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

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

This article walks through carefully selected worked examples that mirror the style and challenge of OCR GCSE Chemistry papers. Each question is broken down with clear reasoning and step-by-step calculations, helping you build confidence with the command words, mathematical demands, and practical scenarios that regularly appear.

本文精选了与 OCR GCSE 化学试卷风格和难度高度一致的典型例题，并逐一详细解析。每道题都配有清晰的推理和分步计算，帮助你熟悉常见指令词、数学要求和实验情景，提升应试信心。

1. Balancing Equations and State Symbols | 配平化学方程式与状态符号

Question: Aluminium reacts with iron(III) oxide in the thermite reaction. Write the balanced chemical equation, including state symbols, for this reaction. (3 marks)

问题： 铝与氧化铁(III)在铝热反应中发生反应。写出该反应的配平化学方程式，并标明状态符号。（3分）

nd iron oxide are both solids at the start, and the reaction produces molten iron (a liquid) and aluminium oxide (solid). Start by balancing the atoms.

未配平的反应为：Al + Fe₂O₃ → Al₂O₃ + Fe。注意反应开始时铝和氧化铁均为固体，反应生成熔融的铁（液体）和氧化铝（固体）。首先配平原子。

e symbols are essential: aluminium (s), iron(III) oxide (s), aluminium oxide (s), iron (l) because the reaction is highly exothermic and melts the iron.

右边有2个Al，所以在左边Al前加上系数2。Fe：左边2个，所以右边Fe前加上系数2。配平的方程式为：2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(l)。状态符号至关重要：铝(s)、氧化铁(III)(s)、氧化铝(s)、铁(l)，因为该反应高度放热，使铁熔化。

2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(l)

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2. Ionic Bonding and Formula of an Ionic Compound | 离子键与离子化合物化学式

Question: Magnesium and chlorine react to form magnesium chloride. Describe, in terms of electrons, what happens when a magnesium atom and two chlorine atoms form ionic bonds. Give the formula of the product. (4 marks)

问题： 镁和氯反应生成氯化镁。从电子角度描述，一个镁原子与两个氯原子如何形成离子键。写出产物的化学式。（4分）

A magnesium atom has electron configuration 2,8,2. It loses its two outer electrons to achieve a stable octet like neon. This forms a Mg²⁺ cation. Each chlorine atom has electron configuration 2,8,7 and needs one electron to achieve a stable octet (argon-like). Each chlorine atom gains one electron to form a Cl⁻ anion.

镁原子的电子排布为2,8,2。它失去两个最外层电子，达到类似氖的稳定八隅体，形成Mg²⁺阳离子。每个氯原子的电子排布为2,8,7，需要获得一个电子才能达到稳定八隅体（类似氩）。每个氯原子获得一个电子，形成Cl⁻阴离子。

Because magnesium loses two electrons and each chlorine gains one electron, two chlorine atoms are needed for every one magnesium atom. The electrostatic attraction between the oppositely charged Mg²⁺ and two Cl⁻ ions holds the ionic lattice together. The formula is MgCl₂.

因为镁失去两个电子，而每个氯只获得一个电子，所以每需要一个镁原子就需要两个氯原子。带相反电荷的Mg²⁺与两个Cl⁻离子之间的静电引力维持着离子晶格。化学式为MgCl₂。

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3. Mole Calculations Using Mass and Mᵣ | 利用质量与相对分子质量进行摩尔计算

Question: 9.2 g of ethanol (C₂H₅OH) is burned completely in oxygen. (Aᵣ: H=1, C=12, O=16) (4 marks)

问题： 计算9.2g乙醇（C₂H₅OH）在氧气中完全燃烧时产生的二氧化碳的质量。（Aᵣ: H=1, C=12, O=16）（4分）

of ethanol: (2×12) + (6×1) + 16 = 46. Moles of ethanol = mass / Mᵣ = 9.2 / 46 = 0.20 mol.

首先写出配平方程式：C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O。接着计算乙醇的相对分子质量：(2×12) + (6×1) + 16 = 46。乙醇的物质的量 = 质量 / 相对分子质量 = 9.2 / 46 = 0.20 mol。

2 = 0.40 mol. Mᵣ of CO₂ = 12 + (2×16) = 44. Mass of CO₂ = moles × Mᵣ = 0.40 × 44 = 17.6 g.

