Case Study Practice in Year 10 CAIE Chemistry | Year 10 CAIE 化学：案例分析实战演练

📚 Case Study Practice in Year 10 CAIE Chemistry | Year 10 CAIE 化学：案例分析实战演练

This article presents a series of real-world case study problems designed for Year 10 CAIE Chemistry students. Each case targets key topics from the IGCSE syllabus, including stoichiometry, electrolysis, energetics, reaction rates, redox titrations, salt preparation, organic synthesis, and equilibrium. Work through each worked example carefully to sharpen your analytical skills and exam technique.

本文为Year 10 CAIE化学学生设计了一系列真实情景的案例分析题。每个案例针对IGCSE大纲的核心主题，涵盖化学计量、电解、能量变化、反应速率、氧化还原滴定、盐类制备、有机合成和化学平衡。请认真研究每一个解题范例，以提升你的分析能力和考试技巧。

1. Case Study 1: Purity of Limestone | 案例1：石灰石的纯度

Problem: me of 480 cm³ at room temperature and pressure (r.t.p). Assuming the molar volume of a gas at r.t.p. is 24.0 dm³/mol, calculate the percentage purity of the limestone sample. (Aᵣ: Ca = 40, C = 12, O = 16)

问题：一名学生将2.00 g不纯的石灰石（主要成分为CaCO₃）与过量的盐酸反应。产生的二氧化碳被收集，在室温和常压下体积为480 cm³。已知气体在r.t.p.下的摩尔体积为24.0 dm³/mol，计算该石灰石样品的纯度百分比。（相对原子质量：Ca=40，C=12，O=16）

Step 1: aq) → CaCl₂(aq) + H₂O(l) + CO₂(g). This shows that 1 mole of CaCO₃ produces 1 mole of CO₂.

步骤1：写出配平的化学方程式：CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)。这表明1摩尔CaCO₃生成1摩尔CO₂。

Step 2: = 0.480 dm³. Moles of CO₂ = volume / molar volume = 0.480 dm³ / 24.0 dm³/mol = 0.0200 mol.

步骤2：计算生成的CO₂的物质的量。将体积转换为dm³：480 cm³ = 0.480 dm³。CO₂的物质的量 = 体积 / 摩尔体积 = 0.480 dm³ / 24.0 dm³/mol = 0.0200 mol。

Step 3: From the 1:1 mole ratio in the equation, the moles of pure CaCO₃ that reacted must also be 0.0200 mol.

步骤3：根据方程式中的1：1摩尔比，参加反应的纯CaCO₃的物质的量也一定是0.0200 mol。

Step 4: 40 + 12 + (16×3) = 100 g/mol. Mass = moles × molar mass = 0.0200 mol × 100 g/mol = 2.00 g.

步骤4：计算纯CaCO₃的质量。CaCO₃的摩尔质量=40+12+(16×3)=100 g/mol。质量=物质的量×摩尔质量=0.0200 mol×100 g/mol=2.00 g。

Step 5: Percentage purity = (mass of pure CaCO₃ / mass of impure sample) × 100% = (2.00 g / 2.00 g) × 100% = 100%. This indicates the sample was actually pure – but in a real scenario impurities would lower the yield of CO₂.

步骤5：纯度百分比=（纯CaCO₃的质量 / 不纯样品的质量）×100% =（2.00 g / 2.00 g）×100% = 100%。这表明该样品实际上是纯净的——但在真实情况下，杂质会降低CO₂的产率。

Common Pitfall: Forgetting to convert cm³ to dm³ or using the wrong molar volume (24.0 dm³ at r.t.p. for IGCSE). Always check units and significant figures.

常见误区：忘记将cm³转换为dm³，或使用错误的摩尔体积（IGCSE中r.t.p.下为24.0 dm³）。务必检查单位和有效数字。

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2. Case Study 2: Electrolysis of Copper(II) Chloride | 案例2：氯化铜溶液的电解

Problem: late the mass of copper deposited on the cathode. (Aᵣ: Cu = 63.5; 1 Faraday = 96500 C/mol)

问题：使用惰性电极电解氯化铜溶液。通入0.50 A的恒定电流达30分钟。计算在阴极上沉积的铜的质量。（Aᵣ: Cu = 63.5；1法拉第 = 96500 C/mol）

Step 1: de: Cu²⁺(aq) + 2e⁻ → Cu(s). This shows that 2 moles of electrons deposit 1 mole of copper.

步骤1：写出阴极半反应方程式：Cu²⁺(aq) + 2e⁻ → Cu(s)。这表明2摩尔电子沉积1摩尔铜。

Step 2: t. First convert time to seconds: 30 min = 30 × 60 = 1800 s. Q = 0.50 A × 1800 s = 900 C.

