📚 Year 10 CAIE Chemistry: Case Study Drills | CAIE 十年级化学：案例分析实战演练
Case studies are a powerful way to connect chemical concepts with real-world applications. In CAIE Year 10 Chemistry, you are expected to apply principles of stoichiometry, acids and bases, electrolysis, energetics, and more to solve practical problems. This article presents ten carefully crafted case studies to sharpen your analytical skills, each followed by a detailed step-by-step solution.
案例学习是将化学概念与实际应用相结合的有效方式。在 CAIE 十年级化学中，你需要运用化学计量、酸碱、电解、能量变化等原理来解决实际问题。本文精选了十个精心设计的案例，旨在提升你的分析能力，每个案例都附有详细的逐步解析。
1. Stoichiometry in Action: Reacting Masses | 化学计量实战：反应质量
A student heated a mixture of 5.60 g of iron filings with excess sulfur. The reaction produced iron(II) sulfide, FeS. She wanted to find the theoretical mass of sulfur that had reacted and the mass of the product formed.
一名学生将 5.60 g 铁屑与过量的硫粉混合加热，反应生成了硫化亚铁 (FeS)。她希望计算理论上参加反应的硫的质量以及生成物的质量。
The balanced equation for the reaction is: Fe + S → FeS. This indicates a 1:1 mole ratio between Fe, S, and FeS.
反应的配平方程式为：Fe + S → FeS。这表明 Fe、S 和 FeS 之间的摩尔比均为 1:1。
n(Fe) = mass / Mᵣ = 5.60 / 56 = 0.100 mol
根据摩尔比，参加反应的硫的物质的量也是 0.100 mol。硫的摩尔质量 Mᵣ(S) = 32，因此所需硫的质量 = 0.100 × 32 = 3.20 g。同理，生成的 FeS 的物质的量为 0.100 mol，Mᵣ(FeS) = 56 + 32 = 88，故 FeS 的质量 = 0.100 × 88 = 8.80 g。
Thus, the mass of sulfur consumed was 3.20 g and the mass of iron(II) sulfide produced was 8.80 g. This case shows how a balanced equation and mole ratios directly yield reacting masses.
因此，消耗的硫的质量为 3.20 g，生成的硫化亚铁质量为 8.80 g。本案例说明如何借助配平方程式和摩尔比直接求得反应质量。
2. Titration Determination of Vinegar Acidity | 滴定法测定食醋酸度
A chemist titrated a 10.0 cm³ sample of vinegar (ethanoic acid solution) against 0.100 mol/dm³ sodium hydroxide solution using phenolphthalein. The average titre volume was 15.20 cm³. She needed to calculate the concentration of ethanoic acid in the vinegar in mol/dm³ and g/dm³.
一位化学家使用酚酞作指示剂，用 0.100 mol/dm³ 的氢氧化钠溶液滴定 10.0 cm³ 的食醋（乙酸溶液），平均滴定体积为 15.20 cm³。她需要计算食醋中乙酸的浓度（单位 mol/dm³ 和 g/dm³）。
The neutralisation reaction: CH₃COOH + NaOH → CH₃COONa + H₂O. Moles of NaOH used = concentration × volume (dm³) = 0.100 × (15.20/1000) = 0.00152 mol.
中和反应为：CH₃COOH + NaOH → CH₃COONa + H₂O。所用 NaOH 的物质的量 = 浓度 × 体积 (dm³) = 0.100 × (15.20/1000) = 0.00152 mol。
le = 0.00152 mol. Hence, concentration of CH₃COOH = 0.00152 / (10.0/1000) = 0.152 mol/dm³.
根据 1:1 的摩尔比，10.0 cm³ 样品中 CH₃COOH 的物质的量也是 0.00152 mol。因此，CH₃COOH 的浓度 = 0.00152 / (10.0/1000) = 0.152 mol/dm³。
6+1 = 60. Mass concentration = 0.152 × 60 = 9.12 g/dm³. A typical vinegar contains about 5% acidity; our result corresponds to roughly 0.9% acidity, indicating a diluted sample.
