IGCSE OCR Chemistry: Worked Examples Explained | IGCSE OCR 化学：典型例题详解

📚 IGCSE OCR Chemistry: Worked Examples Explained | IGCSE OCR 化学：典型例题详解

Mastering IGCSE OCR Chemistry requires not only a solid grasp of the theory but also the ability to apply concepts to exam-style questions. This article presents a series of carefully chosen worked examples that span the core topics of the OCR specification, from atomic structure and bonding to stoichiometry, energetics, and organic chemistry. Each problem is broken down step by step, with clear reasoning to help you build confidence and ace your exams.

要掌握IGCSE OCR化学，不仅需要扎实地理解理论，还要有能力将概念应用到考试题型中。本文精选了一系列典型例题，涵盖了OCR考纲的核心主题，从原子结构、化学键，到化学计量学、能量变化以及有机化学。每道题都进行了逐步拆解，并附有清晰的解题思路，帮助你建立信心，在考试中取得优异成绩。

1. Atomic Structure and Isotopes | 原子结构与同位素

Question: An atom of element X has a mass number of 39 and contains 20 neutrons. Determine the number of protons, electrons, and the identity of the element. Then write the isotopic notation for an isotope of the same element that has 21 neutrons.

题目：某元素 X 的一个原子质量数为 39，含有 20 个中子。请确定其质子数、电子数及元素名称，然后写出该元素另一种含有 21 个中子的同位素的核素符号。

The atomic (proton) number is found by subtracting the number of neutrons from the mass number: 39 − 20 = 19. Therefore the atom has 19 protons and, being neutral, 19 electrons. The element with atomic number 19 is potassium (K). An isotope with 21 neutrons has the same proton number (19) but a mass number of 19 + 21 = 40. Its isotopic notation is ⁴⁰₁₉K.

原子序数（质子数）等于质量数减去中子数：39 − 20 = 19。因此该原子有 19 个质子，由于呈电中性，也有 19 个电子。原子序数为 19 的元素是钾 (K)。含有 21 个中子的同位素具有相同的质子数 (19)，质量数为 19 + 21 = 40，其核素符号为 ⁴⁰₁₉K。

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2. Ionic Bonding and Formula | 离子键与化学式

Question: Explain how the elements magnesium (Mg) and chlorine (Cl) form an ionic compound. Write the formula of the compound and describe the change in electron arrangement for each atom.

题目：请解释镁 (Mg) 和氯 (Cl) 是如何形成离子化合物的。写出该化合物的化学式，并描述每个原子的电子排布变化。

Magnesium has the electron configuration 2,8,2. It loses its two outer electrons to achieve the stable electronic structure of neon (2,8), forming a Mg²⁺ ion. Each chlorine atom has the configuration 2,8,7 and gains one electron to achieve the argon structure (2,8,8), forming a Cl⁻ ion. Two chlorine atoms are needed to accept the two electrons from one magnesium atom, so the formula is MgCl₂.

镁原子的电子排布为 2,8,2。它失去最外层的两个电子，达到氖的稳定电子结构 (2,8)，形成 Mg²⁺ 离子。每个氯原子的电子排布为 2,8,7，得到一个电子后达到氩的结构 (2,8,8)，形成 Cl⁻ 离子。一个镁原子失去的两个电子需要由两个氯原子分别接受，因此化学式为 MgCl₂。

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3. Mole and Mass Calculations | 摩尔与质量计算

Question: Calculate the mass of calcium oxide (CaO) that can be produced when 10.0 g of calcium carbonate (CaCO₃) is heated strongly. The equation is: CaCO₃ → CaO + CO₂. (Relative atomic masses: Ca = 40, C = 12, O = 16)

题目：计算将 10.0 g 碳酸钙 (CaCO₃) 强热分解时，可获得多少质量的氧化钙 (CaO)。反应方程式为：CaCO₃ → CaO + CO₂。（相对原子质量：Ca=40, C=12, O=16）

First, calculate the molar mass of CaCO₃: 40 + 12 + (3 × 16) = 100 g/mol. Molar mass of CaO: 40 + 16 = 56 g/mol. Moles of CaCO₃ used = mass / molar mass = 10.0 / 100 = 0.100 mol. From the equation, the mole ratio of CaCO₃ to CaO is 1:1, so 0.100 mol of CaO forms. Mass of CaO = moles × molar mass = 0.100 × 56 = 5.60 g.

首先计算 CaCO₃ 的摩尔质量：40 + 12 + (3×16) = 100 g/mol。CaO 的摩尔质量为 40 + 16 = 56 g/mol。所用 CaCO₃ 的物质的量 = 质量 / 摩尔质量 = 10.0 / 100 = 0.100 mol。根据方程式，CaCO₃ 与 CaO 的摩尔比为 1:1，因此生成 0.100 mol CaO。CaO 的质量 = 物质的量 × 摩尔质量 = 0.100 × 56 = 5.60 g。

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4. Electrolysis Product Prediction | 电解产物判断

Question: Aqueous copper(II) sulfate is electrolysed using inert graphite electrodes. State the products formed at the cathode and anode, and write the half-equation for the cathode reaction.

