📚 GCSE Chemistry: Worked Examples Explained | GCSE 化学：典型例题详解
Worked examples are one of the most effective ways to master GCSE Chemistry. By following step-by-step solutions, you can understand how to apply concepts, balance equations, perform mole calculations, and interpret experimental data. This article provides ten carefully selected typical exam-style questions with detailed bilingual explanations. Each example targets a core topic and helps build confidence for your revision.
典型例题是掌握 GCSE 化学最有效的方法之一。通过逐步解析解题过程，你可以学会如何应用概念、配平方程式、进行摩尔计算以及分析实验数据。本文精选了十道典型的考试题型，并提供中英双语的详细讲解。每道例题都针对一个核心主题，帮助你在复习中建立信心。
1. Mole and Mass Calculations | 摩尔与质量计算
Question: In the reaction 2Mg + O₂ → 2MgO, what mass of magnesium oxide is produced when 2.43 g of magnesium completely reacts? (Ar: Mg=24.3, O=16.0)
问题：在反应 2Mg + O₂ → 2MgO 中，2.43 g 镁完全反应后生成多少质量的氧化镁？（Ar: Mg=24.3, O=16.0）
Step 1: Calculate moles of Mg. moles = mass / Ar = 2.43 g / 24.3 g mol⁻¹ = 0.100 mol.
步骤一：计算 Mg 的摩尔数。摩尔 = 质量 / Ar = 2.43 g ÷ 24.3 g mol⁻¹ = 0.100 mol。
Step 2: Use the mole ratio from the balanced equation. 2 mol Mg : 2 mol MgO, so mole ratio is 1:1. Therefore moles of MgO = 0.100 mol.
步骤二：利用配平方程式中的摩尔比。2 mol Mg : 2 mol MgO，摩尔比为 1:1。所以 MgO 的摩尔数 = 0.100 mol。
Step 3: Calculate mass of MgO. Mr of MgO = 24.3 + 16.0 = 40.3. mass = moles × Mr = 0.100 mol × 40.3 g mol⁻¹ = 4.03 g.
步骤三：计算 MgO 的质量。Mr(MgO) = 24.3 + 16.0 = 40.3。质量 = 摩尔 × Mr = 0.100 mol × 40.3 g mol⁻¹ = 4.03 g。
Answer: 4.03 g of magnesium oxide is produced.
答案：生成 4.03 g 氧化镁。
2. Balancing Chemical Equations | 配平化学方程式
Problem: Balance the equation: __Fe₂O₃ + __CO → __Fe + __CO₂
问题：配平方程式：__Fe₂O₃ + __CO → __Fe + __CO₂
Step 1: Count atoms on each side. Left: 2 Fe, 3 O from Fe₂O₃, plus from CO: 1 C and 1 O. Right: 1 Fe, 2 O from CO₂.
步骤一：数两边的原子。左边：2 Fe，3 O（来自Fe₂O₃），再加上CO中的1 C和1 O。右边：1 Fe，2 O（来自CO₂）。
Step 2: Balance Fe by placing coefficient 2 in front of Fe. Equation becomes: Fe₂O₃ + CO → 2Fe + CO₂
步骤二：在 Fe 前面加系数 2 平衡铁，方程式变为：Fe₂O₃ + CO → 2Fe + CO₂
Step 3: Balance O and C. We now have 4 O on left (3 from Fe₂O₃ + 1 from CO) and 2 O on right. Try coefficient 3 for CO and 3 for CO₂: Fe₂O₃ + 3CO → 2Fe + 3CO₂. Check: Left: 2 Fe, 6 O, 3 C. Right: 2 Fe, 6 O, 3 C. Balanced.
步骤三：平衡 O 和 C。左边现有4 O (3来自Fe₂O₃ + 1来自CO)，右边2 O。尝试CO系数3、CO₂系数3：Fe₂O₃ + 3CO → 2Fe + 3CO₂。检查：左边：2 Fe, 6 O, 3 C；右边：2 Fe, 6 O, 3 C。已配平。
Balanced equation: Fe₂O₃ + 3CO → 2Fe + 3CO₂
配平后的方程式：Fe₂O₃ + 3CO → 2Fe + 3CO₂
3. Electrolysis of Molten Sodium Chloride | 电解熔融氯化钠
Question: Electrolysis of molten NaCl. What are the products at the anode and cathode? Write half-equations.