换算为 g/dm³，Mᵣ(CH₃COOH) = 60，质量浓度 = 0.152 × 60 = 9.12 g/dm³。通常食醋酸度约为 5%，而计算结果约相当于 0.9%，说明样本已被稀释。
3. Electrolysis of Copper(II) Sulfate Solution | 硫酸铜溶液的电解
In a laboratory, a student set up an electrolytic cell using carbon (graphite) electrodes immersed in aqueous copper(II) sulfate. She observed the formation of a reddish-brown deposit on the cathode and bubbling at the anode. Predict the products at each electrode and write half-equations.
实验中，学生采用石墨电极电解硫酸铜水溶液。她观察到阴极上有红棕色沉积物产生，阳极处有气泡冒出。请预测两极的产物并写出相应的半反应方程式。
Aqueous CuSO₄ contains Cu²⁺, SO₄²⁻, H⁺ and OH⁻ ions. At the cathode, copper is less reactive than hydrogen, so Cu²⁺ ions gain electrons preferentially. Cathode half-equation: Cu²⁺(aq) + 2e⁻ → Cu(s). The reddish-brown deposit is copper metal.
CuSO₄ 溶液中含 Cu²⁺、SO₄²⁻、H⁺ 与 OH⁻ 离子。在阴极，铜的活泼性低于氢，因此 Cu²⁺ 优先得电子。阴极半反应：Cu²⁺(aq) + 2e⁻ → Cu(s)，红棕色沉积物为金属铜。
At the anode, hydroxide ions are discharged in preference to sulfate ions because OH⁻ is easier to oxidise. Anode half-equation: 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻. The gas bubbles are oxygen. The solution gradually becomes more acidic as H⁺ ions accumulate.
在阳极，氢氧根离子比硫酸根离子更容易放电。阳极半反应：4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻，气泡为氧气。随着 H⁺ 富集，溶液酸性逐渐增强。
4. Rate of Reaction: Marble Chips and Acid | 反应速率：大理石与盐酸
A class investigated how the concentration of hydrochloric acid affects the rate of its reaction with marble chips (calcium carbonate). They used 5.0 g of marble chips and 50 cm³ of acid, measuring the volume of carbon dioxide produced over time. Data for two runs are shown below.
某班级探究盐酸浓度如何影响其与大理石（碳酸钙）的反应速率。他们使用 5.0 g 大理石和 50 cm³ 酸，测量不同时间生成的二氧化碳体积。两次实验数据如下。
| Run | [HCl] (mol/dm³) | Time to collect 40 cm³ CO₂ (s) |
|---|---|---|
| 1 | 2.0 | 25 |
| 2 | 1.0 | 48 |
The reaction: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g). Doubling the acid concentration approximately halved the time needed to produce the same volume of CO₂, indicating a faster initial rate. This can be explained by collision theory: higher concentration means more H⁺ ions per unit volume, leading to more frequent effective collisions per unit time.
反应式：CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)。酸浓度加倍，收集相同体积 CO₂ 所需时间约减少一半，表明初始速率更快。这可以用碰撞理论解释：浓度更高时，单位体积内的 H⁺ 数目更多，单位时间内的有效碰撞频率增大。
However, the mass of marble was the same, so the total volume of CO₂ produced (once the reaction is complete) should be the same in both runs, provided acid is in excess. The initial rate is proportional to the concentration of the limiting reactant.
不过，大理石质量相同，因此如果酸过量，反应完全时生成的 CO₂ 总体积在两次实验中应相同。初始速率与限制反应物的浓度成正比。
5. Energetics: Heat of Neutralisation | 能量变化：中和热
A student mixed 50.0 cm³ of 1.0 mol/dm³ HCl with 50.0 cm³ of 1.0 mol/dm³ NaOH in a polystyrene cup. The temperature rose from 20.5 °C to 27.2 °C. Determine the enthalpy change of neutralisation per mole of water formed. Assume the specific heat capacity of the solution is 4.2 J/(g·°C) and the density is 1.0 g/cm³.
一名学生将 50.0 cm³ 1.0 mol/dm³ 的盐酸与 50.0 cm³ 1.0 mol/dm³ 的氢氧化钠在聚苯乙烯杯中混合，温度由 20.5 °C 升至 27.2 °C。计算生成每摩尔水的中和焓变。假设溶液的比热容为 4.2 J/(g·°C)，密度为 1.0 g/cm³。
Total volume = 100 cm³, so mass of solution = 100 g. Temperature change ΔT = 27.2 – 20.5 = 6.7 °C. Heat absorbed by solution: q = mcΔT = 100 × 4.2 × 6.7 = 2814 J ≈ 2.81 kJ.