由方程式可知，1 mol乙醇产生2 mol CO₂。因此CO₂的物质的量 = 0.20 × 2 = 0.40 mol。CO₂的相对分子质量 = 12 + (2×16) = 44。CO₂的质量 = 物质的量 × 相对分子质量 = 0.40 × 44 = 17.6 g。

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4. Electrolysis of Aqueous Solutions: Predicting Products | 水溶液电解：预测产物

Question: A student electrolyses aqueous copper(II) sulfate using inert graphite electrodes. Predict the products at the anode and at the cathode. Include half-equations in your answer. (4 marks)

问题： 学生用惰性石墨电极电解硫酸铜(II)水溶液。预测阳极和阴极的产物，并写出相应的半反应方程式。（4分）

pper is below hydrogen), so Cu²⁺ gains electrons: Cu²⁺ + 2e⁻ → Cu (copper metal deposits).

确定溶液中存在的离子：来自CuSO₄(aq) → Cu²⁺、SO₄²⁻；来自水 → H⁺、OH⁻。在阴极（负极），金属活动性顺序中在氢之后且较容易得电子的阳离子优先放电。Cu²⁺比H⁺更容易放电（铜在氢之后），因此Cu²⁺得到电子：Cu²⁺ + 2e⁻ → Cu（析出金属铜）。

e: 4OH⁻ → O₂ + 2H₂O + 4e⁻. Thus oxygen gas is produced. The sulfate ion stays in solution.

在阳极（正极），易氧化的阴离子优先放电。在SO₄²⁻与OH⁻之间，OH⁻优先放电（除非存在卤离子）。阳极反应：4OH⁻ → O₂ + 2H₂O + 4e⁻。因此产生氧气。硫酸根离子留在溶液中。

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5. Rate of Reaction: Interpreting Graphs and Collision Theory | 反应速率：图像分析与碰撞理论

Question: Magnesium ribbon reacts with excess dilute hydrochloric acid. The volume of hydrogen gas produced is measured every 10 seconds. Explain why the rate of reaction is fastest at the start and then decreases, using ideas from collision theory. (3 marks)

问题： 镁条与过量稀盐酸反应，每10秒测量一次生成氢气的体积。利用碰撞理论解释，为什么反应速率在开始时最快，随后逐渐减慢？（3分）

At the start, the concentration of hydrochloric acid is highest. There are more H⁺ ions per unit volume, so the frequency of successful collisions between H⁺ ions and magnesium atoms is greatest. This gives the fastest rate.

开始时，盐酸浓度最高。单位体积内的H⁺离子数量最多，因此H⁺离子与镁原子之间成功碰撞的频率最高，反应速率最快。

As the reaction proceeds, acid is used up. The concentration of H⁺ ions decreases, leading to fewer collisions per second. The frequency of successful collisions falls, so the rate decreases. Eventually the magnesium is used up or the acid reacts completely, and the reaction stops.

随着反应进行，酸被消耗。H⁺离子浓度降低，导致每秒碰撞次数减少。成功碰撞的频率下降，因此速率减慢。最终镁耗尽或酸完全反应，反应停止。

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6. Energy Changes: Bond Energy Calculations | 能量变化：键能计算

Question: Hydrogen reacts with chlorine: H₂ + Cl₂ → 2HCl. Use the bond energies (H–H 436 kJ/mol, Cl–Cl 242 kJ/mol, H– hange for the reaction. State whether the reaction is exothermic or endothermic. (3 marks)

问题： 氢气与氯气反应：H₂ + Cl₂ → 2HCl。利用以下键能（H–H 436 kJ/mol、Cl–Cl 242 kJ/mol、H–Cl 431 kJ/mol）计算该反应的总能量变化，并说明反应是放热还是吸热。（3分）

Energy required to break bonds: 1 mole H–H bonds (436 kJ) + 1 mole Cl–Cl bonds (242 kJ) = +678 kJ (endothermic). Energy released when making bonds: 2 moles of H–Cl bonds formed, so 2 × 431 = –862 kJ (exothermic).

断裂化学键吸收的能量：1 mol H–H键 (436 kJ) + 1 mol Cl–Cl键 (242 kJ) = +678 kJ（吸热）。形成化学键释放的能量：生成2 mol H–Cl键，即2 × 431 = –862 kJ（放热）。

Overall energy change = energy in − energy out = +678 + (−862) = – gn shows the reaction releases more energy than it absorbs, so the reaction is exothermic.

