¹，密度为 1.00 g cm⁻³。假设锌为限制反应物，计算释放的能量（以 kJ 计）和反应的摩尔焓变。（相对原子质量：Zn = 65.4）

Worked solution: Mass of solution = volume × density = 50 cm³ × 1.00 g cm⁻³ = 50 g. Energy released (q) = mass × specific heat capacity × temperature change = 50 g × 4.18 J g⁻¹ °C⁻¹ × 25.5 °C = 5329.5 J ≈ 5.33 kJ. Moles of Zn used = mass / molar mass = 2.5 g / 65.4 g mol⁻¹ ≈ 0.0382 mol. Molar enthalpy change (ΔH) = q / moles = 5.33 kJ / 0.0382 mol ≈ 139.5 kJ mol⁻¹. Since the temperature rises, the reaction is exothermic, so ΔH = −139.5 kJ mol⁻¹ (to 3 s.f.). Always include the negative sign for exothermic reactions. Common error: using the mass of zinc rather than the mass of solution in the energy calculation, or forgetting to convert J to kJ.

详解： 溶液质量 = 体积 × 密度 = 50 cm³ × 1.00 g cm⁻³ = 50 g。释放的能量 (q) = 质量 × 比热容 × 温度变化 = 50 g × 4.18 J g⁻¹ °C⁻¹ × 25.5 °C = 5329.5 J ≈ 5.33 kJ。所用锌的物质的量 = 质量 / 摩尔质量 = 2.5 g / 65.4 g mol⁻¹ ≈ 0.0382 mol。摩尔焓变 (ΔH) = q / 物质的量 = 5.33 kJ / 0.0382 mol ≈ 139.5 kJ mol⁻¹。由于温度升高，反应放热，所以 ΔH = −139.5 kJ mol⁻¹（保留三位有效数字）。放热反应必须加负号。常见错误：在能量计算中使用锌的质量而非溶液质量，或忘记将 J 转换为 kJ。

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6. Rate of Reaction: Factors and Graphs | 反应速率：因素与图像

Question: The graph shows the volume of hydrogen gas produced over time when 0.50 g of magnesium ribbon reacts with 30 cm³ of 1.0 mol dm⁻³ HCl at 25 °C. On the same axes, sketch the curve you would expect if the experiment were repeated using 0.50 g of magnesium powder instead of ribbon, all other conditions unchanged. Explain your sketch.

题目： 下图表示 0.50 g 镁条与 30 cm³ 1.0 mol dm⁻³ HCl 在 25 °C 下反应时，产生氢气体积随时间的变化。在同一坐标轴上，画出若使用 0.50 g 镁粉代替镁条、其他条件不变时预期的曲线。解释你的作图。

Worked solution: The new curve should start at the origin, rise more steeply than the original curve, and level off at the same final volume of hydrogen. The steeper initial gradient indicates a faster rate of reaction. This is because magnesium powder has a much larger surface area than the same mass of ribbon, leading to more frequent successful collisions between reactant particles. However, the total amount of magnesium is unchanged, so the limiting reactant (Mg) is the same, yielding the same final volume of gas. Do not draw a curve that finishes at a higher volume – that would imply more magnesium was used.

详解： 新曲线应从原点出发，比原曲线更陡峭地上升，最终在相同的氢气体积处趋于平缓。更陡的初始斜率表明反应速率更快。这是因为镁粉比相同质量的镁条具有更大的表面积，导致反应物粒子间成功碰撞的频率增加。但镁的总量不变，因此限制反应物（Mg）相同，最终产生的气体体积相同。切勿画出一条终点体积更高的曲线——那样意味着使用了更多的镁。

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7. Reversible Reactions and Equilibrium (Le Chatelier’s Principle) | 可逆反应与平衡（勒夏特列原理）

Question: The reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is exothermic in the forward direction. Predict the effect of increasing temperature on the yield of SO₃, and explain using Le Chatelier’s principle. State what happens to the value of the equilibrium constant.

题目： 反应 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) 的正反应为放热。预测升高温度对 SO₃ 产率的影响，并用勒夏特列原理解释。说明平衡常数的数值如何变化。

Worked solution: Increasing temperature shifts the equilibrium position in the endothermic direction to absorb the added heat. Since the forward reaction is exothermic, the backward reaction is endothermic. The equilibrium therefore shifts to the left, favouring the decomposition of SO₃ back into SO₂ and O₂. Consequently, the yield of SO₃ decreases. The equilibrium constant K decreases because it is defined as [SO₃]² / ([SO₂]²[O₂]), and a shift to the left reduces the numerator while increasing the denominator. For an exothermic reaction, K decreases with increasing temperature. Common mistake: applying ‘increase temperature favours forward reaction’ without checking whether forward is exothermic or endothermic.

详解： 升高温度会使平衡向吸热方向移动，以吸收新增的热量。由于正反应为放热，逆反应便是吸热。因此平衡向左移动，有利于 SO₃ 分解回 SO₂ 和 O₂。结果是 SO₃ 的产率降低。平衡常数 K 的表达式为 [SO₃]² / ([SO₂]²[O₂])，平衡左移使分子减小、分母增大，故 K 减小。对放热反应而言，K 随温度升高而减小。常见错误：未经核实正反应是放热还是吸热，就直接套用“升高温度有利于正反应”的结论。

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8. Organic Chemistry: Alkanes and Alkenes | 有机化学：烷烃与烯烃

Question: Ethene (C₂H₄) is an unsaturated hydrocarbon. Explain what is meant by ‘unsaturated’. Describe a chemical test to distinguish between ethene and ethane, including the expected observations. Write the equation for the test reaction.

