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

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

This article provides detailed worked examples covering key topics in the GCSE Edexcel Chemistry specification. Each example breaks down the problem-solving process, helping you to understand the application of core concepts and excel in your examinations.

本文提供了覆盖 GCSE Edexcel 化学大纲核心主题的详细例题解析。每个例题都分解了解题过程，帮助你理解核心概念的应用，并在考试中取得优异成绩。

1. Atomic Structure: Calculating Particles | 原子结构：粒子数计算

Example: An atom of sodium has an atomic number of 11 and a mass number of 23. Determine the number of protons, neutrons and electrons in a neutral sodium atom.

例题： 钠原子的原子序数为 11，质量数为 23。计算中性钠原子中的质子数、中子数和电子数。

Solution:

Step 1: Protons = atomic number = 11.

第一步： 质子数 = 原子序数 = 11。

Step 2: For a neutral atom, electrons = protons = 11.

第二步： 中性原子中，电子数 = 质子数 = 11。

Step 3: Neutrons = mass number – atomic number = 23 – 11 = 12.

第三步： 中子数 = 质量数 – 原子序数 = 23 – 11 = 12。

Answer: 11 protons, 11 electrons, 12 neutrons.

答案： 11个质子，11个电子，12个中子。

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

Example: Explain how ionic bonding occurs between sodium and chlorine to form sodium chloride, and write the formula of magnesium oxide.

例题： 解释钠与氯之间如何通过离子键形成氯化钠，并写出氧化镁的化学式。

Solution:

Step 1: Sodium (Na) has one electron in its outer shell. It loses this electron to achieve a stable noble gas electronic configuration, forming a Na⁺ ion.

第一步： 钠（Na）最外层有一个电子。它失去这个电子，获得稳定的惰性气体电子结构，形成 Na⁺ 离子。

Step 2: Chlorine (Cl) has seven outer electrons. It gains one electron to complete its outer shell, forming a Cl⁻ ion.

第二步： 氯（Cl）最外层有七个电子。它获得一个电子填满最外层，形成 Cl⁻ 离子。

Step 3: Electrostatic attraction between the oppositely charged ions forms the ionic bond, giving NaCl with a 1:1 ratio.

第三步： 带相反电荷离子之间的静电引力形成离子键，NaCl 中离子比例为 1:1。

Step 4: For magnesium oxide: Mg loses two electrons to become Mg²⁺; O gains two electrons to become O²⁻. The formula is MgO.

第四步： 对于氧化镁：Mg 失去两个电子变成 Mg²⁺；O 获得两个电子变成 O²⁻。化学式为 MgO。

Answer: NaCl is formed by transfer of one electron from Na to Cl; MgO is formed by transfer of two electrons from Mg to O, giving a 1:1 ratio.

答案： NaCl 由 Na 向 Cl 转移一个电子形成；MgO 由 Mg 向 O 转移两个电子形成，离子比例为 1:1。

3. Covalent Bonding: Dot and Cross Diagrams | 共价键：点叉图

Example: Draw dot and cross diagrams to show the covalent bonding in H₂O, CO₂ and CH₄. Explain why these compounds have the displayed shapes.

例题： 用点叉图表示 H₂O、CO₂ 和 CH₄ 中的共价键，并解释它们为何具有特定的分子形状。

Solution:

Step 1: In H₂O, oxygen shares one electron with each hydrogen atom, forming two single covalent bonds. Oxygen has two lone pairs, which repel, creating a bent shape with a bond angle of about 104.5°.

第一步： 在 H₂O 中，氧原子与每个氢原子共享一个电子，形成两个单共价键。氧原子有两对孤对电子，孤对电子排斥成键电子，使分子成为角形，键角约 104.5°。

Step 2: In CO₂, carbon forms two double bonds with two oxygen atoms (each double bond shares two pairs of electrons). There are no lone pairs on carbon, so the molecule is linear with an O═C═O arrangement.

第二步： 在 CO₂ 中，碳与两个氧原子形成两个双键（每个双键共享两对电子）。碳上没有孤对电子，因此分子呈直线形，O═C═O。

Step 3: In CH₄, carbon shares one electron with each of four hydrogen atoms, forming four single covalent bonds. With no lone pairs, the four bonding pairs repel equally, giving a tetrahedral shape with bond angles of 109.5°.