总能量变化 = 吸收能量 − 释放能量 = +678 + (−862) = –184 kJ。负号表明反应释放的能量多于吸收的能量，因此该反应为放热反应。

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7. Chromatography and Rf Values | 色谱法与Rf值

Question: A student performs paper chromatography on four known dyes (A–D) and an unknown mixture X. The solvent front moved 7.5 cm. Dye C travelled 3.0 cm. Calculate the Rf value for dye C and explain how the student can use the chromatogram to identify which of A–D are present in X. (3 marks)

问题： 学生对四种已知染料（A–D）和一种未知混合物X进行纸色谱分析。溶剂前沿移动了7.5 cm。染料C移动了3.0 cm。计算染料C的Rf值，并解释学生如何利用色谱图确定X中含有A–D中的哪些物质。（3分）

by solvent front = 3.0 / 7.5 = 0.40. The Rf value has no units and is always less than 1.

Rf = 物质移动的距离 / 溶剂前沿移动的距离 = 3.0 / 7.5 = 0.40。Rf值无单位，且始终小于1。

To identify components in X, the student compares the number of spots and their positions. Dyes in X will have Rf values matching one or more of A–D. If a spot from X travels the same distance as dye C (or any known dye) under the same conditions, that dye is present in the mixture.

要鉴定X中的成分，学生需比较斑点的数量及其位置。X中的染料所具有的Rf值将与A–D中的一种或多种匹配。若在相同条件下，X中的一个斑点与染料C（或任一已知染料）移动的距离相同，则该染料存在于混合物中。

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8. Crude Oil and Fractional Distillation | 原油与分馏

Question: Explain how fractional distillation separates crude oil into useful fractions. Refer to boiling points, condensation, and the trend in molecular size. (4 marks)

问题： 解释分馏如何将原油分离成有用的馏分。请涉及沸点、冷凝以及分子大小的变化趋势。（4分）

Crude oil is heated and enters the fractionating column as a mixture of vapours. The column is hottest at the bottom and coolest at the top. Larger hydrocarbon molecules have stronger intermolecular forces and higher boiling points, so they condense lower down the column.

原油被加热，以蒸气混合物的形式进入分馏塔。塔底温度最高，塔顶温度最低。较大的碳氢化合物分子间作用力较强，沸点较高，因此它们在塔的较低位置冷凝。

Smaller hydrocarbon molecules have weaker intermolecular forces and lower boiling points, so they rise higher and condense at the top. Fractions with similar boiling point ranges are collected on different trays, separating the mixture into gases, gasoline, kerosene, diesel, and bitumen.

较小的碳氢化合物分子间作用力较弱，沸点较低，因此它们上升得更高并在塔顶冷凝。沸程相近的各馏分在不同塔盘上收集，从而将混合物分离为气体、汽油、煤油、柴油和沥青等。

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9. Testing for Gases: Describing Observations | 气体检验：描述现象

Question: Describe the chemical test and the expected observation for carbon dioxide gas. Include any necessary reagent or equipment. (2 marks)

问题： 描述二氧化碳气体的化学检验方法及预期现象，包括所需的试剂或仪器。（2分）

Bubble the gas through (or shake it with) limewater (calcium hydroxide solution). If carbon dioxide is present, the limewater turns milky/cloudy because a white precipitate of calcium carbonate is formed.

将气体通入（或与）石灰水（氢氧化钙溶液）振荡。若存在二氧化碳，石灰水变浑浊/呈乳白色，因为生成了白色碳酸钙沉淀。

H₂O(l). This test is linked to the marble chip and acid reaction often seen in practicals.

相关方程式：Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l)。该检验常与实验中大理石与酸的反应相关。

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10. Calculating Percentage Yield | 计算产率

Question: → MgO + Cu. (Aᵣ: Mg=24, Cu=63.5, O=16) Calculate the percentage yield of copper. (3 marks)

问题： 4.8 g镁与过量氧化铜(II)反应，得到10.5 g铜。反应方程式为：Mg + CuO → MgO + Cu。（Aᵣ: Mg=24, Cu=63.5, O=16）计算铜的产率。（3分）

etical moles of Cu = 0.20 mol. Theoretical mass of Cu = moles × Aᵣ = 0.20 × 63.5 = 12.7 g.

所用Mg的物质的量 = 质量 / Aᵣ = 4.8 / 24 = 0.20 mol。由方程式可知，1 mol Mg产生1 mol Cu，因此 Cu理论物质的量 = 0.20 mol。Cu理论质量 = 物质的量 × Aᵣ = 0.20 × 63.5 = 12.7 g。

Percentage yield = (actual mass / theoretical mass) × 100 = (10.5 / 12.7) × 100 ≈ 82.7%. This accounts for potential losses during filtration/transfer or incomplete reaction.

产率 = (实际质量 / 理论质量) × 100 = (10.5 / 12.7) × 100 ≈ 82.7%。这考虑了过滤/转移过程中的损失或反应不完全的情况。

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