📚 IGCSE CCEA Chemistry: Worked Examples Explained | IGCSE CCEA 化学：典型例题详解
Mastering IGCSE CCEA Chemistry requires not only understanding key concepts but also the ability to apply them to exam-style questions. This article walks you through a series of carefully selected worked examples, each targeting a specific topic from the CCEA specification. Step-by-step solutions are provided in clear, bilingual format to help you develop strong problem-solving skills.
要掌握 IGCSE CCEA 化学，不仅需要理解核心概念，还需要能够将其运用到考题中。本文通过一系列精心挑选的典型例题，逐一讲解 CCEA 考纲中的重点主题。每个例题均提供清晰的双语分步解答，帮助培养扎实的解题能力。
1. Atomic Structure and Isotopes | 原子结构与同位素
Example: A sample of neon contains 90% ²⁰Ne and 10% ²²Ne. Calculate the relative atomic mass (Aᵣ) of this neon sample.
To calculate Aᵣ from isotopic abundances, multiply each isotope’s mass number by its percentage, sum the results, and divide by 100. (20 × 90) + (22 × 10) = 1800 + 220 = 2020. Divide by 100 → 20.2. The relative atomic mass is 20.2. This value is not a whole number because it is a weighted average of the isotopes present.
计算相对原子质量时，将每种同位素的质量数乘以其丰度百分比，求和后除以 100。 (20 × 90) + (22 × 10) = 1800 + 220 = 2020。除以 100 → 20.2。相对原子质量为 20.2。这个值不是整数，因为它是存在的同位素的加权平均值。
2. Ionic Bonding and Dot-and-Cross Diagrams | 离子键与点叉图
Example: Draw a dot-and-cross diagram to show the formation of magnesium chloride, MgCl₂. Show only the outer electrons.
Magnesium (group 2) has 2 outer electrons, each chlorine (group 7) has 7 outer electrons. Mg loses its 2 electrons to form Mg²⁺, and each Cl atom gains 1 electron to form Cl⁻. The resulting ions are Mg²⁺ and two Cl⁻. In the diagram, represent Mg electrons with dots, Cl electrons with crosses. Show the Mg²⁺ ion with no outer shell, and each Cl⁻ ion with a full outer shell of 8 electrons (dots and crosses) enclosed in square brackets with the charge outside.
镁（第2族）有2个最外层电子，每个氯（第7族）有7个最外层电子。Mg 失去2个电子形成 Mg²⁺，每个 Cl 原子得到1个电子形成 Cl⁻。生成的离子为 Mg²⁺ 和两个 Cl⁻。在点叉图中，用点表示镁的电子，叉表示氯的电子。画出 Mg²⁺ 无最外层电子，每个 Cl⁻ 离子用方括号括起，最外层有8个电子（点和叉混合），括号外标注电荷。
3. Mole Calculations Using Mass | 运用质量的摩尔计算
Example: What mass of carbon dioxide, CO₂, is produced when 12 g of carbon are burned completely in excess oxygen? (Aᵣ: C = 12, O = 16)
First, write the equation: C + O₂ → CO₂. Moles of C = mass / Aᵣ = 12 / 12 = 1 mol. From the equation, mole ratio C : CO₂ = 1 : 1, so 1 mol of CO₂ is produced. Molar mass of CO₂ = 12 + (16×2) = 44 g/mol. Mass of CO₂ = moles × molar mass = 1 × 44 = 44 g. Always check the balanced equation and use mole ratios correctly.
首先写出方程式：C + O₂ → CO₂。C 的摩尔数 = 质量 / Aᵣ = 12 / 12 = 1 mol。由方程式可知, C 与 CO₂ 的摩尔比为 1 : 1，因此生成 1 mol CO₂。CO₂ 的摩尔质量 = 12 + (16×2) = 44 g/mol。CO₂ 的质量 = 摩尔数 × 摩尔质量 = 1 × 44 = 44 g。务必检查配平的方程式并正确使用摩尔比。
4. Empirical Formula from Percentage Composition | 由百分组成求经验式
Example: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Determine its empirical formula. (Aᵣ: H=1, C=12, O=16)
Assume a 100 g sample, so masses are C: 40.0 g, H: 6.7 g, O: 53.3 g. Divide each mass by the Aᵣ: C: 40.0/12 = 3.33 mol; H: 6.7/1 = 6.7 mol; O: 53.3/16 = 3.33 mol. Divide by the smallest number (3.33) to get ratio C : H : O = 1 : 2 : 1. The empirical formula is CH₂O. This could be methanal or a carbohydrate building block. Always simplify to whole-number ratios.