步骤2：使用Q = I × t计算通过的总电荷量。首先将时间转换为秒：30 min = 30 × 60 = 1800 s。Q = 0.50 A × 1800 s = 900 C。

Step 3: Determine the moles of electrons: moles of e⁻ = Q / F = 900 C / 96500 C/mol ≈ 0.00933 mol.

步骤3：确定电子的物质的量：电子的物质的量 = Q / F = 900 C / 96500 C/mol ≈ 0.00933 mol。

Step 4: o to find moles of copper: moles of Cu = moles of e⁻ / 2 = 0.00933 mol / 2 = 0.004665 mol.

步骤4：利用2：1的比例求铜的物质的量：铜的物质的量 = 电子的物质的量 / 2 = 0.00933 mol / 2 = 0.004665 mol。

Step 5: mass = moles × Aᵣ = 0.004665 mol × 63.5 g/mol ≈ 0.296 g (0.30 g to 2 significant figures).

步骤5：计算铜的质量：质量 = 物质的量 × Aᵣ = 0.004665 mol × 63.5 g/mol ≈ 0.296 g（取两位有效数字为0.30 g）。

Tip: Be careful with unit conversions for time and with the Faraday constant. Also note that inert electrodes must be specified or else anode material could interfere.

技巧：注意时间单位的换算以及法拉第常数的使用。还要注意必须注明惰性电极，否则阳极材料可能会干扰反应。

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3. Case Study 3: Enthalpy Change of Neutralisation | 案例3：中和反应焓变

Problem: 50.0 cm³ of 2.00 mol/dm³ hydrochloric acid is mixed with 50.0 cm³ of 2.00 mol/dm³ sodium hydroxide solution in a polystyrene cup. The temperature rises by 13.5 °C. Calculate 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 1.00 g/cm³)

问题：将50.0 cm³ 2.00 mol/dm³的盐酸与50.0 cm³ 2.00 mol/dm³的氢氧化钠溶液在聚苯乙烯杯中混合。温度上升了13.5 °C。计算每生成1摩尔水时的中和焓变。（假设溶液的比热容为4.2 J/g°C，密度为1.00 g/cm³）

Step 1: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l). 1 mole of acid reacts to produce 1 mole of water.

步骤1：写出中和反应方程式：HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)。1摩尔酸反应生成1摩尔水。

Step 2: Total volume = 50.0 + 50.0 = 100.0 cm³. Since density is 1.00 g/cm³, total mass = 100.0 g.

步骤2：计算溶液的总体积和质量。总体积 = 50.0 + 50.0 = 100.0 cm³。由于密度为1.00 g/cm³，总质量 = 100.0 g。

Step 3: energy released (q) using q = mcΔT. q = 100.0 g × 4.2 J/g°C × 13.5 °C = 5670 J = 5.67 kJ.

步骤3：使用q = mcΔT计算释放的热量。q = 100.0 g × 4.2 J/g°C × 13.5 °C = 5670 J = 5.67 kJ。

Step 4: = 2.00 mol/dm³ × 0.0500 dm³ = 0.100 mol. The same for NaOH, so water produced = 0.100 mol.

步骤4：确定生成水的物质的量。HCl的物质的量 = 浓度 × 体积(dm³) = 2.00 mol/dm³ × 0.0500 dm³ = 0.100 mol。NaOH的物质的量相同，因此生成水0.100 mol。

Step 5: Enthalpy change per mole, ΔH = –q / moles of water = –5.67 kJ / 0.100 mol = –56.7 kJ/mol. The negative sign indicates the reaction is exothermic.

步骤5：每摩尔的焓变，ΔH = –q / 水的物质的量 = –5.67 kJ / 0.100 mol = –56.7 kJ/mol。负号表示反应为放热反应。

Key Point: Always remember to divide by the limiting reactant moles and include the negative sign for exothermic reactions in ΔH.

关键点：务必记得除以限制反应物的物质的量，并在ΔH中为放热反应加上负号。

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4. Case Study 4: Rate of Reaction with Magnesium | 案例4：镁与酸反应的速率

Scenario: A student adds a 3.0 cm strip of magnesium ribbon to 25.0 cm³ of 1.0 mol/dm³ HCl and measures the time taken for the magnesium to disappear completely. The following data were collected for different HCl concentrations:

情景：一名学生将一条3.0 cm的镁带放入25.0 cm³ 1.0 mol/dm³的盐酸中，并测量镁带完全消失所需的时间。在不同HCl浓度下收集了以下数据：

[HCl] (mol/dm³)	Time for Mg to disappear (s)	Initial rate (1/t) (s⁻¹)
1.0	40	0.025
0.8	50	0.020
0.6	67	0.015
0.4	100	0.010

Analysis: dicating that the reaction is first order with respect to HCl. The equation for the reaction is: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g).

