IGCSE AQA Chemistry: Detailed Walkthrough of Typical Exam Questions | IGCSE AQA 化学：典型例题详解

📚 IGCSE AQA Chemistry: Detailed Walkthrough of Typical Exam Questions | IGCSE AQA 化学：典型例题详解

Mastering IGCSE AQA Chemistry requires not only factual knowledge but also the ability to apply concepts to unfamiliar problems. This article breaks down ten classic question types from topics across the specification, providing step-by-step solutions and bilingual commentary to help you understand the examiner’s expectations.

掌握 IGCSE AQA 化学不仅需要记忆知识点，更要能将概念应用到陌生的题目中。本文精选了考纲中十个经典题型，提供逐步解答和中英双语讲解，帮助你理解出题人的考察意图。

1. Mole Calculation | 摩尔计算

Question: Calculate the mass of carbon dioxide produced when 24 g of carbon is completely burnt in excess oxygen. (Aᵣ: C = 12, O = 16)

题目：24 g 碳在过量的氧气中完全燃烧时，生成的二氧化碳的质量是多少？（相对原子质量：C = 12，O = 16）

Step 1: Write the balanced chemical equation. Carbon reacts with oxygen to form carbon dioxide: C + O₂ → CO₂.

步骤 1：写出配平的化学方程式。碳与氧气反应生成二氧化碳：C + O₂ → CO₂。

Step 2: Calculate the number of moles of carbon. moles = mass ÷ Aᵣ = 24 ÷ 12 = 2.0 mol.

步骤 2：计算碳的物质的量。物质的量 = 质量 ÷ 相对原子质量 = 24 ÷ 12 = 2.0 mol。

Step 3: Use the mole ratio from the equation. 1 mol of C produces 1 mol of CO₂, so 2 mol of C will produce 2 mol of CO₂.

步骤 3：利用方程式中的物质的量之比。1 mol C 生成 1 mol CO₂，因此 2 mol C 将生成 2 mol CO₂。

Step 4: Calculate the mass of CO₂. Mᵣ of CO₂ = 12 + (2 × 16) = 44. mass = moles × Mᵣ = 2.0 × 44 = 88 g.

步骤 4：计算 CO₂ 的质量。CO₂ 的相对分子质量 = 12 + (2 × 16) = 44。质量 = 物质的量 × 相对分子质量 = 2.0 × 44 = 88 g。

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2. Concentration Calculation | 浓度计算

Question: 25.0 cm³ of sodium hydroxide solution is neutralised by 20.0 cm³ of 0.100 mol/dm³ hydrochloric acid. Calculate the concentration of the sodium hydroxide solution in mol/dm³. Equation: NaOH + HCl → NaCl + H₂O.

题目：25.0 cm³ 氢氧化钠溶液被 20.0 cm³ 浓度为 0.100 mol/dm³ 的盐酸中和。计算氢氧化钠溶液的浓度（mol/dm³）。方程式：NaOH + HCl → NaCl + H₂O。

Step 1: Convert volumes to dm³. Volume of HCl = 20.0 ÷ 1000 = 0.0200 dm³. Volume of NaOH = 25.0 ÷ 1000 = 0.0250 dm³.

步骤 1：将体积换算成 dm³。HCl 体积 = 20.0 ÷ 1000 = 0.0200 dm³。NaOH 体积 = 25.0 ÷ 1000 = 0.0250 dm³。

Step 2: Calculate moles of HCl used. moles = concentration × volume = 0.100 × 0.0200 = 0.00200 mol.

步骤 2：计算所用 HCl 的物质的量。物质的量 = 浓度 × 体积 = 0.100 × 0.0200 = 0.00200 mol。

Step 3: Use the mole ratio. The equation shows 1 mol of HCl reacts with 1 mol of NaOH, so moles of NaOH = 0.00200 mol.

步骤 3：利用物质的量之比。方程式显示 1 mol HCl 与 1 mol NaOH 反应，因此 NaOH 的物质的量 = 0.00200 mol。

Step 4: Calculate the concentration of NaOH. concentration = moles ÷ volume = 0.00200 ÷ 0.0250 = 0.0800 mol/dm³.

步骤 4：计算 NaOH 的浓度。浓度 = 物质的量 ÷ 体积 = 0.00200 ÷ 0.0250 = 0.0800 mol/dm³。

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3. Electrolysis Half‑Equations | 电解半反应方程式

Question: Write the half‑equations for the reactions occurring at the cathode and anode during the electrolysis of molten lead(II) bromide (PbBr₂).

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。

Step 1: Identify the ions present in the electrolyte. Molten PbBr₂ contains Pb²⁺ cations and Br⁻ anions.

