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

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

Mastering Year 10 Chemistry for the Eduqas specification requires more than recalling facts – it demands the ability to apply knowledge to unfamiliar scenarios. This article guides you through a series of structured case studies, each designed to practise the analytical, mathematical and experimental skills assessed in your exams. You will calculate empirical formulae, interpret rate data, perform titration calculations, analyse electrolysis products, work with bond energies, consider reversible reactions and evaluate practical techniques.

掌握 Year 10 Eduqas 化学并非只是记忆知识点，更需要将知识应用于陌生的情境。本文通过一系列精心设计的案例学习，帮助你训练考试中所要求的分析、计算和实验技能。你将练习计算经验式、解读反应速率数据、完成滴定运算、分析电解产物、运用键能计算、考虑可逆反应以及评价实验技术。

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1. Determining the Empirical Formula of Magnesium Oxide | 案例一：测定氧化镁的经验式

en to allow oxygen to enter. After the magnesium had stopped glowing, the lid was removed and heating continued for a few more minutes. The crucible was left to cool and then weighed again. Results: mass of crucible + lid = 28.50 g; mass of crucible + lid + magnesium = 28.98 g; mass of crucible + lid + magnesium oxide = 29.30 g.

一名学生称量干净坩埚与盖，然后放入卷好的镁条。强热坩埚，微微打开盖子让氧气进入。待镁不再发光后，取下盖子继续加热几分钟。冷却后再次称量。测得数据：坩埚+盖质量 = 28.50 g；坩埚+盖+镁 = 28.98 g；坩埚+盖+氧化镁 = 29.30 g。

culate the empirical formula of magnesium oxide. (Relative atomic masses: Mg = 24, O = 16)

问题：利用数据计算氧化镁的经验式。（相对原子质量：Mg = 24，O = 16）

Step 1: Find the mass of magnesium. Mass of Mg = 28.98 g - 28.50 g = 0.48 g.

步骤1：计算镁的质量。镁的质量 = 28.98 g - 28.50 g = 0.48 g。

sium oxide = 29.30 g - 28.50 g = 0.80 g. Hence, mass of oxygen = 0.80 g - 0.48 g = 0.32 g.

步骤2：计算氧的质量。氧化镁质量 = 29.30 g - 28.50 g = 0.80 g。因此，氧的质量 = 0.80 g - 0.48 g = 0.32 g。

es. Moles of Mg = 0.48 g ÷ 24 g/mol = 0.02 mol. Moles of O = 0.32 g ÷ 16 g/mol = 0.02 mol.

步骤3：将质量转换为物质的量。Mg的物质的量 = 0.48 g ÷ 24 g/mol = 0.02 mol。O的物质的量 = 0.32 g ÷ 16 g/mol = 0.02 mol。

mber of moles. Ratio Mg : O = 0.02/0.02 : 0.02/0.02 = 1 : 1. The empirical formula is MgO.

步骤4：除以最小的物质的量。比例 Mg : O = 0.02/0.02 : 0.02/0.02 = 1 : 1。经验式为 MgO。

Common error: If the lid is closed completely, oxygen cannot enter and some magnesium may not react, leading to a mass of oxide that is too low and an incorrect ratio (more Mg than O).

常见错误：如果盖子完全盖严，氧气无法进入，会导致部分镁未反应，氧化镁质量偏低，得出错误比例（镁多于氧）。

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2. Rate of Reaction: Sodium Thiosulfate and Hydrochloric Acid | 案例二：反应速率：硫代硫酸钠与盐酸

In this case study, a student investigates how the concentration of sodium thiosulfate affects the rate of its reaction with hydrochloric acid. The reaction produces a yellow precipitate of sulfur, which makes the mixture go cloudy. The student measures the time taken for a cross drawn on paper beneath the conical flask to disappear from view.