总体积 = 100 cm³，溶液质量 = 100 g。温度变化 ΔT = 6.7 °C。溶液吸收的热量：q = mcΔT = 100 × 4.2 × 6.7 = 2814 J ≈ 2.81 kJ。
Moles of HCl = 0.050 × 1.0 = 0.050 mol; moles of NaOH = 0.050 mol. Reaction: HCl + NaOH → NaCl + H₂O , moles of water formed = 0.050 mol. The reaction is exothermic, so ΔH per mole of water = –2.81 kJ / 0.050 mol = –56.2 kJ/mol. The accepted value is around –57 kJ/mol; the slight difference is due to heat loss.
HCl 的物质的量 = 0.050 mol，NaOH 的物质的量 = 0.050 mol。反应为 HCl + NaOH → NaCl + H₂O，生成的水的物质的量 = 0.050 mol。反应放热，因此每摩尔水的 ΔH = –2.81 kJ / 0.050 mol = –56.2 kJ/mol。标准值约为 –57 kJ/mol，微小偏差来自热量散失。
6. Identifying Organic Compounds | 有机物的鉴别
Two colourless gases, A and B, are known to be ethane and ethene, but the labels are missing. A technician uses bromine water and a burning test to distinguish them. Describe the expected observations and conclusions.
已知两瓶无色气体 A 和 B 分别为乙烷和乙烯，但标签丢失。技术人员使用溴水和燃烧测试来鉴别它们。请描述预期的现象与结论。
When shaken with a few drops of orange-brown bromine water, ethene (C₂H₄) decolourises it rapidly due to the electrophilic addition reaction: C₂H₄ + Br₂ → C₂H₄Br₂. Ethane (C₂H₆), being an alkane, does not react and the bromine water remains orange.
与几滴橙棕色溴水混合振荡时，乙烯 (C₂H₄) 因发生亲电加成反应 C₂H₄ + Br₂ → C₂H₄Br₂ 而迅速使溴水褪色。乙烷 (C₂H₆) 作为烷烃不发生反应，溴水保持橙棕色。
In the burning test, both gases burn with a blue flame, but ethene produces a slightly smokier flame because of its higher carbon-to-hydrogen ratio, leading to incomplete combustion. This is a supporting observation but not as definitive as the bromine test.
在燃烧测试中，两种气体均呈现蓝色火焰，但乙烯因碳氢比更高，可能产生更明显的黑烟，这是不完全燃烧所致。该现象可作为辅助判断，但不如溴水测试明确。
Thus, the gas that decolourises bromine water is ethene; the one that does not is ethane.
因此，能使溴水褪色的气体为乙烯，反之为乙烷。
7. Metal Extraction: Reduction of Iron Ore | 金属提取：铁矿石的还原
The extraction of iron from hematite (Fe₂O₃) in a blast furnace involves reaction with carbon monoxide. A mining engineer wants to predict the amount of carbon needed to produce 1.0 tonne of iron. The overall reaction can be summarised as: Fe₂O₃ + 3CO → 2Fe + 3CO₂. The carbon monoxide is produced by burning coke: 2C + O₂ → 2CO. Calculate the mass of carbon required.
高炉中用一氧化碳从赤铁矿 (Fe₂O₃) 中提取铁。一位采矿工程师想计算生产 1.0 吨铁所需碳的质量。总反应可表示为：Fe₂O₃ + 3CO → 2Fe + 3CO₂。一氧化碳由焦炭燃烧生成：2C + O₂ → 2CO。计算所需碳的质量。
oles CO. To produce 1.0 tonne (1.0 × 10⁶ g) of Fe, moles Fe = 1.0×10⁶ / 56 = 1.79×10⁴ mol.
由还原方程式可知，生成 2 mol Fe 需要 3 mol CO。要生产 1.0 × 10⁶ g 的铁，n(Fe) = 1.0×10⁶ / 56 = 1.79×10⁴ mol。
combustion equation, 2 moles CO come from 2 moles C, so moles C = moles CO = 2.68×10⁴ mol.