步骤 1：确定电解质中存在的离子。熔融 PbBr₂ 含有 Pb²⁺ 阳离子和 Br⁻ 阴离子。

Step 2: At the cathode (negative electrode), reduction occurs. Pb²⁺ ions gain electrons: Pb²⁺ + 2e⁻ → Pb (lead metal).

步骤 2：在阴极（负极），发生还原反应。Pb²⁺ 离子得到电子：Pb²⁺ + 2e⁻ → Pb（铅金属）。

Step 3: At the anode (positive electrode), oxidation occurs. Br⁻ ions lose electrons: 2Br⁻ → Br₂ + 2e⁻ (bromine gas).

步骤 3：在阳极（正极），发生氧化反应。Br⁻ 离子失去电子：2Br⁻ → Br₂ + 2e⁻（溴气）。

Step 4: Check the balance of atoms and charge. Both half‑equations are balanced for atoms and electrical charge.

步骤 4：检查原子和电荷的配平。两个半反应方程式的原子和电荷均已配平。

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4. Enthalpy Change Calculation | 焓变计算

Question: When 0.50 g of magnesium is added to excess hydrochloric acid, the temperature of 50 cm³ of solution rises from 21.0 °C to 36.5 °C. The specific heat capacity of the solution is 4.2 J/g°C, and its density is 1.0 g/cm³. Calculate the enthalpy change (∆H) per mole of magnesium. (Aᵣ Mg = 24)

题目：将 0.50 g 镁加入过量盐酸中，50 cm³ 溶液的温度从 21.0 °C 升至 36.5 °C。溶液的比热容为 4.2 J/g°C，密度为 1.0 g/cm³。计算每摩尔镁的焓变（∆H）。（相对原子质量 Mg = 24）

Step 1: Calculate the heat energy released (q). Mass of solution = volume × density = 50 cm³ × 1.0 g/cm³ = 50 g. Temperature change ∆T = 36.5 − 21.0 = 15.5 °C. q = m × c × ∆T = 50 × 4.2 × 15.5 = 3255 J ÷ 1000 = 3.255 kJ.

步骤 1：计算释放的热量（q）。溶液质量 = 体积 × 密度 = 50 g。温度变化 ∆T = 36.5 − 21.0 = 15.5 °C。q = 50 × 4.2 × 15.5 = 3255 J ÷ 1000 = 3.255 kJ。

Step 2: Determine the moles of magnesium used. moles = mass ÷ Aᵣ = 0.50 ÷ 24 = 0.02083 mol.

步骤 2：确定镁的物质的量。物质的量 = 质量 ÷ 相对原子质量 = 0.50 ÷ 24 = 0.02083 mol。

Step 3: Calculate ∆H per mole. Since the temperature rose, the reaction is exothermic, so ∆H will be negative. ∆H = −q ÷ moles = −3.255 ÷ 0.02083 ≈ −156 kJ/mol (to 3 significant figures).

步骤 3：计算每摩尔的 ∆H。温度升高表明反应放热，因此 ∆H 为负值。∆H = −q ÷ 物质的量 = −3.255 ÷ 0.02083 ≈ −156 kJ/mol（保留三位有效数字）。

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5. Rate of Reaction Graphs | 反应速率图像

Question: Explain why the gradient of a volume-of-gas vs time graph decreases during a reaction between marble chips and hydrochloric acid.

题目：解释在大理石碎片与盐酸的反应中，气体体积-时间图像的斜率为什么会逐渐减小。

Step 1: Identify what the gradient represents. In a volume‑vs‑time graph, the gradient (steepness) indicates the rate of reaction at that moment.

步骤 1：确定斜率代表的意义。在气体体积-时间图像中，斜率（陡峭程度）表示该时刻的反应速率。

Step 2: State that the reactant particles are used up over time. As the reaction proceeds, the concentration of hydrochloric acid decreases and the surface area of marble chips reduces because they are being consumed.

步骤 2：说明反应物粒子随时间被消耗。随着反应进行，盐酸浓度降低，大理石碎片因被消耗而表面积减小。

Step 3: Link concentration to collision frequency. With fewer acid particles per unit volume, the frequency of successful collisions decreases. A smaller surface area also reduces the number of collisions per second.

步骤 3：将浓度与碰撞频率联系起来。单位体积内酸粒子减少，有效碰撞频率降低。表面积减小也导致每秒碰撞次数减少。

Step 4: Conclude that the rate slows down, making the graph less steep over time until it becomes horizontal when the limiting reactant is used up.