在本案例中，学生研究硫代硫酸钠浓度如何影响其与盐酸反应的速率。反应生成黄色硫沉淀，使混合物变浑浊。学生测量锥形瓶下画在纸上的十字消失所需的时间。

equation for the reaction is: Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl(aq) + SO₂(g) + S(s) + H₂O(l).

反应方程式为：Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl(aq) + SO₂(g) + S(s) + H₂O(l)。

The student records the following results for different volumes of sodium thiosulfate (concentration 0.15 mol/dm³) made up to a total volume of 50 cm³ with water, while keeping the volume of acid (5 cm³ of 2.0 mol/dm³ HCl) constant.

学生记录了以下数据：不同体积的硫代硫酸钠溶液（浓度0.15 mol/dm³），加水配成总体积50 cm³，同时保持盐酸用量不变（5 cm³ 2.0 mol/dm³ HCl）。

Volume of Na₂S₂O₃ (cm³)	Volume of water (cm³)	Time for cross to disappear (s)
50	0	10
40	10	16
30	20	25
20	30	50
10	40	100

Question: For each concentration, calculate a relative rate as 1/time. Then plot a graph of rate against volume of thiosulfate (a measure of concentration). Describe the relationship and explain it using collision theory.

问题：针对每个浓度，计算相对速率 1/时间。然后绘制速率对硫代硫酸钠体积（代表浓度）的曲线。描述两者关系，并用碰撞理论解释。

Step: The concentration of thiosulfate is proportional to its volume because the total volume is constant. The rate (1/t) values are: 0.100, 0.0625, 0.040, 0.020, 0.010 s⁻¹.

步骤：由于总体积恒定，硫代硫酸钠的浓度与其体积成正比。相对速率（1/t）分别为：0.100, 0.0625, 0.040, 0.020, 0.010 s⁻¹。

Plotting the data gives a straight line passing through the origin, showing that rate is directly proportional to the concentration of Na₂S₂O₃.

绘制数据可得一条过原点的直线，表明速率与 Na₂S₂O₃ 的浓度成正比。

Explanation: Increasing the concentration increases the number of reactant particles per unit volume, so collisions between Na₂S₂O₃ and HCl particles become more frequent, raising the rate of successful collisions that lead to product formation.

解释：增加浓度使单位体积内的反应物粒子数增多，Na₂S₂O₃ 与 HCl 粒子之间的碰撞更加频繁，发生有效碰撞的频率增加，从而提高了反应速率。

The student notices that the cross disappears when a fixed amount of sulfur has been produced. This means that measuring the time for the cross to disappear is equivalent to measuring the time taken for a fixed amount of product to form.

学生注意到，当生成一定量的硫时十字就会消失。这说明测量十字消失的时间，即相当于测定了生成固定量产物所需的时间。

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3. Acid–Base Titration Calculation | 案例三：酸碱滴定计算

A student uses a pipette to transfer 25.0 cm³ of sodium hydroxide solution of unknown concentration into a conical flask. A few drops of phenolphthalein indicator are added. The student fills a burette with 0.100 mol/dm³ sulfuric acid and titrates the alkali until the solution just turns from pink to colourless. The mean titre is found to be 20.00 cm³.

学生用移液管吸取 25.0 cm³ 未知浓度的氢氧化钠溶液放入锥形瓶，加入几滴酚酞指示剂。在滴定管中装入 0.100 mol/dm³ 的硫酸，滴定碱液，直至溶液刚好由粉红色变为无色。平均滴定体积为 20.00 cm³。

The balanced equation is: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.

反应方程式为：H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O。

Question: Calculate the concentration of the sodium hydroxide solution in mol/dm³.

问题：计算氢氧化钠溶液的浓度，单位为 mol/dm³。

SO₄ used. Moles = concentration × volume = 0.100 mol/dm³ × (20.00/1000) dm³ = 0.00200 mol.