所需 CO 的物质的量 = (3/2) × 1.79×10⁴ = 2.68×10⁴ mol。由碳的燃烧方程式，2 mol C 生成 2 mol CO，因此 n(C) = n(CO) = 2.68×10⁴ mol。
. In reality, excess carbon is used and some is lost, so the actual requirement is higher.
碳的质量 = 2.68×10⁴ × 12 = 3.22×10⁵ g = 0.322 吨。实际生产中会使用过量的碳，且会有损耗，因此实际需求更高。
8. Acid Rain Formation and Impact | 酸雨的形成与影响
A power station burns coal containing 2.0% sulfur by mass. In one day, it burns 1000 tonnes of coal. Assuming all sulfur is converted to sulfuric acid via SO₂ and SO₃, determine the mass of H₂SO₄ potentially released into the atmosphere. Also discuss the environmental effects.
某发电站用含硫 2.0%（质量分数）的煤作燃料，每天燃烧 1000 吨煤。假设所有的硫最终均转化为硫酸（经 SO₂ 和 SO₃），计算可能排入大气的 H₂SO₄ 质量，并讨论环境效应。
Mass of sulfur in coal = 2.0% H₂SO₄, each step with 1:1 sulfur conservation. Thus, moles H₂SO₄ = moles S = 6.25×10⁵ mol.
煤中硫的质量 = 2.0% × 1000 = 20 吨 = 2.0×10⁷ g。n(S) = 2.0×10⁷ / 32 = 6.25×10⁵ mol。转化路径 S → SO₂ → SO₃ → H₂SO₄ 中硫元素守恒，因此 n(H₂SO₄) = 6.25×10⁵ mol。
id rain lowers the pH of soil and lakes, damages buildings (especially limestone) by reaction: CaCO₃ + H₂SO₄ → CaSO₄ + CO₂ + H₂O, and harms aquatic life.
Mᵣ(H₂SO₄) = 98，H₂SO₄ 质量 = 6.25×10⁵ × 98 = 6.125×10⁷ g = 61.25 吨。酸雨会降低土壤与湖泊的 pH 值，腐蚀建筑（尤其是石灰石），反应为：CaCO₃ + H₂SO₄ → CaSO₄ + CO₂ + H₂O，并危害水生生物。
9. Chemical Tests for Ions | 离子的化学检验
A sample of wastewater is suspected to contain chloride ions (Cl⁻) and sulfate ions (SO₄²⁻). A technician acidifies portions with dilute nitric acid and then adds silver nitrate solution to one portion, and barium chloride solution to another. Describe the expected observations and the rationale behind acidification.
某废水样本疑含有氯离子 (Cl⁻) 和硫酸根离子 (SO₄²⁻)。技术人员用稀硝酸分别将两份试样酸化，然后一份加入硝酸银溶液，另一份加入氯化钡溶液。请描述预期的现象以及酸化的理由。
aSO₄(s), also a white precipitate. Acidification with nitric acid removes carbonate ions (CO₃²⁻) which would otherwise form precipitates with both reagents, causing false positives.
检验氯离子：Ag⁺(aq) + Cl⁻(aq) → AgCl(s)，生成白色沉淀。检验硫酸根：Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)，也是白色沉淀。用稀硝酸酸化是为了去除碳酸根离子 (CO₃²⁻)，因为后者会与两种试剂均产生沉淀，导致误判。
Upon addition of AgNO₃ to the first portion, a white precipitate confirms Cl⁻. Adding BaCl₂ to the second portion, a white precipitate confirms SO₄²⁻. If no precipitate forms in either test, the corresponding ions are absent.
向第一份中加入 AgNO₃ 后出现白色沉淀，证明含 Cl⁻。第二份中加入 BaCl₂ 后出现白色沉淀，证明含 SO₄²⁻。若某一份无沉淀生成，则对应离子不存在。
10. Redox in Displacement Reactions | 置换反应中的氧化还原
A student places a strip of zinc metal into blue copper(II) sulfate solution. Over time, the blue colour fades, a reddish-brown deposit coats the zinc, and the solution becomes colourless. Identify the oxidising and reducing agents and write the ionic half
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