问题：电解熔融 NaCl，阳极和阴极的产物分别是什么？写出半反应方程式。
Cathode (reduction): Na⁺ + e⁻ → Na (liquid sodium metal)
阴极（还原）：Na⁺ + e⁻ → Na（液态金属钠）
Anode (oxidation): 2Cl⁻ → Cl₂ + 2e⁻ (chlorine gas)
阳极（氧化）：2Cl⁻ → Cl₂ + 2e⁻（氯气）
Overall: 2NaCl(l) → 2Na(l) + Cl₂(g)
总反应：2NaCl(l) → 2Na(l) + Cl₂(g)
Remember: In molten salts, metal is produced at the cathode and non-metal at the anode.
记住：在熔融盐中，阴极产生金属，阳极产生非金属单质。
4. Bond Energy Calculations | 键能计算
Question: Calculate the enthalpy change for H₂ + Cl₂ → 2HCl using the following bond energies: H–H: 436 kJ mol⁻¹, Cl–Cl: 242 kJ mol⁻¹, H–Cl: 431 kJ mol⁻¹.
问题：用下列键能计算 H₂ + Cl₂ → 2HCl 的焓变：H–H: 436 kJ mol⁻¹，Cl–Cl: 242 kJ mol⁻¹，H–Cl: 431 kJ mol⁻¹。
Energy required to break bonds: 1×H–H + 1×Cl–Cl = 436 + 242 = 678 kJ
断裂化学键吸收的能量：1×H–H + 1×Cl–Cl = 436 + 242 = 678 kJ
Energy released when new bonds form: 2×H–Cl = 2 × 431 = 862 kJ
形成新化学键释放的能量：2×H–Cl = 2 × 431 = 862 kJ
ΔH = energy absorbed – energy released = 678 – 862 = –184 kJ mol⁻¹
ΔH = 吸收的能量 – 释放的能量 = 678 – 862 = –184 kJ mol⁻¹
The reaction is exothermic because more energy is released forming bonds than is used to break them.
反应是放热的，因为成键释放的能量大于断键吸收的能量。
5. Rate of Reaction Calculations | 反应速率计算
Question: Magnesium ribbon reacts with hydrochloric acid, and the volume of hydrogen gas is collected. In the first 60 seconds, 45 cm³ of H₂ is produced. Calculate the average rate of reaction in cm³ s⁻¹.
问题：镁条与盐酸反应，收集氢气体积。前60秒产生45 cm³ H₂，计算平均反应速率（cm³ s⁻¹）。
Average rate = volume of gas produced / time taken = 45 cm³ / 60 s = 0.75 cm³ s⁻¹
平均速率 = 产生气体的体积 / 时间 = 45 cm³ ÷ 60 s = 0.75 cm³ s⁻¹
If the volume is measured at 20-second intervals, you could also calculate the rate for each interval to see how it decreases as the reaction proceeds.
如果每20秒记录一次体积，还可以计算各时间段的速率，观察速率如何随反应进行而下降。
6. Titration Calculation | 滴定计算
Question: 25.0 cm³ of sodium hydroxide solution is neutralised by 20.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid. Calculate the concentration of the NaOH solution. NaOH + HCl → NaCl + H₂O
问题：25.0 cm³ 氢氧化钠溶液被 20.0 cm³ 0.100 mol dm⁻³ 盐酸中和。计算 NaOH 溶液的浓度。NaOH + HCl → NaCl + H₂O
Step 1: Moles of HCl = concentration × volume (dm³) = 0.100 × (20.0/1000) = 0.00200 mol
步骤一：HCl 的摩尔数 = 浓度 × 体积 (dm³) = 0.100 × (20.0/1000) = 0.00200 mol
Step 2: Mole ratio from equation is 1:1, so moles of NaOH = 0.00200 mol
步骤二：方程式中摩尔比为1:1，所以 NaOH 摩尔数 = 0.00200 mol
Step 3: Concentration of NaOH = moles / volume (dm³) = 0.00200 / (25.0/1000) = 0.0800 mol dm⁻³
步骤三：NaOH 浓度 = 摩尔数 / 体积 (dm³) = 0.00200 ÷ (25.0/1000) = 0.0800 mol dm⁻³
Always convert cm³ to dm³ by dividing by 1000 before using in concentration calculations.
