📚 IGCSE Edexcel Chemistry: Worked Examples Explained | IGCSE Edexcel 化学：典型例题详解
In IGCSE Edexcel Chemistry, mastering exam questions is as important as understanding theory. This revision guide walks you through ten carefully selected worked examples, covering the most common topics and question types. Each example is broken down into a clear method, followed by a detailed explanation in both English and Chinese. By studying these models, you will sharpen your problem-solving skills and build confidence for the final examination.
在IGCSE Edexcel化学中，掌握考题与理解理论同样重要。这份复习指南通过十个精心挑选的典型例题，带你一步步攻克最常见的知识点和题型。每道例题都采用清晰的解题方法，并用中英双语详细解释。通过学习这些范例，你将提升解题能力，为最终的考试树立信心。
1. Atomic Structure and Isotopes | 原子结构与同位素
Example: An atom has 18 electrons, 18 protons and 22 neutrons. (a) Write the full isotopic symbol for this atom. (b) Explain why this atom is neutral.
例题：某原子有18个电子、18个质子和22个中子。(a) 写出该原子的完整同位素符号。(b) 解释为什么这个原子是电中性的。
(a) The atomic number Z = number of protons = 18, so the element is argon (Ar). The mass number A = protons + neutrons = 18 + 22 = 40. The isotopic symbol is 4018Ar. (b) The negative charge of 18 electrons exactly balances the positive charge of 18 protons. Therefore, the overall charge is zero and the atom is neutral.
(a) 原子序数 Z = 质子数 = 18，该元素为氩（Ar）。质量数 A = 质子数 + 中子数 = 18 + 22 = 40。同位素符号是4018Ar。(b) 18个电子的负电荷与18个质子的正电荷恰好平衡。因此，净电荷为零，原子呈电中性。
2. Ionic and Covalent Bonding | 离子键与共价键
Example: Sodium chloride (NaCl) has a high melting point (801 °C), while chlorine (Cl₂) is a gas at room temperature. Explain this difference in terms of bonding and structure.
例题：氯化钠（NaCl）具有高熔点（801 °C），而氯气（Cl₂）在室温下为气体。请从成键和结构的角度解释这一差异。
NaCl is an ionic compound with a giant ionic lattice. Strong electrostatic forces of attraction between oppositely charged Na⁺ and Cl⁻ ions require a large amount of energy to overcome, leading to a high melting point. Cl₂ is a simple molecular substance consisting of diatomic molecules. The molecules are held together by weak intermolecular forces (van der Waals’ forces), which need very little energy to break, giving it a low boiling point and gaseous state at room temperature.
NaCl是离子化合物，具有巨型离子晶格。带相反电荷的Na⁺和Cl⁻离子之间存在强烈的静电吸引力，需要大量能量才能克服，因此熔点高。Cl₂是简单的分子物质，由双原子分子组成。分子之间通过微弱的分子间作用力（范德华力）结合，打破这些作用力所需能量极少，因此沸点低，室温下呈气态。
3. Moles and Calculations | 摩尔与化学计算
Example: 5.40 g of an unknown metal reacts completely with 0.200 mol of chlorine gas (Cl₂). Determine the molar mass of the metal and suggest its identity, given that the metal forms a chloride with the formula MCl₃.
例题：5.40 g某未知金属与0.200 mol氯气（Cl₂）完全反应。已知该金属形成化学式为MCl₃的氯化物，求该金属的摩尔质量并推测其名称。
Equation: 2M + 3Cl₂ → 2MCl₃. Moles of Cl₂ = 0.200 mol. From the equation, 3 mol Cl₂ react with 2 mol M. So moles of M = (2/3) × 0.200 = 0.1333 mol. Mass of M used = 5.40 g. Molar mass M = mass / moles = 5.40 / 0.1333 = approximately 40.5 g/mol. This value is close to 40 g/mol, suggesting the metal is calcium (Ca), but calcium forms CaCl₂, not MCl₃. 40.5 suggests atomic mass near 40 could be calcium but valency mismatch. Better check: If M forms MCl₃, the group would be 3, element aluminium has Ar ≈ 27. But 0.1333 mol gives 40.5. Let me recalc: 5.40 g and 0.200mol Cl₂, ratio 2M:3Cl₂, moles M = (2/3)*0.2 = 0.13333, M = 5.4/0.13333=40.5. This is close to calcium (40.1) but calcium chloride is CaCl₂. Perhaps scandium? ScCl₃, Sc Ar = 45.0. Could be a misprint. Actually a typical example uses 5.40 g of Al to react with Cl₂ to form AlCl₃. Al molar mass 27. 2Al + 3Cl₂ → 2AlCl₃. Check: if Al is 27, moles Al = 5.4/27 = 0.2 mol, reacting with Cl₂: (3/2)*0.2 = 0.3 mol Cl₂, not 0.2. So not that. For MCl₃, metal is likely aluminium, but to get 5.4g and 0.2 mol Cl₂, molar mass must be 40.5, which is not an exact match but could be an approximation. Let’s adjust example to a metal that forms MCl₂ for better match, but question states MCl₃. Keep as is, it’s a calculation exercise. We’ll just calculate and note it could be calcium but valency mismatch, then suggest aluminium with different numbers? I’ll rewrite to avoid confusion: Use a metal that forms MCl₂. Let’s change the formula to MCl₂ and use Ca. Example: 4.00 g metal reacts with 0.100 mol Cl₂ forming MCl₂. Then 2M + Cl₂? Actually M + Cl₂ → MCl₂, 1 mol Cl₂ reacts with 1 mol M. Moles M = 0.100, molar mass = 4.00/0.100=40 g/mol, Ca. So I can change to MCl₂. That is simpler. I’ll edit: unknown metal reacts with 0.100 mol Cl₂ to form MCl₂. 4.00 g metal used. Perfect. So I’ll update the example to: “4.00 g of an unknown metal reacts completely with 0.100 mol of chlorine gas (Cl₂) to form a chloride of formula MCl₂. Determine the molar mass and suggest its identity.” Then equation: M + Cl₂ → MCl₂. moles M = moles Cl₂ = 0.100 mol, molar mass = 4.00/0.100 = 40 g/mol, calcium. This works. Will replace. Good.
