Year 10 Edexcel Chemistry: Case Study Practical Drills | 案例分析实战演练

📚 Year 10 Edexcel Chemistry: Case Study Practical Drills | 案例分析实战演练

Case study questions in Year 10 Edexcel Chemistry require you to connect theory with real-world situations. This article presents a series of practical drills covering key topics such as water analysis, yield calculations, titrations, electrolysis, corrosion, reversible reactions, energy changes and reaction rates. Each case is broken down into clear explanations and worked steps, helping you build confidence in applying chemical principles.

Year 10 Edexcel 化学的案例分析题要求你将理论与实际情景联系起来。本文提供一系列实战演练，涵盖水质分析、产率计算、滴定、电解、腐蚀、可逆反应、能量变化和反应速率等关键主题。每个案例都配有清晰的解释和分步计算，帮助你建立应用化学原理的信心。

1. Water Purity Analysis | 水质纯度分析

A student collects a sample of river water and performs simple tests to check its purity. She finds that the river water boils at 102 °C and freezes at −2 °C, while pure water boils at 100 °C and freezes at 0 °C. She also carries out ion tests on the water.

一位学生采集了一份河水样品，并进行了简单的纯度检测。她发现河水在 102 °C 沸腾、在 −2 °C 结冰，而纯水在 100 °C 沸腾、0 °C 结冰。她还对水样进行了离子检测。

The raised boiling point and lowered freezing point indicate the presence of dissolved impurities, typically mineral salts. These particles interfere with the escape of water molecules during boiling and disrupt the formation of the regular ice lattice, causing boiling point elevation and freezing point depression.

沸点升高和凝固点降低说明水中含有溶解的杂质，通常是矿物盐。这些粒子干扰了沸腾时水分子的逸出，并破坏了结冰时规则晶格的形成，导致沸点上升和凝固点下降。

To test for chloride ions, a few drops of acidified silver nitrate solution are added. A white precipitate of silver chloride (AgCl) confirms the presence of Cl⁻ ions: Ag⁺(aq) + Cl⁻(aq) → AgCl(s). For sulfate ions, acidified barium chloride solution is used; a white precipitate of barium sulfate (BaSO₄) indicates SO₄²⁻: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s).

检验氯离子时，加入几滴酸化的硝酸银溶液。生成白色氯化银 (AgCl) 沉淀即证明有 Cl⁻ 离子：Ag⁺(aq) + Cl⁻(aq) → AgCl(s)。检验硫酸根离子时，使用酸化的氯化钡溶液；白色硫酸钡 (BaSO₄) 沉淀表示 SO₄²⁻ 离子：Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)。

The student evaporates 100 cm³ of the filtered river water and obtains 0.45 g of solid residue. The concentration of dissolved solids can be calculated as: concentration = mass of solute / volume of solution = 0.45 g / 100 cm³ = 0.0045 g/cm³, or 4.5 g/dm³.

该学生将 100 cm³ 过滤后的河水蒸发，得到 0.45 g 固体残留物。溶解固体的浓度可以这样计算：浓度 = 溶质质量 / 溶液体积 = 0.45 g / 100 cm³ = 0.0045 g/cm³，即 4.5 g/dm³。

To obtain pure water from the sample, simple distillation is used. The impure water is heated until it boils, the steam is cooled and condensed in a condenser, and the pure liquid is collected. The dissolved solids remain in the flask because they do not turn into vapour at 100 °C.

要从样品中获得纯水，可以使用简单蒸馏。将不纯水加热至沸腾，产生的蒸气在冷凝管中冷却凝结，收集到的液体即为纯水。溶解的固体会留在烧瓶中，因为它们在 100 °C 时不会变为蒸气。

2. Preparing Copper Sulfate Crystals: Yield Calculation | 制备硫酸铜晶体：产率计算

A student reacts 1.60 g of black copper(II) oxide (CuO) with an excess of warm dilute sulfuric acid. The resulting blue solution is filtered, heated to evaporate some water, and left to crystallise. The final yield of hydrated copper(II) sulfate crystals, CuSO₄·5H₂O, is 3.80 g. The goal is to calculate the percentage yield.

某学生将 1.60 g 黑色氧化铜 (CuO) 与过量的温热稀硫酸反应。将得到的蓝色溶液过滤、加热蒸发部分水分后，冷却结晶。最终获得五水合硫酸铜晶体 CuSO₄·5H₂O 共 3.80 g。需要计算其产率。

The balanced equation for the neutralisation is: CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l). On crystallisation, CuSO₄ + 5H₂O → CuSO₄·5H₂O. Therefore, 1 mol of CuO produces 1 mol of CuSO₄·5H₂O.

中和反应的化学方程式为：CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l)。结晶时，CuSO₄ + 5H₂O → CuSO₄·5H₂O。因此，1 mol CuO 生成 1 mol CuSO₄·5H₂O。

Relative atomic masses: Cu = 63.5, S = 32, O = 16, H = 1. Mᵣ of CuO = 63.5 + 16 = 79.5. Mᵣ of CuSO₄·5H₂O = 63.5 + 32 + (4 × 16) + 5 × (2 × 1 + 16) = 63.5 + 32 + 64 + 90 = 249.5.

相对原子质量：Cu = 63.5，S = 32，O = 16，H = 1。CuO 的相对分子质量 Mᵣ = 63.5 + 16 = 79.5。CuSO₄·5H₂O 的 Mᵣ = 63.5 + 32 + (4 × 16) + 5 × (2 × 1 + 16) = 249.5。

First, find the number of moles of the limiting reactant CuO used: moles = mass / Mᵣ = 1.60 / 79.5 ≈ 0.02013 mol. Theoretical mass of CuSO₄·5H₂O = moles × Mᵣ = 0.02013 × 249.5 ≈ 5.02 g. The actual mass obtained is 3.80 g.

首先，计算限制反应物 CuO 的物质的量：物质的量 = 质量 / Mᵣ = 1.60 / 79.5 ≈ 0.02013 mol。CuSO₄·5H₂O 的理论产量 = 物质的量 × Mᵣ = 0.02013 × 249.5 ≈ 5.02 g。实际获得的晶体质量为 3.80 g。

Percentage yield = (actual yield / theoretical yield) × 100% = (3.80 / 5.02) × 100% ≈ 75.7%

产率 = (实际产量 / 理论产量) × 100% = (3.80 / 5.02) × 100% ≈ 75.7%

The yield is less than 100% mainly because some product is lost during filtration (left on filter paper), some remains in solution after crystallisation, and some may be spilled when transferring between containers.

产率低于 100% 主要是因为过滤时有部分产物留在滤纸上、结晶后溶液中仍有部分残留，以及转移过程中可能有溅出损失。

3. Acid-Base Titration: Finding Concentration | 酸碱滴定：测定浓度

In a titration, 25.0 cm³ of sodium hydroxide solution is transferred into a conical flask using a pipette. A few drops of phenolphthalein indicator are added, giving a pink colour. Hydrochloric acid of concentration 0.100 mol/dm³ is added from a burette until the solution just turns colourless. The final burette reading shows that 23.5 cm³ of

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