分析：反应的初始速率可以估算为1/时间。以速率对[HCl]作图，图形为一条过原点的直线，表明该反应对HCl是一级反应。反应方程式为：Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)。

Discussion: Why does the graph pass through the origin? rs, giving a rate of zero. From the gradient, one can determine the rate constant k. This simple approach assumes the reaction rate is constant until the Mg is consumed, which is an approximation.

讨论：为什么图形通过原点？因为如果[HCl]=0，不发生反应，速率为零。根据斜率可以求出速率常数k。这种简单方法假设在Mg耗尽前反应速率恒定，这只是一个近似。

Exam Tip: When describing how to measure rate, always state how a variable is controlled: mass of Mg, length of ribbon, surface area, temperature. Use appropriate apparatus like a gas syringe or measuring cylinder over water for H₂ collection.

考试提示：在描述如何测量速率时，一定要说明如何控制变量：镁的质量、镁带的长度、表面积、温度。使用合适的仪器，如气体注射器或排水集气法收集H₂。

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5. Case Study 5: Redox Titration of Iron(II) | 案例5：铁(II)的氧化还原滴定

Problem: . Calculate the concentration of Fe²⁺ in the original solution. The ionic equation is: MnO₄⁻(aq) + 5Fe²⁺(aq) + 8H⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l)

问题：取25.0 cm³的硫酸亚铁溶液酸化后，用0.0200 mol/dm³的高锰酸钾溶液滴定，滴定体积为24.0 cm³。计算原溶液中Fe²⁺的浓度。离子方程式为：MnO₄⁻(aq) + 5Fe²⁺(aq) + 8H⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l)

Step 1: ntration × volume (in dm³) = 0.0200 mol/dm³ × 0.0240 dm³ = 0.000480 mol (4.80 × 10⁻⁴ mol).

步骤1：计算所用的MnO₄⁻的物质的量：物质的量 = 浓度 × 体积(dm³) = 0.0200 mol/dm³ × 0.0240 dm³ = 0.000480 mol (4.80 × 10⁻⁴ mol)。

Step 2: e of MnO₄⁻ reacts with 5 moles of Fe²⁺. So moles of Fe²⁺ = 5 × 0.000480 mol = 0.00240 mol.

步骤2：由方程式可知，1摩尔MnO₄⁻与5摩尔Fe²⁺反应。因此Fe²⁺的物质的量 = 5 × 0.000480 mol = 0.00240 mol。

Step 3: concentration of Fe²⁺ = moles / volume (dm³) = 0.00240 mol / 0.0250 dm³ = 0.0960 mol/dm³.

步骤3：这物质的量存在于25.0 cm³的Fe²⁺溶液中。计算浓度：Fe²⁺浓度 = 物质的量 / 体积(dm³) = 0.00240 mol / 0.0250 dm³ = 0.0960 mol/dm³。

Common Error: Forgetting to multiply by 5 or using the wrong mole ratio. Some students incorrectly use a 1:1 ratio. Always check the balanced half-equations before combining.

常见错误：忘记乘以5或使用错误的摩尔比。有些学生错误地使用1:1的比例。在合并半反应前一定要检查配平情况。

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6. Case Study 6: Preparation of an Insoluble Salt | 案例6：不溶性盐的制备

Scenario: itate is filtered, washed and dried, giving a mass of 3.10 g. (Aᵣ: Ba = 137, S = 32, O = 16) Calculate the percentage yield.

情景：一名学生将30.0 cm³ 0.50 mol/dm³的氯化钡溶液与过量的硫酸钠溶液混合来制备硫酸钡。沉淀经过滤、洗涤和干燥后，质量为3.10 g。（Aᵣ: Ba=137，S=32，O=16）计算产率百分比。

Step 1: n: BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq). The mole ratio of BaCl₂ to BaSO₄ is 1:1.

步骤1：方程式：BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)。BaCl₂与BaSO₄的摩尔比为1:1。

Step 2: dm³ = 0.015 mol. This is the limiting reactant, so theoretical moles of BaSO₄ = 0.015 mol.

步骤2：BaCl₂的物质的量 = 浓度 × 体积 = 0.50 mol/dm³ × 0.030 dm³ = 0.015 mol。这是限制反应物，因此理论BaSO₄的物质的量 = 0.

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