重新设计了例题：4.00 g某未知金属与0.100 mol氯气（Cl₂）完全反应，形成MCl₂的氯化物。求其摩尔质量并推测其名称。反应：M + Cl₂ → MCl₂，因此金属的物质的量 = Cl₂的物质的量 = 0.100 mol。摩尔质量 = 4.00 g / 0.100 mol = 40.0 g/mol。该金属是钙（Ca），因为钙的相对原子质量为40.1。
Equation: M + Cl₂ → MCl₂. From the equation, moles of M = moles of Cl₂ = 0.100 mol. Molar mass = mass / moles = 4.00 / 0.100 = 40.0 g/mol. The relative atomic mass of calcium is approximately 40.1, so the unknown metal is calcium.
反应式：M + Cl₂ → MCl₂。根据反应式，M的物质的量= Cl₂的物质的量=0.100 mol。摩尔质量=质量/物质的量=4.00/0.100=40.0 g/mol。钙的相对原子质量约为40.1，因此该未知金属是钙。
4. Electrolysis | 电解
Example: Aqueous copper(II) sulfate is electrolysed using inert graphite electrodes. Write the half-equations for the reactions at the cathode and anode, and describe the observable changes.
例题：使用惰性石墨电极电解硫酸铜水溶液。写出阴极和阳极反应的半方程式，并描述可观察到的变化。
At the cathode: Cu²⁺(aq) + 2e⁻ → Cu(s). Copper ions are reduced to copper metal, which deposits as a pink-brown solid on the cathode. At the anode: 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻. Hydroxide ions are oxidised to oxygen gas; bubbles of a colourless gas are seen. The blue colour of the solution fades as Cu²⁺ ions are discharged and replaced by H⁺ ions from water oxidation, eventually forming sulfuric acid.
阴极：Cu²⁺(aq) + 2e⁻ → Cu(s)。铜离子被还原为金属铜，在阴极上沉积出粉棕色固体。阳极：4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻。氢氧根离子被氧化为氧气；可看到无色气体气泡。随着Cu²⁺离子放电并被水氧化产生的H⁺离子取代，溶液蓝色逐渐褪去，最终形成硫酸。
5. Rates of Reaction | 反应速率
Example: Marble chips (calcium carbonate) react with dilute hydrochloric acid. A student investigates how the surface area of marble chips affects the rate of reaction. Explain why using powdered marble makes the reaction faster, using collision theory.
例题：大理石碎块（碳酸钙）与稀盐酸反应。一位学生研究大理石的表面积如何影响反应速率。请用碰撞理论解释为什么使用大理石粉末会使反应更快。
Powdered marble has a much larger total surface area than the same mass of larger chips. This means more calcium carbonate particles are exposed to the acid at any instant. According to collision theory, a greater surface area increases the frequency of successful collisions between reactant particles per unit time. With more frequent effective collisions, the rate of reaction increases significantly.
大理石粉末的总表面积远大于相同质量的较大颗粒。这意味着在同一时刻有更多的碳酸钙粒子暴露在酸中。根据碰撞理论，表面积增大提高了单位时间内反应物粒子之间成功碰撞的频率。由于有效碰撞更加频繁，反应速率显著增加。
6. Dynamic Equilibrium | 动态平衡
Example: For the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ/mol. Explain the effect of decreasing the temperature on the position of equilibrium and the rate of attainment.
例题：对于哈伯法：N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ/mol。解释降低温度对平衡位置和达到平衡速率的影响。
The forward reaction is exothermic (ΔH negative). Decreasing temperature favours the exothermic direction to release heat and partially counteract the change. Therefore, the equilibrium position shifts to the right, increasing the yield of ammonia. However, lowering the temperature also decreases the kinetic energy of particles, so the rate at which equilibrium is reached becomes slower. Industrially, a compromise temperature (e.g., 450 °C) is used to achieve a reasonable rate while maintaining a good yield.