第三步： 在 CH₄ 中，碳与四个氢原子各共享一个电子，形成四个单共价键。没有孤对电子，四对成键电子均等排斥，形成正四面体形，键角 109.5°。

Dot and cross description: In diagrams, dots represent electrons from one atom, crosses represent electrons from the other atom. Overlapping circles show shared pairs.

点叉图描述： 在图中，点代表一个原子的电子，叉代表另一个原子的电子。重叠区域表示共享电子对。

4. Balancing Equations and Moles | 方程式配平与摩尔

Example: Balance the equation Fe + Cl₂ → FeCl₃. Then calculate the amount in moles of FeCl₃ produced when 0.5 mol of Fe reacts with excess chlorine.

例题： 配平方程式 Fe + Cl₂ → FeCl₃。然后计算 0.5 mol 铁与过量氯气反应时生成的 FeCl₃ 的摩尔量。

Solution:

Step 1: Count atoms: LHS 1 Fe, 2 Cl; RHS 1 Fe, 3 Cl. To balance Cl, find the lowest common multiple of 2 and 3, which is 6. Put 3 in front of Cl₂ and 2 in front of FeCl₃: Fe + 3Cl₂ → 2FeCl₃.

第一步： 计算原子数：左边 1 Fe, 2 Cl；右边 1 Fe, 3 Cl。为配平氯原子，取 2 和 3 的最小公倍数 6。在 Cl₂ 前加 3，在 FeCl₃ 前加 2：Fe + 3Cl₂ → 2FeCl₃。

Step 2: Now Fe is unbalanced: 1 Fe on left, 2 on right. Add coefficient 2 before Fe: 2Fe + 3Cl₂ → 2FeCl₃.

第二步： 现在铁原子不平衡：左边 1 Fe，右边 2 Fe。在 Fe 前加系数 2：2Fe + 3Cl₂ → 2FeCl₃。

Step 3: Mole ratio: 2 mol Fe produces 2 mol FeCl₃ (1:1). Therefore, 0.5 mol Fe produces 0.5 mol FeCl₃.

第三步： 摩尔比：2 mol Fe 生成 2 mol FeCl₃（1:1）。因此 0.5 mol Fe 生成 0.5 mol FeCl₃。

Answer: Balanced equation: 2Fe + 3Cl₂ → 2FeCl₃. Amount of FeCl₃ = 0.5 mol.

答案： 配平方程式：2Fe + 3Cl₂ → 2FeCl₃。生成的 FeCl₃ 量为 0.5 mol。

5. Reacting Mass Calculations | 反应质量计算

Example: Calcium carbonate decomposes on heating: CaCO₃ → CaO + CO₂. Calculate the mass of calcium oxide produced when 50 g of calcium carbonate is heated completely. (Aᵣ values: Ca=40, C=12, O=16)

例题： 碳酸钙加热分解：CaCO₃ → CaO + CO₂。计算完全加热 50 g 碳酸钙时生成的氧化钙质量。（Aᵣ 值：Ca=40, C=12, O=16）

Solution:

Step 1: Calculate the relative formula mass (Mᵣ) of CaCO₃: 40 + 12 + (3 × 16) = 100 g/mol.

第一步： 计算 CaCO₃ 的相对分子质量（Mᵣ）：40 + 12 + (3 × 16) = 100 g/mol。

Step 2: Mᵣ of CaO: 40 + 16 = 56 g/mol.

第二步： CaO 的 Mᵣ：40 + 16 = 56 g/mol。

Step 3: Mole ratio from equation: 1 mol CaCO₃ → 1 mol CaO. Moles of CaCO₃ used = mass / Mᵣ = 50 / 100 = 0.5 mol.

第三步： 方程式摩尔比：1 mol CaCO₃ 生成 1 mol CaO。已用 CaCO₃ 的摩尔数 = 质量 / Mᵣ = 50 / 100 = 0.5 mol。

Step 4: Moles of CaO produced = 0.5 mol. Mass of CaO = moles × Mᵣ = 0.5 × 56 = 28 g.

