📚 Year 11 CAIE Chemistry: Case Study Practical Drills | 案例分析实战演练
Welcome to this intensive case-study revision session for Year 11 CAIE Chemistry. Each scenario below mimics the style of real practical-based exam questions, combining theoretical concepts with data analysis, calculations, and evaluative thinking. Work through them carefully to sharpen your problem-solving skills and deepen your understanding of the CAIE curriculum.
欢迎来到 Year 11 CAIE 化学案例实战演练。每一个案例都模拟了真实的实验类考题风格，将理论概念与数据分析、计算和评价思维相结合。请认真练习，提升解题能力，加深对 CAIE 课程的理解。
1. Stoichiometry: Determining the Formula of a Hydrated Salt | 化学计量：确定水合盐的化学式
A student heated 3.12 g of hydrated barium chloride, BaCl₂·xH₂O, to constant mass. After heating, the mass of anhydrous BaCl₂ was 2.66 g. Determine the value of x and write the formula of the hydrated salt.
一名学生将 3.12 g 水合氯化钡 BaCl₂·xH₂O 加热至恒重。加热后，无水 BaCl₂ 的质量为 2.66 g。求 x 值并写出该水合盐的化学式。
Mass of water driven off = 3.12 g – 2.66 g = 0.46 g.
失去的水质量 = 3.12 g – 2.66 g = 0.46 g。
Moles of BaCl₂ = 2.66 g / Mᵣ(BaCl₂) = 2.66 / 208.3 = 0.0128 mol. (Mᵣ: Ba=137.3, Cl=35.5×2)
BaCl₂ 的物质的量 = 2.66 g / 208.3 g mol⁻¹ = 0.0128 mol。（相对分子质量：BaCl₂ = 137.3 + 71.0 = 208.3）
Moles of H₂O = 0.46 g / 18.0 g mol⁻¹ = 0.0256 mol.
H₂O 的物质的量 = 0.46 g / 18.0 g mol⁻¹ = 0.0256 mol。
Ratio n(H₂O) : n(BaCl₂) = 0.0256 : 0.0128 = 2 : 1, therefore x = 2. The formula is BaCl₂·2H₂O.
物质的量之比 n(H₂O) : n(BaCl₂) = 0.0256 : 0.0128 = 2 : 1，因此 x = 2，化学式为 BaCl₂·2H₂O。
The sample must be heated to constant mass to ensure all water of crystallisation is removed. Repeated heating and weighing eliminate errors from incomplete dehydration.
必须加热至恒重，以确保所有结晶水都已去除。反复加热和称量可消除脱水不完全带来的误差。
2. Moles and Gas Volumes: Reaction of Magnesium with Excess Acid | 摩尔与气体体积：镁与过量酸的反应
A 0.060 g strip of magnesium ribbon is added to excess dilute hydrochloric acid. Calculate the volume of hydrogen gas produced at room temperature and pressure (r.t.p., molar volume = 24 dm³ mol⁻¹).
将一条 0.060 g 的镁带加入过量稀盐酸中。计算在室温常压下产生的氢气体积（r.t.p., 摩尔体积 = 24 dm³ mol⁻¹）。
Moles of Mg = mass / Aᵣ = 0.060 g / 24.3 g mol⁻¹ = 0.00247 mol.
Mg 的物质的量 = 0.060 g / 24.3 g mol⁻¹ = 0.00247 mol。
Equation: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). 1 mol Mg produces 1 mol H₂.
反应方程式：Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)。1 mol Mg 生成 1 mol H₂。
Moles of H₂ = 0.00247 mol. Volume = moles × 24 dm³ mol⁻¹ = 0.00247 × 24 = 0.0593 dm³ = 59.3 cm³.
H₂ 的物质的量 = 0.00247 mol。体积 = 物质的量 × 24 dm³ mol⁻¹ = 0.00247 × 24 = 0.0593 dm³ = 59.3 cm³。
If the experiment were carried out at standard temperature and pressure (s.t.p., 22.4 dm³ mol⁻¹), the volume would be slightly lower, highlighting the importance of specifying conditions.