假设样品为 100 g，则各元素质量分别为 C: 40.0 g，H: 6.7 g，O: 53.3 g。各除以 Aᵣ： C: 40.0/12 = 3.33 mol；H: 6.7/1 = 6.7 mol；O: 53.3/16 = 3.33 mol。除以最小值 (3.33) 得比例 C : H : O = 1 : 2 : 1。经验式为 CH₂O。这可能是甲醛或碳水化合物的基本单元。记住要化简为最简整数比。
5. Concentration and Titration Calculations | 浓度与滴定计算
Example: 25.0 cm³ of sodium hydroxide solution required 20.0 cm³ of 0.100 mol/dm³ hydrochloric acid for neutralisation. Find the concentration of the NaOH solution.
Equation: HCl + NaOH → NaCl + H₂O. Moles of HCl = concentration × volume (in dm³) = 0.100 × (20.0/1000) = 0.00200 mol. Mole ratio HCl : NaOH = 1 : 1, so moles of NaOH = 0.00200 mol. Volume of NaOH = 25.0/1000 = 0.0250 dm³. Concentration of NaOH = moles / volume = 0.00200 / 0.0250 = 0.0800 mol/dm³. Always convert cm³ to dm³ by dividing by 1000.
方程式：HCl + NaOH → NaCl + H₂O。HCl 的摩尔数 = 浓度 × 体积 (dm³) = 0.100 × (20.0/1000) = 0.00200 mol。摩尔比 HCl : NaOH = 1 : 1，因此 NaOH 的摩尔数 = 0.00200 mol。NaOH 体积 = 25.0/1000 = 0.0250 dm³。NaOH 浓度 = 摩尔数 / 体积 = 0.00200 / 0.0250 = 0.0800 mol/dm³。务必先除以 1000 将 cm³ 换算为 dm³。
6. Rates of Reaction – Interpreting Graphs | 反应速率 – 图像解读
Example: The graph shows the volume of gas produced against time for the reaction of magnesium with excess dilute hydrochloric acid. Explain why the curve becomes less steep over time and eventually levels off.
The steepness (gradient) of the curve indicates the rate of reaction. Initially, reactant concentration is high, so the rate is fast. As the magnesium reacts, its mass decreases and the acid concentration falls, so the frequency of successful collisions decreases, making the reaction slower. The curve levels off when all the magnesium (the limiting reactant) has been used up, so no more gas is produced. The final volume represents the total gas produced from that mass of Mg.
曲线的陡度（斜率）代表反应速率。开始时反应物浓度高，速率快。随着镁不断反应，其质量减少，酸浓度下降，因此有效碰撞频率降低，反应变慢。当所有镁（限制反应物）被消耗完后，曲线趋于水平，不再产生气体。最终体积反映了该质量 Mg 所能产生的气体总量。
7. Energy Changes and Bond Energies | 能量变化与键能
Example: Using the bond energies (kJ/mol): H–H = 436, Cl–Cl = 242, H–Cl = 431, calculate the energy change (ΔH) for the reaction H₂ + Cl₂ → 2HCl.
Energy needed to break bonds: 1 mol H–H (436) + 1 mol Cl–Cl (242) = +678 kJ. Energy released forming bonds: 2 × (H–Cl) = 2 × 431 = −862 kJ. Overall ΔH = +678 + (−862) = −184 kJ. The negative sign indicates the reaction is exothermic, releasing 184 kJ per mole of equation (as written for 2 mol HCl). Always show working and state whether the reaction is exothermic or endothermic.
断裂化学键吸收能量：1 mol H–H (436) + 1 mol Cl–Cl (242) = +678 kJ。形成化学键释放能量：2 × (H–Cl) = 2 × 431 = −862 kJ。总 ΔH = +678 + (−862) = −184 kJ。负号表示反应放热，按所写方程式（生成 2 mol HCl）释放 184 kJ 热量。务必写清楚过程并指出反应是放热还是吸热。
8. Electrolysis of Aqueous Solutions | 水溶液的电解
Example: Predict the products at the anode and cathode during the electrolysis of concentrated aqueous sodium chloride using inert electrodes. Write the half-equations.
In concentrated NaCl(aq), ions present: Na⁺, Cl⁻, H⁺ (from water), OH⁻ (from water). At the cathode, hydrogen ions are discharged in preference to sodium ions because H⁺ is lower in the reactivity series: 2H⁺ + 2e⁻ → H₂(g). At the anode, chloride ions are discharged in preference to hydroxide ions in a concentrated solution: 2Cl⁻ → Cl₂(g) + 2e⁻. The overall products are hydrogen at cathode and chlorine at anode, leaving NaOH in solution.