正向反应是放热反应（ΔH为负）。降低温度有利于放热方向，释放热量以部分抵消这一变化。因此，平衡位置向右移动，氨的产率增加。然而，降低温度也减小了粒子的动能，因此达到平衡的速率变慢。工业上采用折中温度（例如450 °C），在保持较好产率的同时获得合理的反应速率。
7. Acids, Bases and Salts | 酸、碱和盐
Example: Describe a method to prepare pure, dry crystals of copper(II) sulfate pentahydrate (CuSO₄·5H₂O) from copper(II) oxide and dilute sulfuric acid.
例题：描述以氧化铜和稀硫酸为原料制备纯净干燥的五水合硫酸铜晶体（CuSO₄·5H₂O）的方法。
Warm dilute sulfuric acid in a beaker. Add copper(II) oxide powder in small portions while stirring, until no more dissolves (until it is in excess). Filter the hot mixture to remove unreacted copper(II) oxide. Heat the filtrate gently to evaporate some water, then set aside to cool and crystallise. Filter off the blue crystals and dry them between pieces of filter paper.
在烧杯中加热稀硫酸。边搅拌边分批加入氧化铜粉末，直到不再溶解（即过量）。趁热过滤混合物，去除未反应的氧化铜。将滤液温和加热蒸发部分水分，然后静置冷却结晶。过滤得到蓝色晶体，用滤纸吸干。
8. Organic Chemistry: Alkanes and Alkenes | 有机化学：烷烃与烯烃
Example: Ethene (C₂H₄) undergoes an addition reaction with bromine water. (a) Write the chemical equation. (b) State the colour change observed and explain why alkanes do not give this reaction.
例题：乙烯（C₂H₄）与溴水发生加成反应。(a) 写出化学方程式。(b) 描述观察到的颜色变化，并解释为什么烷烃不发生该反应。
(a) C₂H₄ + Br₂ → C₂H₄Br₂ (1,2-dibromoethane). (b) Bromine water is orange-brown; it turns colourless as the bromine adds across the double bond. Alkanes are saturated hydrocarbons containing only C–C single bonds. Unlike alkenes, they lack a reactive C=C double bond, so they cannot undergo addition reactions with bromine under ordinary conditions.
(a) C₂H₄ + Br₂ → C₂H₄Br₂ (1,2-二溴乙烷)。(b) 溴水为橙棕色；随着溴加成到双键上，溶液变为无色。烷烃是饱和烃，只含C–C单键。与烯烃不同，它们没有活泼的C=C双键，因此在普通条件下不能与溴发生加成反应。
9. Extraction of Metals and Reactivity | 金属提取与活泼性
Example: Zinc can be extracted from its ore by heating with carbon, but aluminium must be extracted by electrolysis. Explain this difference using the reactivity series.
例题：锌可通过用碳加热矿石提取，而铝必须用电解法提取。请用金属活泼性顺序解释这一差异。
Zinc is below carbon in the reactivity series. Carbon can displace zinc from its oxide because carbon is more reactive and will reduce zinc oxide to zinc metal. Aluminium is above carbon in the reactivity series; it is more reactive than carbon, so carbon cannot displace it. Therefore, electrolysis of molten aluminium oxide (in cryolite) is necessary to reduce aluminium ions to the metal.
锌在金属活泼性顺序中排在碳之后。碳可以从氧化锌中置换出锌，因为碳更活泼，能将氧化锌还原为金属锌。铝在活泼性顺序中排在碳之前，其活泼性高于碳，因此碳无法将其置换。所以必须电解熔融的氧化铝（在冰晶石中）来将铝离子还原为金属。
10. Experimental Skills and Data Interpretation | 实验技能与数据解读
Example: A student measures the temperature change when different volumes of hydrochloric acid are added to 25 cm³ of sodium hydroxide solution. The results show a maximum temperature rise when exactly 25.0 cm³ of acid is used. Explain how these results illustrate neutralisation stoichiometry and suggest a safety precaution.
例题：一名学生测量了将不同体积的盐酸加入25 cm³氢氧化钠溶液中时的温度变化。结果显示，当恰好使用25.0 cm³ 酸时，温度升高最大。解释这些结果如何体现中和反应的化学计量关系，并提出一项安全预防措施。
The balanced equation is HCl + NaOH → NaCl + H₂O, a 1:1 molar ratio. The maximum temperature rise occurs at the exact volume where the moles of acid equal the moles of alkali, i.e., the endpoint of neutralisation. Before this point, the limiting reagent is acid; after it, the acid is in excess, and no further neutralisation occurs, so the temperature falls. A safety precaution: wear safety goggles and a lab coat, as both acid and alkali are corrosive.
反应方程式为 HCl + NaOH → NaCl + H₂O，摩尔比为1:1。最大温升出现在酸的体积恰好使酸与碱的物质的量相等时，即中和反应的终点。在此点之前，限制试剂是酸；超过此点后，酸过量，不再发生中和反应，因此温度下降。安全预防措施：佩戴护目镜和实验服，因为酸和碱均有腐蚀性。
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