📚 GCSE WJEC Chemistry: Worked Examples of Typical Exam Questions | GCSE WJEC 化学：典型例题详解
This article provides detailed, step-by-step solutions to common question types in the GCSE WJEC Chemistry exam. Mastering these worked examples will help you apply key concepts and achieve top marks.
本文为GCSE WJEC化学考试中常见的题型提供详细的分步解答。掌握这些典型例题将帮助你运用核心概念，取得优异成绩。
1. Balancing Chemical Equations | 配平化学方程式例题
Question: Balance the equation for the complete combustion of ethane, C₂H₆.
例题：配平乙烷完全燃烧的方程式：C₂H₆ + O₂ → CO₂ + H₂O。
Step 1: Count atoms on each side. Left: 2 C, 6 H, 2 O. Right: 1 C, 2 H, 3 O (from CO₂ and H₂O).
第一步：数原子。左边：2个C，6个H，2个O。右边：1个C，2个H，3个O（来自CO₂和H₂O）。
Step 2: Balance carbon by placing a 2 before CO₂: C₂H₆ + O₂ → 2CO₂ + H₂O.
第二步：在CO₂前加系数2平衡碳：C₂H₆ + O₂ → 2CO₂ + H₂O。
Step 3: Balance hydrogen by placing a 3 before H₂O: C₂H₆ + O₂ → 2CO₂ + 3H₂O. Now 6 H on both sides.
第三步：在H₂O前加系数3平衡氢：C₂H₆ + O₂ → 2CO₂ + 3H₂O。现在两边都有6个H。
Step 4: Recount oxygen: right side has (2×2) + 3 = 7 O atoms. Place 7/2 before O₂: C₂H₆ + 3.5O₂ → 2CO₂ + 3H₂O.
第四步：重数氧原子：右边(2×2)+3=7个O。在O₂前放7/2：C₂H₆ + 3.5O₂ → 2CO₂ + 3H₂O。
Step 5: Multiply all coefficients by 2 to eliminate fractions: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O.
第五步：所有系数乘以2去掉分数：2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O。
Final: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
最终方程式：2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
2. Moles and Mass Calculations | 摩尔与质量计算
Question: 10.0 g of calcium carbonate, CaCO₃, is heated strongly. Calculate the mass of calcium oxide, CaO, produced. Aᵣ: Ca = 40, C = 12, O = 16.
例题：10.0 g碳酸钙受热分解，计算生成氧化钙的质量。相对原子质量：Ca=40，C=12，O=16。
Step 1: Write the equation: CaCO₃ → CaO + CO₂.
第一步：写方程式：CaCO₃ → CaO + CO₂。
Step 2: Mᵣ of CaCO₃ = 40 + 12 + (3×16) = 100. Moles CaCO₃ = mass / Mᵣ = 10.0 / 100 = 0.100 mol.
第二步：CaCO₃相对分子质量=100。摩尔数=10.0/100=0.100 mol。
Step 3: 1:1 mole ratio, so 0.100 mol CaO forms. Mᵣ of CaO = 40 + 16 = 56.
第三步：摩尔比1:1，生成0.100 mol CaO。CaO相对分子质量=56。
Step 4: Mass CaO = moles × Mᵣ = 0.100 × 56 = 5.60 g.
第四步：CaO质量=0.100×56=5.60 g。
3. Predicting Products of Electrolysis | 预测电解产物
Question: Molten lead(II) bromide, PbBr₂, is electrolysed using inert electrodes. State the products at the cathode and anode, and give half-equations.
例题：用惰性电极电解熔融溴化铅。写出阴、阳极产物及半方程式。
In molten PbBr₂, ions are Pb²⁺ and Br⁻. At the cathode (negative electrode), Pb²⁺ ions gain electrons.
熔融PbBr₂中存在Pb²⁺和Br⁻离子。阴极（负极）上Pb²⁺得电子。
Cathode half-equation: Pb²⁺ + 2e⁻ → Pb (liquid lead)
阴极半方程式：Pb²⁺ + 2e⁻ → Pb（液态铅）
At the anode (positive electrode), Br⁻ ions lose electrons. Bromine gas is produced.
阳极（正极）上Br⁻失电子，生成溴气。
Anode half-equation: 2Br⁻ → Br₂ + 2e⁻ (bromine gas)
阳极半方程式：2Br⁻ → Br₂ + 2e⁻（溴气）
Observation: silvery liquid at cathode, brown gas at anode.