步骤 4：得出结论：反应速率减慢，图像斜率随时间变小，直至限制性反应物耗尽时曲线变平。

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6. Bonding and Structure Explanations | 化学键与结构解释

Question: Explain why sodium chloride (NaCl) has a high melting point but hydrogen chloride (HCl) has a low melting point.

题目：解释为什么氯化钠（NaCl）的熔点高，而氯化氢（HCl）的熔点低。

Step 1: Identify the types of structure and bonding. NaCl is an ionic compound with a giant ionic lattice. HCl is a simple molecular (covalent) compound.

步骤 1：确定结构和键合类型。NaCl 是离子化合物，具有巨型离子晶格结构。HCl 是简单分子（共价）化合物。

Step 2: Describe the forces in NaCl. In solid NaCl, strong electrostatic forces of attraction hold oppositely charged Na⁺ and Cl⁻ ions together throughout the lattice. A large amount of heat energy is required to overcome these strong ionic bonds.

步骤 2：描述 NaCl 中的作用力。在固态 NaCl 中，强大的静电引力将整个晶格中带相反电荷的 Na⁺ 和 Cl⁻ 离子结合在一起。需要大量热量才能克服这些强的离子键。

Step 3: Describe the forces in HCl. HCl molecules are held together by weak intermolecular forces (van der Waals’ forces). Only a small amount of energy is needed to overcome these weak forces between molecules. The covalent bonds within the molecules are not broken during melting.

步骤 3：描述 HCl 中的作用力。HCl 分子之间仅靠弱的分子间作用力（范德华力）聚集。只需少量能量即可克服分子间的微弱作用力。熔化时分子内的共价键并未被破坏。

Step 4: Contrast the two. Therefore, NaCl has a high melting point because of strong ionic bonding throughout the structure, whereas HCl has a low melting point due to weak intermolecular forces.

步骤 4：进行对比。因此，NaCl 因整个结构存在强离子键而熔点高，而 HCl 因分子间作用力微弱而熔点低。

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7. Percentage Yield | 产率计算

Question: 4.0 g of calcium carbonate (CaCO₃) is decomposed by heating, producing 1.4 g of calcium oxide (CaO). Calculate the percentage yield. (Mᵣ: CaCO₃ = 100, CaO = 56)

题目：4.0 g 碳酸钙（CaCO₃）通过加热分解，生成了 1.4 g 氧化钙（CaO）。计算产率。（相对分子质量：CaCO₃ = 100，CaO = 56）

Step 1: Write the equation: CaCO₃ → CaO + CO₂. Determine the mole ratio: 1 mol CaCO₃ produces 1 mol CaO.

步骤 1：写出方程式：CaCO₃ → CaO + CO₂。确定物质的量之比：1 mol CaCO₃ 生成 1 mol CaO。

Step 2: Calculate the expected (theoretical) moles of CaO. Moles of CaCO₃ used = mass ÷ Mᵣ = 4.0 ÷ 100 = 0.040 mol. Therefore, theoretical moles of CaO = 0.040 mol.

步骤 2：计算预期（理论）CaO 的物质的量。所用 CaCO₃ 的物质的量 = 4.0 ÷ 100 = 0.040 mol。因此理论 CaO 的物质的量 = 0.040 mol。

Step 3: Calculate the theoretical mass of CaO. theoretical mass = moles × Mᵣ = 0.040 × 56 = 2.24 g.

步骤 3：计算理论 CaO 的质量。理论质量 = 物质的量 × 相对分子质量 = 0.040 × 56 = 2.24 g。

Step 4: Calculate percentage yield. % yield = (actual mass ÷ theoretical mass) × 100 = (1.4 ÷ 2.24) × 100 ≈ 62.5%.

步骤 4：计算产率。产率 = (实际质量 ÷ 理论质量) × 100 = (1.4 ÷ 2.24) × 100 ≈ 62.5%。

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8. Flame Test for Cations | 阳离子的火焰试验

Question: Describe how you would carry out a flame test to distinguish between copper(II) sulfate and potassium sulfate, and state the expected observations.

题目：描述如何进行火焰试验以区分硫酸铜（II）和硫酸钾，并说明预期观察结果。

Step 1: Clean a nichrome or platinum wire loop by dipping it in concentrated hydrochloric acid and heating it in a roaring Bunsen flame until no distinctive colour is observed.

步骤 1：清洁镍铬或铂丝环，将其浸入浓盐酸后在本生灯的蓝色火焰中灼烧，直到观察不到特殊颜色为止。

Step 2: Dip the clean loop into the solid sample, or a concentrated solution of the sample, moistened with hydrochloric acid. Then place the loop in the edge of the non‑luminous Bunsen flame.