步骤1：计算所用硫酸的物质的量。物质的量 = 浓度 × 体积 = 0.100 mol/dm³ × (20.00/1000) dm³ = 0.00200 mol。

ion. 1 mol H₂SO₄ reacts with 2 mol NaOH. So moles of NaOH = 2 × 0.00200 mol = 0.00400 mol.

步骤2：根据方程式中的物质的量之比计算。1 mol H₂SO₄ 与 2 mol NaOH 反应。因此 NaOH 的物质的量 = 2 × 0.00200 mol = 0.00400 mol。

n of NaOH. Concentration = moles ÷ volume = 0.00400 mol ÷ (25.0/1000) dm³ = 0.160 mol/dm³.

步骤3：计算 NaOH 的浓度。浓度 = 物质的量 ÷ 体积 = 0.00400 mol ÷ (25.0/1000) dm³ = 0.160 mol/dm³。

If the student had used methyl orange as the indicator instead, the colour change would be from yellow to orange/red. The titre would remain the same because both reactants are strong.

如果学生改用甲基橙作指示剂，颜色会由黄色变为橙色/红色。滴定体积不变，因为两者均为强电解质。

Common mistake: forgetting to multiply the acid moles by 2, leading to half the correct concentration of NaOH.

常见错误：忘记将酸的物质的量乘以2，导致求出的 NaOH 浓度仅为正确值的一半。

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4. Electrolysis of Copper(II) Chloride Solution | 案例四：电解氯化铜溶液

A student sets up an electrolysis cell using copper(II) chloride solution (CuCl₂) as the electrolyte and two inert graphite electrodes. The electrodes are connected to a DC power supply. After a few minutes, the student observes bubbles forming at the anode and a pink-brown solid depositing on the cathode.

学生搭建一个电解池，以氯化铜溶液（CuCl₂）为电解质，两根惰性石墨电极与直流电源相连。几分钟后，观察到阳极有气泡产生，阴极有粉棕色固体析出。

Question: Identify the products at the anode and cathode. Write the half-equations occurring at each electrode and explain how the ions move during electrolysis.

问题：判断阳极和阴极的产物。写出每个电极的半反应方程式，并解释电解过程中离子如何移动。

de. Cu²⁺ ions gain two electrons to form copper metal: Cu²⁺ + 2e⁻ → Cu. This is reduction.

在阴极（负极）：Cu²⁺ 离子移向阴极，因为它们被负电极吸引。Cu²⁺ 得到两个电子形成铜金属：Cu²⁺ + 2e⁻ → Cu，属于还原反应。

de. They lose one electron each to form chlorine gas: 2Cl⁻ → Cl₂ + 2e⁻. This is oxidation.

在阳极（正极）：Cl⁻ 离子移向阳极，每个离子失去一个电子，生成氯气：2Cl⁻ → Cl₂ + 2e⁻，属于氧化反应。

l₂. The pink-brown solid is copper; the bubbles at the anode are chlorine gas, which has a sharp smell and bleaches damp litmus paper.

总反应为：CuCl₂ → Cu + Cl₂。粉棕色固体是铜；阳极产生的气泡是氯气，有刺激性气味，能使湿润的蓝色石蕊试纸褪色。

If the electrodes were replaced with copper electrodes, the anode would dissolve, giving Cu²⁺ ions, and copper would still deposit on the cathode. This process is used in electroplating and refining copper.

若将电极换为铜电极，阳极会溶解生成 Cu²⁺，而阴极仍会析出铜。该过程应用于电镀和电解精炼铜。

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5. Bond Energy Calculations for an Exothermic Reaction | 案例五：放热反应的键能计算

chlorine to form hydrogen chloride is exothermic. The student is given bond energy data: H–H bond energy = 436 kJ/mol, Cl–Cl bond energy = 243 kJ/mol, H–Cl bond energy = 432 kJ/mol.

氢气与氯气生成氯化氢的反应是放热的。给出的键能数据：H–H 键能 = 436 kJ/mol，Cl–Cl 键能 = 243 kJ/mol，H–Cl 键能 = 432 kJ/mol。

Balanced equation: H₂ + Cl₂ → 2HCl.