如果实验在标准状况下进行（s.t.p., 22.4 dm³ mol⁻¹），体积会略小，这说明明确条件的重要性。
3. Electrolysis: Predicting Products in Aqueous Copper(II) Sulfate | 电解：预测硫酸铜溶液产物
Aqueous copper(II) sulfate is electrolysed using inert graphite electrodes. Predict the products at the anode and cathode, and calculate the mass of copper deposited when a current of 0.50 A flows for 1930 seconds. (1 Faraday = 96500 C mol⁻¹, Aᵣ Cu = 63.5)
用惰性石墨电极电解硫酸铜溶液。预测阳极和阴极产物，并计算 0.50 A 电流通电 1930 秒后析出的铜的质量。（1 法拉第 = 96500 C mol⁻¹，Aᵣ Cu = 63.5）
Cathode: Cu²⁺ ions are discharged because copper is less reactive than hydrogen. Cu²⁺ + 2e⁻ → Cu(s). Anode: OH⁻ ions are discharged in preference to SO₄²⁻, forming oxygen gas: 4OH⁻ → 2H₂O + O₂ + 4e⁻.
阴极：放电的是 Cu²⁺ 离子，因为铜的活动性比氢弱。Cu²⁺ + 2e⁻ → Cu(s)。阳极：OH⁻ 离子优先于 SO₄²⁻ 放电，生成氧气：4OH⁻ → 2H₂O + O₂ + 4e⁻。
Quantity of charge Q = I × t = 0.50 A × 1930 s = 965 C.
电荷量 Q = I × t = 0.50 A × 1930 s = 965 C。
Moles of electrons = Q / 96500 = 965 / 96500 = 0.0100 mol.
电子的物质的量 = 965 C / 96500 C mol⁻¹ = 0.0100 mol。
From the half-equation, 2 mol electrons deposit 1 mol Cu. Moles of Cu = 0.0100 / 2 = 0.00500 mol.
根据半反应，2 mol 电子析出 1 mol Cu。Cu 的物质的量 = 0.00500 mol。
Mass of Cu = 0.00500 mol × 63.5 g mol⁻¹ = 0.318 g.
铜的质量 = 0.00500 mol × 63.5 g mol⁻¹ = 0.318 g。
The blue colour of the electrolyte fades as Cu²⁺ ions are removed, and the solution becomes acidic due to the formation of sulfuric acid.
随着 Cu²⁺ 离子被消耗，电解液蓝色变浅，同时生成硫酸使得溶液变酸性。
4. Energetics: Determining the Enthalpy Change of Neutralisation | 能量学：测定中和反应焓变
50.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH in a polystyrene cup. The temperature rises from 21.5 °C to 28.0 °C. Calculate the enthalpy change of neutralisation, ΔH, assuming the solution has a specific heat capacity of 4.2 J g⁻¹ °C⁻¹ and density 1.0 g cm⁻³.
在聚苯乙烯杯中混合 50.0 cm³ 1.00 mol dm⁻³ HCl 和 50.0 cm³ 1.00 mol dm⁻³ NaOH。温度从 21.5 °C 升至 28.0 °C。假设溶液比热容为 4.2 J g⁻¹ °C⁻¹，密度 1.0 g cm⁻³，计算中和反应焓变 ΔH。
Total volume = 100 cm³ → mass of solution = 100 g. Temperature rise ΔT = 28.0 – 21.5 = 6.5 °C.
总体积 = 100 cm³ → 溶液质量 = 100 g。温度升高 ΔT = 6.5 °C。
Heat energy released q = m × c × ΔT = 100 g × 4.2 J g⁻¹ °C⁻¹ × 6.5 °C = 2730 J = 2.73 kJ.
释放的热量 q = m × c × ΔT = 100 × 4.2 × 6.5 = 2730 J = 2.73 kJ。
Moles of HCl used = concentration × volume = 1.00 mol dm⁻³ × 0.050 dm³ = 0.050 mol. Moles of NaOH = 0.050 mol, so water formed = 0.050 mol.
HCl 的物质的量 = 1.00 × 0.050 = 0.050 mol。NaOH 同样是 0.050 mol，生成水 0.050 mol。
ΔH per mole of water formed = –q / n = –2.73 kJ / 0.050 mol = –54.6 kJ mol⁻¹. The negative sign indicates an exothermic reaction.
每摩尔水的 ΔH = –q / n = –2.73 kJ / 0.050 mol = –54.6 kJ mol⁻¹。负号表示放热反应。
Heat loss to the surroundings is a major source of error; a lid and insulated cup help reduce this systematic error.
散热是主要误差来源；使用杯盖和隔热杯有助于减少这一系统误差。
5. Rates of Reaction: Interpreting Volume-Time Data from Marble Chips and Acid | 反应速率：解读大理石与酸反应的体积-时间数据
A student reacts excess marble chips (CaCO₃) with 50 cm³ of 2.0 mol dm⁻³ HCl and records the volume of CO₂ gas collected every 30 seconds. The initial rate of reaction is found to be 4.0 cm³ s⁻¹. After 120 seconds, the rate drops to 0.8 cm³ s⁻¹. Explain why the rate decreases and calculate the average rate of CO₂ production over the first two minutes.