在浓 NaCl 溶液中，存在的离子有：Na⁺、Cl⁻、H⁺（来自水）、OH⁻（来自水）。在阴极，H⁺ 比 Na⁺ 优先放电，因为氢在金属活动性顺序中位置更低：2H⁺ + 2e⁻ → H₂(g)。在阳极，浓溶液中 Cl⁻ 比 OH⁻ 优先放电：2Cl⁻ → Cl₂(g) + 2e⁻。总产物为阴极产生氢气，阳极产生氯气，溶液中留下 NaOH。
9. Organic Chemistry – Cracking and Alkanes | 有机化学 – 裂化与烷烃
Example: The alkane C₁₀H₂₂ undergoes catalytic cracking to produce octane (C₈H₁₈) and one other hydrocarbon. Write a balanced equation and suggest why cracking is important industrially.
Cracking breaks a long-chain alkane into a shorter-chain alkane and an alkene (or another alkane). For C₁₀H₂₂ → C₈H₁₈ + X, balancing carbon and hydrogen gives X = C₂H₄ (ethene). Equation: C₁₀H₂₂ → C₈H₁₈ + C₂H₄. Cracking is important because it converts less useful long-chain hydrocarbons from crude oil into shorter-chain alkanes (higher demand fuels) and alkenes (used as feedstock for polymers and chemicals).
裂化将长链烷烃分解为短链烷烃和烯烃（或其他烷烃）。C₁₀H₂₂ → C₈H₁₈ + X，平衡碳和氢原子后得 X = C₂H₄（乙烯）。方程式：C₁₀H₂₂ → C₈H₁₈ + C₂H₄。裂化的重要性在于，它将原油中需求较低的长链烃转化为更有价值的短链烷烃（高需求燃料）和烯烃（用作聚合物和化学品的原料）。
10. Reversible Reactions and Equilibrium | 可逆反应与平衡
Example: The reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) has ΔH = −197 kJ/mol. Predict the effect of increasing temperature on the equilibrium position and explain your reasoning using Le Chatelier’s principle.
The forward reaction is exothermic (ΔH negative). According to Le Chatelier’s principle, if a system at equilibrium is subjected to a change in temperature, the equilibrium position moves to oppose the change. Increasing temperature favours the endothermic reaction, which is the backward reaction (absorbing heat). Therefore the equilibrium shifts to the left, decreasing the yield of SO₃. Industrially, a compromise temperature is used to balance rate and yield.
正向反应放热（ΔH 为负）。根据勒夏特列原理，如果改变平衡系统的温度，平衡会向减弱该改变的方向移动。升高温度有利于吸热反应，这里是逆向反应（吸收热量）。因此平衡向左移动，SO₃ 的产量下降。工业上采用折中温度以平衡速率与产率。
11. Acid-Base Theory and pH | 酸碱理论与 pH
Example: Explain why a solution of hydrogen chloride in water has a low pH but a solution of hydrogen chloride in methylbenzene does not conduct electricity and has no effect on blue litmus.
In water, HCl dissociates completely into H⁺ and Cl⁻ ions, making it a strong acid with a high concentration of H⁺ ions, thus a low pH. In methylbenzene (a non-polar solvent), HCl does not ionise; it remains as covalent molecules. Without mobile ions, the solution cannot conduct electricity, and with no H⁺ ions present, it shows no acidic properties such as turning blue litmus red. This illustrates the difference between aqueous solutions and non-aqueous solutions for acidic behaviour.
在水中，HCl 完全电离为 H⁺ 和 Cl⁻ 离子，成为强酸，H⁺ 离子浓度高，因此 pH 低。在甲基苯（非极性溶剂）中，HCl 不电离，保持共价分子形态。没有可移动的离子，溶液不能导电；没有 H⁺ 存在，不表现出酸性，如不能使蓝色石蕊试纸变红。这体现了酸的行为依赖于溶剂的性质。
12. Identification of Gases and Ions | 气体与离子的鉴别
Example: A student adds dilute acid to an unknown solid; a gas is produced that turns limewater milky. When aqueous sodium hydroxide is added to another portion of the solid, no ammonia smell is detected. Identify the anion present and write an ionic equation for the reaction with acid.
The gas that turns limewater milky is carbon dioxide (CO₂). This indicates the solid contains a carbonate ion (CO₃²⁻). No ammonia on adding NaOH means ammonium ions are absent. The reaction of carbonate with acid: CO₃²⁻(s) + 2H⁺(aq) → H₂O(l) + CO₂(g). The observation of limewater turning milky confirms CO₂. This is a classic test for carbonates and hydrogencarbonates. A follow-up test for the cation (e.g., flame test) could be used.
使石灰水变浑浊的气体是二氧化碳 (CO₂)，证明固体中含有碳酸根离子 (CO₃²⁻)。加氢氧化钠无氨味说明不含铵根离子。碳酸盐与酸反应的离子方程式：CO₃²⁻(s) + 2H⁺(aq) → H₂O(l) + CO₂(g)。石灰水变浑浊这一现象确证了 CO₂。这是碳酸盐和碳酸氢盐的经典检验方法。可进一步通过焰色反应鉴定阳离子。
Published by TutorHao | IGCSE Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导