第四步： 生成的 CaO 摩尔数 = 0.5 mol。CaO 质量 = 摩尔数 × Mᵣ = 0.5 × 56 = 28 g。

Answer: 28 g of calcium oxide.

答案： 28 g 氧化钙。

6. Energy Changes: Exothermic and Endothermic | 能量变化：放热与吸热

Example: Using the bond energies below, calculate the enthalpy change for the reaction H₂ + Cl₂ → 2HCl and state whether the reaction is exothermic or endothermic.

例题： 使用下列键能数据，计算反应 H₂ + Cl₂ → 2HCl 的焓变，并说明该反应是放热还是吸热。

Bond	键	Bond energy (kJ/mol)	键能 (kJ/mol)
H−H	氢氢键	436	436
Cl−Cl	氯氯键	243	243
H−Cl	氢氯键	432	432

Solution:

Step 1: Energy absorbed to break bonds: 1 × (H−H) + 1 × (Cl−Cl) = 436 + 243 = 679 kJ.

第一步： 断裂化学键吸收的能量：1 × (H−H) + 1 × (Cl−Cl) = 436 + 243 = 679 kJ。

Step 2: Energy released when new bonds form: 2 × (H−Cl) = 2 × 432 = 864 kJ.

第二步： 形成新化学键释放的能量：2 × (H−Cl) = 2 × 432 = 864 kJ。

Step 3: ΔH = energy in – energy out = 679 – 864 = −185 kJ/mol.

第三步： ΔH = 输入能量 – 输出能量 = 679 – 864 = −185 kJ/mol。

Step 4: A negative ΔH means the reaction is exothermic (gives out heat).

第四步： 负的 ΔH 表明反应放热。

Answer: ΔH = −185 kJ/mol; the reaction is exothermic.

答案： ΔH = −185 kJ/mol；反应是放热的。

7. Rates of Reaction: Collision Theory | 反应速率：碰撞理论

Example: Magnesium ribbon reacts with dilute hydrochloric acid. Suggest two ways to increase the rate of reaction and explain them using collision theory. Describe the expected change in the shape of a graph of volume of hydrogen gas produced against time if the acid concentration is doubled.

例题： 镁带与稀盐酸反应。提出两种提高反应速率的方法，并用碰撞理论解释。描述如果将酸浓度加倍，氢气体积随时间变化的曲线形状将如何改变。

Solution:

Method 1: Increase the temperature. Particles have more kinetic energy, so they move faster and collide more frequently. More particles have energy greater than the activation energy, leading to more successful collisions per second.

方法 1： 升高温度。粒子具有更高的动能，运动更快，碰撞更频繁。更多粒子具有超过活化能的能量，导致每秒成功碰撞次数增加。

Method 2: Increase the concentration of the acid (or use a more reactive metal / increase surface area of magnesium). A higher concentration means more reactant particles per unit volume, increasing collision frequency.

方法 2： 增加酸的浓度（或使用更活泼的金属/增大镁的表面积）。更高浓度意味着单位体积内反应物粒子增多，碰撞频率增加。

Graph shape: With double the concentration, the initial gradient of the graph is steeper, and the final volume of gas is produced in a shorter time. The total volume of hydrogen stays the same because the same mass of magnesium is used; the reaction reaches completion faster.

图线形状： 浓度加倍时，图线的初始斜率更陡，最终气体体积在更短时间内产生。由于使用相同质量的镁，氢气总体积不变；反应更快达到终点。

8. Equilibrium and Le Chatelier’s Principle | 平衡与勒夏特列原理

Example: The formation of ammonia is a reversible reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ/mol. Explain the effect on the position of equilibrium and on the yield of ammonia if: (a) the pressure is increased, (b) the temperature is increased, (c) more nitrogen is added.

例题： 氨的生成是一个可逆反应：N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ/mol。解释以下操作对平衡位置和氨的产率的影响：(a) 增大压强，(b) 升高温度，(c) 加入更多氮气。

Solution:

(a) Increasing pressure: The equilibrium shifts to the side with fewer moles of gas to oppose the change. Left side has 4 moles (1+3), right side has 2 moles. Position shifts to the right, increasing the yield of NH₃.