学生将过量大理石碎块 (CaCO₃) 与 50 cm³ 2.0 mol dm⁻³ HCl 反应，每 30 秒记录 CO₂ 气体体积。初始反应速率为 4.0 cm³ s⁻¹。120 秒后速率降至 0.8 cm³ s⁻¹。解释速率下降的原因并计算前两分钟内 CO₂ 的平均生成速率。
The rate decreases because the concentration of HCl falls as it is consumed. Lower concentration means fewer collisions per unit time between reactant particles, so collision frequency decreases.
速率下降是因为 HCl 浓度随着反应消耗而减小。浓度降低意味着反应物粒子在单位时间内的碰撞次数减少，碰撞频率下降。
Surface area of marble is roughly constant because it is in excess, so the main factor is the decreasing acid concentration.
大理石过量所以表面积基本不变，因此主要影响因素是酸浓度的降低。
Average rate = total volume of CO₂ produced / total time. To find total volume, we need to integrate; but a simpler approach: initial and final rates are given. At constant deceleration, average rate can be estimated, but more precisely, total volume produced in 120 s can be deduced if we assume a linear drop? Not accurate. Instead, we need experimental total volume. The question implies we know initial rate and rate after 120 s, but not total volume. This is a modelling case: suppose the rate drops uniformly, average rate = (4.0 + 0.8) / 2 = 2.4 cm³ s⁻¹, but real kinetics are exponential. For an average, more data is required. In this example, we can state that in a typical data set, if total volume in 120 s was 280 cm³, then average rate = 280/120 ≈ 2.33 cm³ s⁻¹, aligning with the concept.
平均速率 = 总 CO₂ 体积 / 总时间。要获得总体积需要积分；在匀速降低的简单模型中，平均速率 = (4.0 + 0.8) / 2 = 2.4 cm³ s⁻¹。实际反应往往是指数下降，实验中若收集到总体积 280 cm³，平均速率 ≈ 2.33 cm³ s⁻¹，与估算值吻合。通过实时曲线下的面积可以更精确地确定平均速率。
6. Equilibrium: Applying Le Chatelier’s Principle to the Haber Process | 平衡：运用勒夏特列原理分析哈伯法
The Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ mol⁻¹. At a certain temperature, an equilibrium mixture contains 0.50 mol dm⁻³ N₂, 1.50 mol dm⁻³ H₂ and 0.20 mol dm⁻³ NH₃. Calculate Kc and predict the effect of increasing pressure on the equilibrium yield of ammonia.
哈伯法：N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ mol⁻¹。在某温度下，平衡混合物含 0.50 mol dm⁻³ N₂、1.50 mol dm⁻³ H₂ 和 0.20 mol dm⁻³ NH₃。计算 Kc 并预测增大压力对氨平衡产率的影响。
Kc = [NH₃]² / ([N₂] × [H₂]³) = (0.20)² / (0.50 × 1.50³) = 0.040 / (0.50 × 3.375) = 0.040 / 1.6875 = 0.0237 (units dm⁶ mol⁻²).
Kc = [NH₃]² / ([N₂] × [H₂]³) = (0.20)² / (0.50 × 1.50³) = 0.040 / 1.6875 = 0.0237（单位 dm⁶ mol⁻²）。
Increasing pressure shifts the equilibrium to the side with fewer gas molecules. The forward reaction produces 2 moles of gas from 4 moles, so higher pressure favours the forward reaction, increasing the yield of NH₃.
增大压力平衡向气体分子数较少的一侧移动。正反应由 4 摩尔气体生成 2 摩尔气体，所以高压有利于正反应，提高 NH₃ 产率。
Although high pressure improves yield, it is expensive to build plants that withstand very high pressures. A compromise of 200 atm is typically used along with a temperature of about 450 °C and an iron catalyst.
尽管高压提高产率，但建造耐受极高压力的设备成本高昂。实际生产通常采用约 200 atm、450 °C 和铁催化剂的折中条件。
7. Organic Chemistry: Cracking and Tests for Unsaturation | 有机化学：裂化与不饱和性测试
A long-chain alkane, C₁₄H₃₀, is cracked to produce a shorter alkane and an alkene. Write a possible equation and describe a chemical test to distinguish the alkene product from the original alkane.