题目：用惰性石墨电极电解硫酸铜 (CuSO₄) 水溶液。请说出阴极和阳极的产物，并写出阴极反应的半反应方程式。

In the solution, Cu²⁺ and H⁺ ions are attracted to the cathode, while SO₄²⁻ and OH⁻ ions go to the anode. Copper is less reactive than hydrogen, so copper is discharged at the cathode: Cu²⁺ + 2e⁻ → Cu. At the anode, where oxidation occurs, OH⁻ ions are discharged in preference to SO₄²⁻ because the hydroxide ion is easier to oxidise. The anode half-equation is: 4OH⁻ → O₂ + 2H₂O + 4e⁻. The products are copper metal at the cathode and oxygen gas at the anode.

溶液中，Cu²⁺ 和 H⁺ 离子移向阴极，SO₄²⁻ 和 OH⁻ 离子移向阳极。铜比氢更不活泼，因此阴极析出铜：Cu²⁺ + 2e⁻ → Cu。阳极发生氧化反应，OH⁻ 离子优先于 SO₄²⁻ 放电，因为氢氧根更容易被氧化。阳极半反应为：4OH⁻ → O₂ + 2H₂O + 4e⁻。产物分别为阴极的金属铜和阳极的氧气。

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5. Bond Energy and Enthalpy Change | 键能与焓变

Question: Using the bond energies given, calculate the enthalpy change (ΔH) for the reaction: H₂ + Cl₂ → 2HCl. Bond energies (kJ/mol): H−H = 436, Cl−Cl = 242, H−Cl = 431.

题目：利用所给的键能数据，计算反应 H₂ + Cl₂ → 2HCl 的焓变 (ΔH)。键能 (kJ/mol)：H−H=436，Cl−Cl=242，H−Cl=431。

Energy is absorbed to break bonds (endothermic) and released when bonds form (exothermic). Bonds broken: 1 × H−H = 436 kJ, 1 × Cl−Cl = 242 kJ; total energy in = 436 + 242 = 678 kJ. Bonds made: 2 × H−Cl = 2 × 431 = 862 kJ; total energy out = 862 kJ. ΔH = energy in − energy out = 678 − 862 = −184 kJ. The negative sign shows the reaction is exothermic.

断键吸收能量（吸热），成键释放能量（放热）。断裂的键：1 个 H−H = 436 kJ，1 个 Cl−Cl = 242 kJ；吸收总能量 = 436 + 242 = 678 kJ。生成的键：2 个 H−Cl = 2 × 431 = 862 kJ；释放总能量 = 862 kJ。ΔH = 吸收能量 − 释放能量 = 678 − 862 = −184 kJ。负号表明反应为放热反应。

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6. Titration Calculation | 滴定计算

Question: 25.0 cm³ of sodium hydroxide solution of unknown concentration is neutralised by 20.0 cm³ of 0.100 mol/dm³ hydrochloric acid. Calculate the concentration of the NaOH solution in mol/dm³ and g/dm³. (Na = 23, O = 16, H = 1)

题目：25.0 cm³ 未知浓度的氢氧化钠溶液被 20.0 cm³ 0.100 mol/dm³ 的盐酸完全中和。计算 NaOH 溶液的浓度，分别以 mol/dm³ 和 g/dm³ 表示。（Na=23, O=16, H=1）

The neutralisation equation is: NaOH + HCl → NaCl + H₂O. Mole ratio NaOH:HCl = 1:1. Moles of HCl = concentration × volume (in dm³) = 0.100 × (20.0/1000) = 0.00200 mol. Therefore moles of NaOH = 0.00200 mol. Concentration of NaOH = moles / volume in dm³ = 0.00200 / (25.0/1000) = 0.0800 mol/dm³. Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Concentration in g/dm³ = 0.0800 × 40 = 3.20 g/dm³.

中和反应方程式：NaOH + HCl → NaCl + H₂O。NaOH 与 HCl 的摩尔比为 1:1。HCl 的物质的量 = 浓度 × 体积 (dm³) = 0.100 × (20.0/1000) = 0.00200 mol，因此 NaOH 的物质的量也为 0.00200 mol。NaOH 的浓度 = 物质的量 / 体积（dm³）= 0.00200 / (25.0/1000) = 0.0800 mol/dm³。NaOH 的摩尔质量 = 23 + 16 + 1 = 40 g/mol，质量浓度 = 0.0800 × 40 = 3.20 g/dm³。

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7. Rate of Reaction and Collision Theory | 反应速率与碰撞理论

Question: Explain, using collision theory, why increasing the concentration of a reactant increases the rate of reaction. Support your answer with a particle diagram description.