(a) 增大压强： 平衡向气体分子总数较少的方向移动，以抵消改变。左边 4 摩尔气体，右边 2 摩尔。平衡右移，NH₃ 产率提高。

(b) Increasing temperature: This favours the endothermic direction (reverse reaction here) to absorb the added heat. Position shifts to the left, decreasing the yield of NH₃.

(b) 升高温度： 升温有利于吸热方向（此处为逆反应）以吸收添加的热量。平衡左移，NH₃ 产率降低。

(c) Adding more nitrogen: The system tries to use up the added reactant by shifting to the right. Position shifts to the right, increasing the yield of NH₃.

(c) 加入更多氮气： 系统通过向右移动来消耗添加的反应物。平衡右移，NH₃ 产率提高。

9. Organic Chemistry: Alkanes and Alkenes | 有机化学：烷烃与烯烃

Example: Describe a chemical test to distinguish between ethane (C₂H₆) and ethene (C₂H₄). Write an equation for the reaction that occurs and state the type of reaction.

例题： 描述一个区分乙烷（C₂H₆）和乙烯（C₂H₄）的化学测试。写出发生反应的方程式，并指出反应类型。

Solution:

Test: Add bromine water (orange-brown) to each gas. With ethane, there is no reaction and the bromine water stays orange-brown. With ethene, the bromine water is decolourised immediately, turning from orange-brown to colourless.

测试： 向每种气体中加入溴水（橙棕色）。乙烷不发生反应，溴水保持橙棕色。乙烯会使溴水立即褪色，从橙棕色变为无色。

Equation: C₂H₄ + Br₂ → C₂H₄Br₂ (1,2-dibromoethane).

方程式： C₂H₄ + Br₂ → C₂H₄Br₂（1,2-二溴乙烷）。

Type of reaction: Electrophilic addition (addition reaction). The double bond opens up to add the bromine atoms.

反应类型： 亲电加成（加成反应）。双键打开，加上溴原子。

10. Acids, Bases and Titration Calculations | 酸、碱与滴定计算

Example: 25.0 cm³ of sulfuric acid (H₂SO₄) of unknown concentration is neutralised by 30.0 cm³ of 0.10 mol/dm³ sodium hydroxide (NaOH) solution. The equation is: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the sulfuric acid in mol/dm³.

例题： 25.0 cm³ 未知浓度的硫酸（H₂SO₄）被 30.0 cm³ 0.10 mol/dm³ 的氢氧化钠（NaOH）溶液中和。反应方程式为：H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O。计算硫酸的浓度（mol/dm³）。

Solution:

Step 1: Convert volumes to dm³: NaOH volume = 30.0 / 1000 = 0.0300 dm³; H₂SO₄ volume = 25.0 / 1000 = 0.0250 dm³.

第一步： 将体积转换为 dm³：NaOH 体积 = 30.0 / 1000 = 0.0300 dm³；H₂SO₄ 体积 = 25.0 / 1000 = 0.0250 dm³。

Step 2: Moles of NaOH = concentration × volume = 0.10 × 0.0300 = 0.0030 mol.

第二步： NaOH 的摩尔数 = 浓度 × 体积 = 0.10 × 0.0300 = 0.0030 mol。

Step 3: From the equation, 1 mol H₂SO₄ reacts with 2 mol NaOH. Moles of H₂SO₄ = 0.0030 / 2 = 0.0015 mol.

第三步： 由方程式知，1 mol H₂SO₄ 与 2 mol NaOH 反应。H₂SO₄ 的摩尔数 = 0.0030 / 2 = 0.0015 mol。

Step 4: Concentration of H₂SO₄ = moles / volume = 0.0015 / 0.0250 = 0.060 mol/dm³.

第四步： H₂SO₄ 的浓度 = 摩尔数 / 体积 = 0.0015 / 0.0250 = 0.060 mol/dm³。

Answer: 0.060 mol/dm³.

答案： 0.060 mol/dm³。

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