步骤 2：将清洁的金属丝环沾取固体样品或用盐酸润湿的浓溶液，然后将环放在无色本生灯火焰的边缘。

Step 3: Observe the flame colour. Copper(II) compounds produce a blue‑green flame. Potassium compounds give a lilac flame (which can be masked by sodium impurities; a blue cobalt glass filter helps see the lilac colour).

步骤 3：观察火焰颜色。铜（II）化合物产生蓝绿色火焰。钾化合物呈现淡紫色火焰（可能被钠杂质掩盖；用蓝色钴玻璃滤光片有助于看到淡紫色）。

Step 4: Record the results. A blue‑green flame indicates copper(II) sulfate, while a lilac flame (viewed through cobalt glass) indicates potassium sulfate.

步骤 4：记录结果。蓝绿色火焰表明是硫酸铜（II），而（通过钴玻璃观察到的）淡紫色火焰表明是硫酸钾。

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9. Organic Nomenclature and Functional Groups | 有机命名与官能团

Question: Draw the displayed formula of propanoic acid and name its functional group. Write a balanced equation for its reaction with ethanol in the presence of an acid catalyst.

题目：画出丙酸的展结构式并写出其官能团名称。写出在酸催化剂存在下丙酸与乙醇反应的配平方程式。

Step 1: Propanoic acid is a carboxylic acid with three carbon atoms. Its displayed formula can be drawn as: CH₃–CH₂–COOH. The functional group is –COOH, called the carboxyl group.

步骤 1：丙酸是一种含有三个碳原子的羧酸。其展结构式可画为：CH₃–CH₂–COOH。官能团为 –COOH，称为羧基。

Step 2: Identify the reaction type. The reaction with an alcohol (ethanol) in the presence of an acid catalyst (e.g. concentrated sulfuric acid) is esterification. It produces an ester and water.

步骤 2：确定反应类型。在酸催化剂（如浓硫酸）存在下与醇（乙醇）的反应为酯化反应，生成酯和水。

Step 3: Write the equation using molecular formulas. Propanoic acid = C₂H₅COOH (or C₃H₆O₂). Ethanol = C₂H₅OH. The ester formed is ethyl propanoate (C₂H₅COOC₂H₅). The balanced equation: C₂H₅COOH + C₂H₅OH ⇌ C₂H₅COOC₂H₅ + H₂O. (Concentrated H₂SO₄ is written over the equilibrium arrow as a catalyst.)

步骤 3：写出使用分子式的方程式。丙酸 = C₂H₅COOH（或 C₃H₆O₂）。乙醇 = C₂H₅OH。生成的酯为丙酸乙酯（C₂H₅COOC₂H₅）。配平方程式：C₂H₅COOH + C₂H₅OH ⇌ C₂H₅COOC₂H₅ + H₂O。（浓硫酸作为催化剂，可写在可逆箭头上方。）

Step 4: Name the ester product. The product is ethyl propanoate, an ester with the functional group –COO–.

步骤 4：命名酯产物。产物为丙酸乙酯，是一种含有 –COO– 官能团的酯。

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10. Equilibrium and Le Chatelier’s Principle | 平衡与勒夏特列原理

Question: The Haber process is represented by: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) , ∆H = −92 kJ/mol. Predict and explain the effect of increasing temperature on the equilibrium yield of ammonia.

题目：哈伯法反应为：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，∆H = −92 kJ/mol。预测并解释升高温度对平衡氨产率的影响。

Step 1: Identify the nature of the forward reaction. The negative ∆H means the forward reaction is exothermic (releases heat). The reverse reaction is therefore endothermic.

步骤 1：确定正反应的性质。∆H 为负值表明正反应是放热反应（释放热量）。因此逆反应是吸热反应。

Step 2: Apply Le Chatelier’s principle. If the temperature is increased, the equilibrium shifts in the direction that tends to reduce the temperature – it favours the endothermic reaction, which absorbs the extra heat.

步骤 2：应用勒夏特列原理。如果升高温度，平衡将向降低温度的方向移动——即倾向于吸热反应，以吸收多余的热量。

Step 3: Determine the shift. The endothermic reaction is the reverse reaction (breaking down NH₃ into N₂ and H₂). Therefore, the equilibrium position shifts to the left.

步骤 3：判断移动方向。吸热反应是逆反应（NH₃ 分解为 N₂ 和 H₂）。因此，平衡位置向左移动。

Step 4: Conclude the effect on yield. Shifting to the left reduces the amount of ammonia at equilibrium, so the yield of ammonia decreases.

步骤 4：得出对产率的影响。向左移动减少了平衡时氨的含量，因此氨的产率降低。

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