题目： 乙烯（C₂H₄）是一种不饱和烃。解释“不饱和”的含义。描述一个区分乙烯和乙烷的化学检验方法，包括预期的现象。写出该检验反应的方程式。

Worked solution: ‘Unsaturated’ means the molecule contains at least one carbon–carbon double bond (C=C). The test uses bromine water (an orange‑brown solution). When shaken with ethene in the dark, the bromine water is decolourised (turns colourless) because ethene undergoes an addition reaction across the double bond. Ethane, being saturated, does not react with bromine water under these conditions, so the solution remains orange‑brown. The equation: C₂H₄ + Br₂ → C₂H₄Br₂ (1,2‑dibromoethane). Important: always mention the colour change – addition of bromine can also produce a colourless organic product; mark schemes often require ‘bromine water turns from orange/brown to colourless’.

详解： “不饱和”表示分子中至少含有一个碳碳双键（C=C）。检验方法使用溴水（橙棕色溶液）。在黑暗中与乙烯一起振荡，溴水会褪色（变为无色），因为乙烯在双键处发生了加成反应。乙烷是饱和烃，在该条件下不与溴水反应，因此溶液保持橙棕色。反应方程式：C₂H₄ + Br₂ → C₂H₄Br₂（1,2‑二溴乙烷）。重要提示：必须描述颜色变化——溴的加成也会生成无色有机产物；评分标准通常要求写出“溴水由橙色/棕色变为无色”。

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9. Chemical Analysis: Testing for Ions | 化学分析：离子检验

Question: A sample of an unknown ionic solid is tested. When a few drops of dilute nitric acid followed by silver nitrate solution are added, a white precipitate forms. When a flame test is carried out, a brick‑red flame is observed. Identify the anion and the cation present, and write the formula of the compound.

题目： 测试一种未知离子固体样品。加入几滴稀硝酸，然后加入硝酸银溶液，生成白色沉淀。进行焰色试验时，观察到砖红色火焰。鉴定存在的阴离子和阳离子，并写出该化合物的化学式。

Worked solution: The white precipitate with acidified silver nitrate indicates the presence of chloride ions (Cl⁻). Acidification with nitric acid removes carbonate or sulfite ions that could interfere. Other halides give different coloured precipitates: bromide produces cream, iodide produces yellow. The brick‑red flame test result is characteristic of calcium ions (Ca²⁺). Therefore the compound is calcium chloride, with formula CaCl₂. Common error: confusing flame test colours – sodium is yellow, potassium is lilac, lithium is crimson, barium is green. For the anion test, always add dilute nitric acid first to rule out carbonate interference.

详解： 加酸化的硝酸银生成白色沉淀，表明存在氯离子（Cl⁻）。用硝酸酸化是为了排除可能干扰的碳酸根或亚硫酸根离子。其他卤化物会生成不同颜色的沉淀：溴化物为奶油色，碘化物为黄色。砖红色焰色是钙离子（Ca²⁺）的特征。因此化合物是氯化钙，化学式为 CaCl₂。常见错误：混淆焰色——钠为黄色，钾为淡紫色，锂为深红色，钡为绿色。阴离子检验时，务必先加稀硝酸以排除碳酸根的干扰。

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10. Using Resources: Life Cycle Assessment (LCA) | 资源利用：生命周期评估

Question: A company is choosing between a paper bag and a plastic (polyethene) bag for packaging. Outline the main stages of a life cycle assessment (LCA) that should be considered. Suggest one advantage and one disadvantage of the plastic bag in terms of environmental impact, based on a typical LCA.

题目： 一家公司要在纸袋和塑料（聚乙烯）袋之间选择包装材料。概述生命周期评估（LCA）应考虑的主要阶段。根据典型的 LCA，从环境影响的角度分别提出塑料袋的一个优点和一个缺点。

Worked solution: The main stages of an LCA are: (1) extracting and processing raw materials; (2) manufacturing and packaging; (3) using the product; (4) disposal (including transport at each stage). An LCA also considers the use of energy, water, and the production of waste and emissions at each stage. Advantage of plastic bag: Plastic bags are lightweight and durable, often reuseable, and require less energy to transport. Their production may emit fewer greenhouse gases compared to paper bags. Disadvantage: Plastic bags are non‑biodegradable, persist in the environment, can cause harm to wildlife, and are made from finite crude oil resources. Paper bags, while biodegradable, have high water and energy consumption in production. A complete LCA would weigh these factors, though the relative importance can differ subjectively.

详解： LCA 的主要阶段包括：(1) 原材料的获取与加工；(2) 制造与包装；(3) 产品的使用；(4) 废弃物处理（含各阶段的运输）。LCA 还会评估每一阶段的能源、水资源消耗以及废物和排放物的产生。塑料袋的优点：重量轻、耐用、常可重复使用，运输能耗较低；其生产过程与纸袋相比温室气体排放可能较少。缺点：塑料袋不可生物降解，会长期存留在环境中，可能危害野生动物，且原料来自有限的石油资源。纸袋虽可生物降解，但生产过程水耗和能耗很高。全面的 LCA 需权衡这些因素，但各因素的相对重要性可能因主观而异。

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