在浓度计算中，务必先将 cm³ 除以1000转换为 dm³。
7. Identifying Bonding Type | 识别化学键类型
Question: Substance X has a melting point of 801 °C. It does not conduct electricity when solid but conducts when molten. Suggest the type of bonding present and explain your answer.
问题：物质 X 的熔点为 801 °C，固态时不导电，但熔融状态可导电。推测其化学键类型并解释。
High melting point indicates strong forces between particles. Conduction only when molten but not solid is characteristic of ionic compounds: ions are fixed in the solid lattice but free to move when melted.
高熔点表明粒子间作用力强。只有熔融态导电、固态不导电是离子化合物的特征：固态晶格中离子固定，熔融时离子可自由移动。
Therefore, substance X has ionic bonding. (Example: sodium chloride.)
因此，物质 X 具有离子键。（例如：氯化钠。）
If it conducted as solid and had a very high melting point, metallic bonding would be likely. Simple molecular substances have low melting points and do not conduct.
如果固态可导电且熔点极高，则可能是金属键。简单分子物质熔点低且不导电。
8. Le Chatelier’s Principle | 勒夏特列原理
Question: Consider the equilibrium: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ mol⁻¹. Predict the shift of equilibrium when (a) pressure is increased, (b) temperature is increased, (c) ammonia is removed.
问题：考虑平衡：N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ mol⁻¹。分别预测下列条件下的平衡移动方向：(a) 增大压强，(b) 升高温度，(c) 移走氨气。
(a) Increase pressure: There are 4 moles of gas on the left and 2 moles on the right. The system shifts to reduce pressure by favouring the side with fewer gas molecules, so equilibrium shifts right (towards NH₃).
(a) 增大压强：左边4 mol气体，右边2 mol气体。体系会向气体分子数减少的方向移动以减弱压强，因此平衡向右（生成 NH₃）移动。
(b) Increase temperature: The forward reaction is exothermic, so the reverse is endothermic. Adding heat causes the system to favour the endothermic direction to absorb heat; equilibrium shifts left (towards N₂ and H₂).
(b) 升高温度：正反应放热，逆反应吸热。升高温度，体系向吸热方向移动以吸收热量，平衡向左（生成 N₂ 和 H₂）移动。
(c) Remove NH₃: Removing a product reduces its concentration. The system shifts to replace it, so equilibrium shifts right to produce more NH₃.
(c) 移走 NH₃：移走生成物降低了其浓度，体系会向补充生成物的方向移动，因此平衡向右移动产生更多 NH₃。
9. Simple Distillation | 简单蒸馏
Question: Describe how pure water can be obtained from seawater using simple distillation. Explain why the thermometer is placed at a specific point.
问题：描述如何利用简单蒸馏从海水中获得纯水，并解释温度计应放置的位置及其原因。
In simple distillation, seawater is heated in a distillation flask. Water evaporates, leaving dissolved salts behind. The water vapour passes into a condenser where it cools and condenses back into liquid, collected as pure distillate.
简单蒸馏中，海水在蒸馏烧瓶中加热。水蒸发，溶解的盐类留在瓶内。水蒸气进入冷凝管冷却并冷凝回液态，收集到的馏出液即为纯水。
The thermometer bulb must be placed level with the T-junction (side arm) of the distillation flask, where the vapour exits. This measures the boiling point of the vapour exactly as it condenses, ensuring the reading reflects the pure substance’s boiling point.
温度计的水银球应放置在蒸馏烧瓶支管口（T形接合处）同一水平线上，这里是蒸气逸出的位置。这样可测量蒸气刚好冷凝时的沸点，确保读数代表纯物质的沸点。
10. Calorimetry: Heat of Neutralisation
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