反应方程式：H₂ + Cl₂ → 2HCl。

Question: Use the bond energies to calculate the overall energy change for the reaction. State whether the reaction is exothermic or endothermic and explain what this tells us about the energy of the products and reactants.

问题：利用键能计算反应的总能量变化。判断反应是放热还是吸热，并解释这说明了产物和反应物的能量高低。

Step 1: Energy required to break bonds (reactants). Breaking 1 mole of H–H bonds uses 436 kJ; breaking 1 mole of Cl–Cl bonds uses 243 kJ. Total energy in = 436 + 243 = 679 kJ.

步骤1：断键所需的能量（反应物）。断裂 1 mol H–H 键吸收 436 kJ；断裂 1 mol Cl–Cl 键吸收 243 kJ。总吸收能量 = 436 + 243 = 679 kJ。

Step 2: Energy released forming bonds (products). Forming 2 moles of H–Cl bonds releases 2 × 432 = 864 kJ. Total energy out = 864 kJ.

步骤2：成键释放的能量（生成物）。形成 2 mol H–Cl 键释放 2 × 432 = 864 kJ。总释放能量 = 864 kJ。

Step 3: Overall energy change = Energy in – Energy out = 679 – 864 = –185 kJ per mole of reaction as written.

步骤3：总能量变化 = 吸收能量 – 释放能量 = 679 – 864 = –185 kJ（按方程式的反应进度）。

The negative sign indicates that the reaction is exothermic. Energy is given out to the surroundings; the products are at a lower energy level than the reactants, and the temperature of the reaction mixture would rise.

负值表明反应为放热反应。能量散发到周围环境中；生成物的能量低于反应物，反应混合物的温度会升高。

In contrast, an endothermic reaction, such as photosynthesis, would have a positive overall energy change because more energy is absorbed to break bonds than is released forming them.

相反，吸热反应（如光合作用）的总能量变化为正值，因为断键吸收的能量大于成键释放的能量。

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6. Reduction of Copper(II) Oxide: Finding the Percentage Yield | 案例六：还原氧化铜——计算产率

n is: CuO + H₂ → Cu + H₂O. After the reaction, the tube is cooled while hydrogen still flows, and the reddish-brown copper produced is weighed. The mass of copper obtained is 1.45 g.

学生称取 2.00 g 黑色氧化铜粉末，放入硬质玻璃管中。加热时通入氢气。方程式：CuO + H₂ → Cu + H₂O。反应结束后，在氢气气氛中冷却，称得红褐色铜的质量为 1.45 g。

om 2.00 g CuO, and hence determine the percentage yield of copper. (Ar: Cu = 63.5, O = 16)

问题：计算 2.00 g CuO 理论上可得到的铜的质量，并求出铜的产率。（相对原子质量：Cu = 63.5，O = 16）

Mr = 63.5 + 16 = 79.5 g/mol. Moles of CuO = 2.00 ÷ 79.5 = 0.02516 mol (approx 0.0252 mol).

步骤1：计算 CuO 的摩尔质量。Mr = 63.5 + 16 = 79.5 g/mol。CuO 的物质的量 = 2.00 ÷ 79.5 = 0.02516 mol（约 0.0252 mol）。

Cu. Theoretical moles of Cu = 0.0252 mol. Theoretical mass of Cu = 0.0252 × 63.5 = 1.60 g.

步骤2：由方程式，1 mol CuO 生成 1 mol Cu。理论 Cu 的物质的量 = 0.0252 mol，理论 Cu 质量 = 0.0252 × 63.5 = 1.60 g。

3: Percentage yield = (actual mass / theoretical mass) × 100 = (1.45 / 1.60) × 100 = 90.6%.

步骤3：产率 = (实际质量 / 理论质量) × 100 = (1.45 / 1.60) × 100 = 90.6%。

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