长链烷烃 C₁₄H₃₀ 被裂化生成一个较短烷烃和一个烯烃。写出一条可能的方程式，并描述区分烯烃产物与原来烷烃的化学测试。
One possible equation: C₁₄H₃₀ → C₁₀H₂₂ + C₄H₈. The alkene could be but-1-ene, but-2-ene, or 2-methylpropene.
一条可能的方程式：C₁₄H₃₀ → C₁₀H₂₂ + C₄H₈。烯烃可能是 1-丁烯、2-丁烯或 2-甲基丙烯。
To test for unsaturation, add a few drops of orange bromine water to each hydrocarbon in the dark. The alkene will rapidly decolourise the bromine water (orange → colourless) due to an addition reaction, while the alkane will not react and the colour persists.
检验不饱和性：在避光条件下分别向两种烃中加入几滴橙色溴水。烯烃会因加成反应使溴水迅速褪色（橙色→无色），而烷烃不反应，溴水颜色保持不变。
The reaction with bromine: C₄H₈ + Br₂ → C₄H₈Br₂. Safety note: bromine water is corrosive and must be handled with care.
与溴的反应：C₄H₈ + Br₂ → C₄H₈Br₂。安全提示：溴水有腐蚀性，操作须小心。
8. Acids and Bases: Titration Calculations and Indicator Selection | 酸碱：滴定计算和指示剂选择
25.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid is titrated with sodium hydroxide solution of unknown concentration. The average titre is 22.50 cm³. Calculate the concentration of NaOH and name a suitable indicator for this strong acid–strong base titration.
用 25.0 cm³ 0.100 mol dm⁻³ 盐酸滴定未知浓度的氢氧化钠溶液，平均滴定体积为 22.50 cm³。计算 NaOH 的浓度，并说出适合此强酸–强碱滴定的指示剂名称。
Equation: HCl + NaOH → NaCl + H₂O. Mole ratio 1:1. Moles of HCl = 0.100 mol dm⁻³ × 0.0250 dm³ = 0.00250 mol.
反应方程式：HCl + NaOH → NaCl + H₂O。物质的量比 1:1。HCl 物质的量 = 0.100 × 0.0250 = 0.00250 mol。
Moles of NaOH in titre = 0.00250 mol. Concentration of NaOH = 0.00250 mol / 0.02250 dm³ = 0.111 mol dm⁻³.
滴定所用 NaOH 物质的量 = 0.00250 mol。NaOH 浓度 = 0.00250 mol / 0.02250 dm³ = 0.111 mol dm⁻³。
Suitable indicators: phenolphthalein (colourless in acid, pink in alkali) or methyl orange (red in acid, yellow in alkali). Both show a sharp colour change at the equivalence point near pH 7.
合适的指示剂：酚酞（酸性无色，碱性粉红）或甲基橙（酸性红色，碱性黄色）。两者在接近 pH 7 的滴定终点都有敏锐的颜色变化。
Rinsing the burette with water instead of the titrant solution would dilute the NaOH, making the calculated concentration lower than the true value.
若滴定前只用蒸馏水润洗滴定管而未用待测 NaOH 溶液润洗，会稀释碱液，导致计算浓度偏低。
9. Redox: Oxidation Numbers and Manganate(VII) Titration | 氧化还原：氧化数与高锰酸盐滴定
A 1.20 g sample of iron wire containing iron(II) ions is dissolved in dilute sulfuric acid and titrated with 0.0200 mol dm⁻³ potassium manganate(VII). The titre is 21.60 cm³. Work out the oxidation number of manganese in MnO₄⁻ and calculate the percentage of iron in the wire.
将 1.20 g 含铁(II)离子的铁丝样品溶于稀硫酸，用 0.0200 mol dm⁻³ 高锰酸钾滴定，消耗 21.60 cm³。计算 MnO₄⁻ 中锰的氧化数及铁丝中铁的质量分数。
In MnO₄⁻, let oxidation state of Mn be x: x + 4(–2) = –1 → x – 8 = –1 → x = +7. Manganese is in the +7 oxidation state.
MnO₄⁻ 中，设 Mn 氧化数为 x：x + 4(–2) = –1 → x = +7。锰的氧化数为 +7。
The redox equation in acid medium: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. 1 mol MnO₄⁻ reacts with 5 mol Fe²⁺.
酸性介质中的氧化还原反应：MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺。1 mol MnO₄⁻ 与 5 mol Fe²⁺ 反应。
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