题目：运用碰撞理论解释，为什么增大反应物的浓度会提高反应速率。请结合粒子图示的描述加以说明。

For a reaction to occur, reactant particles must collide with energy equal to or greater than the activation energy. Increasing concentration means there are more particles per unit volume. This increases the frequency of collisions. With more frequent collisions, the number of successful collisions per unit time also rises, leading to a higher rate of reaction. A particle diagram would show a closer packing of particles in a given volume, reducing the distance between them and making collisions more likely.

反应发生需要反应物粒子发生碰撞，且碰撞能量大于或等于活化能。增大浓度意味着单位体积内粒子数增多，碰撞频率随之增加。由于碰撞更频繁，单位时间内有效碰撞的次数也增多，从而导致反应速率提高。粒子图将显示相同体积内粒子排列更紧密，粒子间距离缩短，碰撞更易发生。

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8. Group 1 Reactivity Trend | 第 1 族反应活性趋势

Question: Describe and explain the trend in reactivity as you go down Group 1 of the Periodic Table. Use the reaction of potassium with water as an example, including the word and symbol equations.

题目：描述并解释第 1 族元素从上到下反应活性的变化趋势。以钾与水的反应为例，写出文字方程式和符号方程式。

Reactivity increases down Group 1. This is because the outer electron becomes further from the nucleus as the number of shells increases. The increased distance and greater shielding by inner electrons reduce the electrostatic attraction between the nucleus and the outer electron, so the electron is more easily lost. Potassium reacts vigorously with water, floating, fizzing, and igniting with a lilac flame. Word equation: potassium + water → potassium hydroxide + hydrogen. Symbol equation: 2K + 2H₂O → 2KOH + H₂.

随着第 1 族从上到下，反应活性逐渐增强。这是因为原子层数增加，最外层电子离核更远。距离增大以及内层电子的屏蔽作用增强，削弱了原子核与最外层电子之间的静电引力，使得电子更易失去。钾与水剧烈反应，浮在水面、冒泡，并产生淡紫色火焰。文字方程式：钾 + 水 → 氢氧化钾 + 氢气。符号方程式：2K + 2H₂O → 2KOH + H₂。

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9. Organic Chemistry – Addition Reaction | 有机化学 – 加成反应

Question: Ethene (C₂H₄) reacts with bromine (Br₂) in an addition reaction. Draw the structural formula of the product and explain why this reaction can be used to distinguish alkanes from alkenes.

题目：乙烯 (C₂H₄) 与溴 (Br₂) 发生加成反应。画出产物的结构式，并解释为什么该反应可用于区分烷烃和烯烃。

Ethene has a carbon–carbon double bond (C=C). In the addition reaction, the double bond opens up and each carbon atom joins to a bromine atom. The product is 1,2-dibromoethane, with the structural formula CH₂Br−CH₂Br. Alkanes, which have only single bonds, do not react with bromine under normal conditions. If bromine water (orange-brown) is added to an alkene, it is decolourised immediately; with an alkane, the colour remains. This makes it a simple chemical test for unsaturation.

乙烯含有碳碳双键 (C=C)。在加成反应中，双键打开，每个碳原子连接一个溴原子。产物为 1,2-二溴乙烷，结构式为 CH₂Br−CH₂Br。烷烃仅含单键，在通常条件下不与溴反应。若将溴水（橙棕色）加入烯烃中，溴水会立即褪色；而加入烷烃中，颜色保持不变。这使它成为一种检验不饱和键的简单化学方法。

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10. Gas Volume and Molar Volume | 气体体积与摩尔体积

Question: At room temperature and pressure (RTP), one mole of any gas occupies 24 dm³. Calculate the volume of carbon dioxide gas produced when 6.0 g of ethanoic acid (CH₃COOH) is completely burned in excess oxygen. (Mᵣ of ethanoic acid = 60)

题目：在常温常压 (RTP) 下，1 摩尔任何气体的体积约为 24 dm³。计算 6.0 g 乙酸 (CH₃COOH) 在过量氧气中完全燃烧时产生的二氧化碳气体体积。（乙酸的相对分子质量 = 60）

First, write the balanced combustion equation: CH₃COOH + 2O₂ → 2CO₂ + 2H₂O. Moles of ethanoic acid = mass / Mᵣ = 6.0 / 60 = 0.10 mol. From the equation, the mole ratio of acid to CO₂ is 1:2, so moles of CO₂ = 0.10 × 2 = 0.20 mol. At RTP, volume of CO₂ = moles × 24 dm³/mol = 0.20 × 24 = 4.8 dm³.

首先写出配平的燃烧方程式：CH₃COOH + 2O₂ → 2CO₂ + 2H₂O。乙酸的物质的量 = 质量 / 相对分子质量 = 6.0 / 60 = 0.10 mol。根据方程式，乙酸与 CO₂ 的摩尔比为 1:2，因此 CO₂ 的物质的量 = 0.10 × 2 = 0.20 mol。在 RTP 下，CO₂ 的体积 = 物质的量 × 24 dm³/mol = 0.20 × 24 = 4.8 dm³。

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Published by TutorHao | Chemistry Revision Series | aleveler.com

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