📚 Year 11 CIE Chemistry: Case Study Practice Drill | 11年级CIE化学：案例分析实战演练
CIE IGCSE Chemistry exams increasingly use case studies that integrate practical skills, data analysis, and core chemical principles. This drill provides ten realistic scenarios that mirror the style of Paper 4 and Paper 6 questions. Working through them systematically will strengthen your ability to interpret observations, perform calculations, and justify scientific conclusions. Each case is presented with step-by-step bilingual explanations to help you master the reasoning expected in high-stakes assessments.
CIE IGCSE化学考试越来越多地使用案例分析题，这些题目综合考查实验技能、数据分析和核心化学原理。本演练提供了十个贴近真题风格的场景，涵盖Paper 4和Paper 6的常见题型。通过系统练习每个案例，你将提高解读现象、完成计算和论证科学结论的能力。每个案例都配有逐步的中英双语解析，帮助你掌握高分答案所需的推理逻辑。
1. Identifying an Unknown Salt | 鉴定未知盐
A student is given a white solid X. It dissolves readily in water to form a colourless solution. When dilute hydrochloric acid is added to the solution, brisk effervescence occurs and the gas produced turns limewater milky. The flame test on solid X shows a persistent bright yellow colour. Addition of barium chloride solution acidified with dilute HCl yields no precipitate. Identify solid X and explain your reasoning.
某学生拿到一种白色固体X，它易溶于水形成无色溶液。向溶液中加入稀盐酸，立即产生大量气泡，该气体能使石灰水变浑浊。对固体X进行焰色反应，看到持久的亮黄色。在经稀盐酸酸化的X溶液中加入氯化钡溶液，没有沉淀生成。请鉴定固体X并解释推理过程。
Flame test golden yellow indicates the presence of sodium ions, Na⁺. The colour is intense and persistent, confirming sodium rather than a fleeting impurity.
焰色反应呈金黄色表明存在钠离子Na⁺。颜色强烈且持久，证实是钠元素，而非短暂的杂质焰色。
Gas turns limewater milky: the gas is carbon dioxide, CO₂. This proves the solid contains a carbonate or hydrogencarbonate, which reacts with acid to release CO₂.
气体使石灰水变浑浊：该气体为二氧化碳CO₂。这证明固体中含有碳酸盐或碳酸氢盐，遇酸会释放CO₂。
No precipitate with acidified BaCl₂ rules out sulfate ions, SO₄²⁻. If sulfate were present, a white precipitate of BaSO₄ would form.
加入酸化的氯化钡溶液无沉淀，排除了硫酸根离子SO₄²⁻。若存在硫酸根，会生成硫酸钡白色沉淀。
ormula is Na₂CO₃, sodium carbonate. Ionic equation: CO₃²⁻(aq) + 2H⁺(aq) → CO₂(g) + H₂O(l).
综合所有线索：钠离子与碳酸根离子，无硫酸根。最简化学式为Na₂CO₃，即碳酸钠。离子方程式：CO₃²⁻(aq) + 2H⁺(aq) → CO₂(g) + H₂O(l)。
2. Calculating Percentage Yield | 计算产率百分比
A student heats 20.0 g of calcium carbonate strongly. After thermal decomposition, the mass of the remaining solid is 8.8 g of calcium oxide. Calculate the percentage yield of calcium oxide.
某学生强热20.0 g碳酸钙。热分解后，剩余固体氧化钙的质量为8.8 g。计算氧化钙的产率。
alues: CaCO₃ = 40.1 + 12.0 + (16.0 × 3) = 100.1 g mol⁻¹; CaO = 40.1 + 16.0 = 56.1 g mol⁻¹.
配平方程式：CaCO₃(s) → CaO(s) + CO₂(g)。摩尔质量：CaCO₃ = 40.1 + 12.0 + (16.0 × 3) = 100.1 g mol⁻¹；CaO = 40.1 + 16.0 = 56.1 g mol⁻¹。
l⁻¹ ≈ 0.1998 mol. According to the 1:1 molar ratio, theoretical moles of CaO = 0.1998 mol.
碳酸钙的摩尔数 = 20.0 g / 100.1 g mol⁻¹ ≈ 0.1998 mol。根据1:1摩尔比，氧化钙的理论摩尔数 = 0.1998 mol。
eoretical mass of CaO = 0.1998 mol × 56.1 g mol⁻¹ ≈ 11.21 g. Actual mass obtained = 8.8 g.
氧化钙的理论质量 = 0.1998 mol × 56.1 g mol⁻¹ ≈ 11.21 g。实际得到的质量 = 8.8 g。
Percentage yield = (8.8 / 11.21) × 100% ≈ 78.5%
产率 = (8.8 / 11.21) × 100% ≈ 78.5%
Reasons for a yield less than 100%: incomplete reaction, loss of product during transfer, or some CaO reacting with moisture in the air.
产率低于100%的原因：反应不完全、转移过程中产品损失，或部分氧化钙与空气中水分发生反应。
3. Rate of Reaction Data Analysis | 反应速率数据分析
Marble chips (CaCO₃) react with an excess of dilute hydrochloric acid in a conical flask connected to a gas syringe. The volume of CO₂ gas is recorded every 10 seconds.
大理石碎片（CaCO₃）与过量稀盐酸在连接气体注射器的锥形瓶中反应，每10秒记录一次CO₂气体的体积。
| Time / s | Volume of CO₂ / cm³ |
|---|---|
| 0 | 0 |
| 10 | 30 |
| 20 | 50 |
| 30 | 65 |
| 40 | 75 |
| 50 | 80 |
Calculate the average rate of reaction between 10 s and 30 s. Explain why the rate decreases over time.
计算10秒至30秒间的平均反应速率，并解释速率为何随时间递减。
Average rate = change in volume / change in time = (65 – 30) cm³ / (30 – 10) s = 35 cm³ / 20 s = 1.75 cm³ s⁻¹.
平均速率 = 体积变化量 / 时间变化量 = (65 – 30) cm³ / (30 – 10) s = 35 cm³ / 20 s = 1.75 cm³ s⁻¹。
The rate decreases because the concentration of hydrochloric acid falls as it is consumed. Additionally, the surface area of marble chips reduces as they react, exposing fewer reactive sites.
速率下降是因为盐酸浓度随着反应消耗而降低。同时大理石碎片表面积减小，暴露的反应位点变少。
If the experiment were repeated with powdered CaCO₃ instead of chips, the initial rate would be much higher due to larger surface area, but the final volume of CO₂ would remain the same.
若将碎片换成粉末状碳酸钙，初始速率将因更大的表面积而显著提升，但最终CO₂体积不变。
4. Titration Calculation | 滴定计算
25.0 cm³ of sodium hydroxide solution (0.100 mol dm⁻³) is titrated against sulfuric acid of unknown concentration. The equivalence point is reached after adding 20.0 cm³ of the acid. Determine the concentration of the sulfuric acid.
用未知浓度的硫酸滴定25.0 cm³氢氧化钠溶液（0.100 mol dm⁻³），终点时消耗硫酸20.0 cm³。计算硫酸的浓度。
d equation: 2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l). Mole ratio NaOH : H₂SO₄ = 2 : 1.
配平方程式：2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)。摩尔比 NaOH：H₂SO
Published by TutorHao | Year 11 Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图