现象：阴极有银白色液体，阳极有红棕色气体。
4. Interpreting Rate of Reaction Graphs | 反应速率图像解读
Question: The graph below shows the volume of hydrogen gas produced when excess zinc reacts with dilute sulfuric acid. (Graph description: steep rise, then level off.) Explain how the graph shows the reaction is complete, and calculate the mean rate in the first 30 seconds if 45 cm³ gas was collected.
例题：下图表示过量锌与稀硫酸反应产生氢气的体积（描述：曲线先陡后平）。解释如何从图像判断反应完成，并计算前30秒的平均速率（收集到45 cm³气体）。
The curve becomes horizontal, meaning no more gas is produced – the reaction is finished.
曲线变为水平，说明不再有气体产生——反应已完成。
Mean rate = total volume / time = 45 cm³ / 30 s = 1.5 cm³/s.
平均速率=总体积/时间=45 cm³/30 s=1.5 cm³/s。
The steeper the slope, the faster the reaction. At the start the slope is steepest because reactant concentrations are highest.
斜率越大，反应越快。开始时斜率最大，因为反应物浓度最高。
5. Acid–Base Titration Calculations | 酸碱滴定计算
Question: 25.0 cm³ of sodium hydroxide solution required 20.0 cm³ of 0.100 mol/dm³ sulfuric acid for neutralisation. Find the concentration of NaOH in mol/dm³. Equation: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.
例题：25.0 cm³氢氧化钠溶液完全中和20.0 cm³ 0.100 mol/dm³硫酸。求NaOH的浓度。方程式：2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O。
Step 1: Moles of H₂SO₄ = c × V = 0.100 × (20.0/1000) = 0.00200 mol.
第一步：硫酸摩尔数= c×V = 0.100×(20.0/1000)=0.00200 mol。
Step 2: From the equation, 2 mol NaOH react with 1 mol H₂SO₄, so moles NaOH = 2 × 0.00200 = 0.00400 mol.
第二步：根据方程式，2 mol NaOH与1 mol H₂SO₄反应，故NaOH摩尔数=2×0.00200=0.00400 mol。
Step 3: Concentration NaOH = moles / volume = 0.00400 / (25.0/1000) = 0.160 mol/dm³.
第三步：NaOH浓度= 0.00400/(25.0/1000)= 0.160 mol/dm³。
6. Bond Energy Calculations | 键能计算
Question: Calculate the energy change for the reaction H₂ + Cl₂ → 2HCl using the bond energies: H–H 436 kJ/mol, Cl–Cl 242 kJ/mol, H–Cl 431 kJ/mol. State if it is exothermic or endothermic.
例题：使用键能H–H 436 kJ/mol，Cl–Cl 242 kJ/mol，H–Cl 431 kJ/mol计算反应H₂ + Cl₂ → 2HCl的能量变化，并判断吸热或放热。
Energy absorbed to break bonds: 1×436 + 1×242 = +678 kJ.
断键吸收能量：1×436 + 1×242 = +678 kJ。
Energy released when new bonds form: 2×431 = –862 kJ.
成键释放能量：2×431 = –862 kJ。
Overall energy change = +678 + (–862) = –184 kJ/mol (exothermic).
总能量变化=678–862= –184 kJ/mol（放热反应）。
Because more energy is given out making bonds than is taken in breaking bonds.
因为成键释放的能量大于断键吸收的能量。
7. Explaining Giant Covalent Structures | 解释巨型共价结构
Question: Diamond and graphite are both allotropes of carbon. Explain why diamond is very hard and does not conduct electricity, while graphite is soft and conducts electricity.
例题：金刚石和石墨都是碳的同素异形体。解释为什么金刚石很硬且不导电，而石墨又软又导电。
Diamond: each carbon atom forms four strong covalent bonds in a tetrahedral network; all outer shell electrons are localised in bonds. No delocalised electrons → no conductivity.
金刚石：每个碳原子与四个碳原子形成强共价键，呈四面体网络；所有外层电子都定域在键中。无离域电子→不导电。
Graphite: each carbon forms three covalent bonds in layers. The fourth outer electron becomes delocalised and can move along layers, conducting electricity.
石墨：每个碳原子形成三个共价键，呈层状结构。第四个外层电子离域，可在层间移动，因此导电。
Graphite is soft because weak forces between layers allow them to slide over each other.
石墨软是因为层间的弱作用力使它们容易滑动。
8. Cracking and Alkenes | 裂化与烯烃
Question: Decane (C₁₀H₂₂) is cracked to produce ethene and one other hydrocarbon. Write a balanced equation and describe a chemical test to show ethene is unsaturated.
例题：癸烷裂化生成乙烯和另一种烃。写出配平方程式，并描述一种证明乙烯是不饱和烃的化学检验。
Possible equation: C₁₀H₂₂ → C₂H₄ + C₈H₁₈ (octane).
可能的方程式：C₁₀H₂₂ → C₂H₄ + C₈H₁₈（辛烷）。
To test for unsaturation: shake ethene gas with orange bromine water. The bromine water turns colourless (addition reaction).
检验不饱和性：将乙烯气体通入橙色溴水中振荡，溴水褪色（发生加成反应）。
C₂H₄ + Br₂ → C₂H₄Br₂ (colourless).
C₂H₄ + Br₂ → C₂H₄Br₂（无色）。
9. Testing for Ions | 离子鉴定
Question: Describe the tests, including any observations and equations, to identify carbonate ions (CO₃²⁻), sulfate ions (SO₄²⁻) and chloride ions (Cl⁻) in aqueous solutions.
例题：描述鉴定溶液中碳酸根离子、硫酸根离子和氯离子的方法，包括现象和方程式。
Carbonate test: add dilute hydrochloric acid. Bubbles of gas produced. Bubble the gas through limewater; limewater turns milky.
碳酸根检验：加入稀盐酸，产生气泡。将气体通入石灰水，石灰水变浑浊。
Equation: CO₃²⁻ + 2H⁺ → CO₂ + H₂O. Then Ca(OH)₂ + CO₂ → CaCO₃ + H₂O.
方程式：CO₃²⁻ + 2H⁺ → CO₂ + H₂O。然后Ca(OH)₂ + CO₂ → CaCO₃ + H₂O。
Sulfate test: add dilute HCl, then barium chloride solution. White precipitate forms (barium sulfate).
硫酸根检验：先加稀盐酸，再加氯化钡溶液，产生白色沉淀（硫酸钡）。
Equation: Ba²⁺ + SO₄²⁻ → BaSO₄(s).
方程式：Ba²⁺ + SO₄²⁻ → BaSO₄(s)。
Chloride test: add dilute nitric acid, then silver nitrate solution. White precipitate (silver chloride) forms, soluble in dilute ammonia.
氯离子检验：加稀硝酸，再加硝酸银溶液，产生白色沉淀（氯化银），可溶于稀氨水。
Equation: Ag⁺ + Cl⁻ → AgCl(s).
方程式：Ag⁺ + Cl⁻ → AgCl(s)。
10. Dynamic Equilibrium and Le Chatelier’s Principle | 动态平衡与勒夏特列原理
Question: In the Haber process N₂ + 3H₂ ⇌ 2NH₃ (ΔH=−92 kJ/mol), explain the effect on the equilibrium yield of ammonia when (a) pressure is increased, (b) temperature is decreased.
例题：哈伯法N₂ + 3H₂ ⇌ 2NH₃ (ΔH=−92 kJ/mol)，解释增加压强和降低温度对氨平衡产率的影响。
(a) Increased pressure: equilibrium shifts to the side with fewer gas molecules. Left: 4 moles; right: 2 moles. Position moves right, increasing NH₃ yield.
(a) 增加压强：平衡向气体分子数少的方向移动。左边4 mol，右边2 mol。平衡右移，NH₃产率提高。
(b) The forward reaction is exothermic. Decreasing temperature shifts equilibrium in the exothermic direction to produce more heat, so position moves right, increasing NH₃ yield.
(b) 正反应放热。降低温度使平衡向放热方向移动以产生更多热量，因此平衡右移，NH₃产率提高。
Note: In practice, a compromise temperature (≈450 °C) is used because low temperature gives a high yield but very slow rate.
注意：实际采用折中温度（约450°C），因为低温产率高但速率极慢。
Published by TutorHao | Chemistry Revision Series | aleveler.com
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