📚 Year 10 AQA Chemistry: Case Study Practical Exercises | AQA 10年级化学：案例分析实战演练
Welcome to this case study workshop, where we will link theory to real-world chemical decision-making. You will compare two methods for preparing copper(II) sulfate crystals, a common salt with uses in agriculture and electroplating. This exercise mirrors the type of application questions you will meet in your AQA exams — testing your ability to calculate, evaluate, and justify chemical choices.
欢迎来到本次案例分析实战演练。我们将把理论知识与真实世界中的化学决策联系起来。你将比较两种制备硫酸铜晶体的方法，这是一种在农业和电镀中常用的盐。这项练习模拟了你在 AQA 考试中会遇到的应用类问题——考查你计算、评估和论证化学选择的能力。
1. Case Scenario Overview | 案例情景概览
A small chemical company plans to produce 100 kg of pure copper(II) sulfate pentahydrate crystals, CuSO₄·5H₂O. The production team can choose between two synthetic routes. Your task is to analyse each route and recommend the more sustainable and cost-effective option.
一家小型化工企业计划生产 100 kg 纯净的五水合硫酸铜晶体 CuSO₄·5H₂O。生产团队可在两条合成路线之间选择。你的任务是分析每条路线，并推荐更具可持续性和成本效益的方案。
Method 1: React black copper(II) oxide powder with warm dilute sulfuric acid, then crystallise the product.
Method 2: React copper metal turnings with hot concentrated sulfuric acid, then crystallise the product.
方法一：将黑色氧化铜粉末与温热的稀硫酸反应，然后结晶得到产物。
方法二：将铜金属屑与热的浓硫酸反应，然后结晶得到产物。
2. Chemical Principles Behind Method 1 | 方法一的化学原理
Method 1 is a classic neutralisation between a metal oxide and an acid. Copper(II) oxide is a base, which reacts with sulfuric acid to form a soluble salt and water. The reaction is carried out with gentle warming until the black powder disappears, leaving a clear blue solution. Because no toxic gases are formed and the acid is dilute, the procedure is relatively safe for a school or small-scale industrial setting.
方法一是金属氧化物与酸之间典型的“中和”反应。氧化铜是一种碱，它与硫酸反应生成可溶性盐和水。反应在微微加热下进行，直至黑色粉末消失，留下清澈的蓝色溶液。由于不产生有毒气体，且使用的是稀酸，该过程在学校或小型工业环境中相对安全。
- Word equation: copper(II) oxide + sulfuric acid → copper(II) sulfate + water
- 文字表达式：氧化铜 + 硫酸 → 硫酸铜 + 水
- Balanced symbol equation: CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l)
- 配平符号方程式：CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l)
3. Chemical Principles Behind Method 2 | 方法二的化学原理
Method 2 involves a redox reaction between copper and concentrated sulfuric acid. The acid acts as an oxidising agent, not just an acid. Copper is oxidised to Cu²⁺ ions, while sulfuric acid is reduced to sulfur dioxide, a toxic gas with a pungent odour. The reaction requires heating and must be performed in a fume cupboard. This route is less atom-efficient and poses safety and environmental challenges.
方法二涉及铜与浓硫酸之间的氧化还原反应。这里的硫酸不仅是一种酸，还充当了氧化剂。铜被氧化为 Cu²⁺ 离子，而硫酸被还原为二氧化硫，一种具有刺激性气味的有毒气体。该反应需要加热，必须在通风橱中进行。这条路线原子效率较低，并带来安全和环境方面的挑战。
Balanced symbol equation: Cu(s) + 2H₂SO₄(aq) → CuSO₄(aq) + SO₂(g) + 2H₂O(l)
配平符号方程式：Cu(s) + 2H₂SO₄(aq) → CuSO₄(aq) + SO₂(g) + 2H₂O(l)
4. Writing and Balancing Equations for the Routes | 为两条路线书写和配平方程式
Correctly balanced equations are essential for quantitative analysis. Always check that the number of atoms of each element is equal on both sides. For Method 1, one mole of CuO reacts with one mole of H₂SO₄. For Method 2, one mole of Cu requires two moles of H₂SO₄. This immediately tells us that Method 2 uses twice as much sulfuric acid per mole of copper sulfate produced.
正确配平的方程式对于定量分析至关重要。务必检查每种元素在反应两侧的原子数是否相等。对于方法一，每摩尔 CuO 消耗一摩尔 H₂SO₄。对于方法二，每摩尔 Cu 需要两摩尔 H₂SO₄。这立刻告诉我们，每生成一摩尔硫酸铜，方法二消耗的硫酸量是方法一的两倍。
e formation of SO₂(g) signals a gaseous waste product that must be captured or neutralised.
状态符号提供了额外细节：(s) 表示固体，(l) 表示液体，(g) 表示气体，(aq) 表示水溶液。在方法二中，SO₂(g) 的生成表明有一种气态废弃物需要被捕获或中和。
5. Calculating Moles of Reactants | 计算反应物的物质的量
reful: 5 water molecules add 5 × 18.0 = 90.0) Wait, CuSO₄ is Mᵣ = 63.5 + 32.1 + (4 × 16.0) = 159.6. Then add 5H₂O: 5 × 18.0 = 90.0. Total Mᵣ = 249.6 (to 1 decimal place).
我们需要制备 100 kg 的 CuSO₄·5H₂O。首先，计算该水合盐的相对分子质量 Mᵣ。 .5 (Cu) + 32.1 (S) + (4 × 16.0) + 5 × (2 × 1.0 + 16.0) = 63.5 + 32.1 + 64.0 + 90.0 = 249.6（保留一位小数）。
oles = 100,000 ÷ 249.6 ≈ 400.6 mol. So we need to produce roughly 401 moles of CuSO₄·5H₂O.
目标产物的物质的量 = 质量 (g) ÷ Mᵣ。100 kg = 100,000 g。物质的量 = 100,000 ÷ 249.6 ≈ 400.6 mol。因此我们需要制备约 401 mol 五水合硫酸铜。
need 400.6 mol CuO and 400.6 mol H₂SO₄. For Method 2, 1 mol Cu gives 1 mol CuSO₄, so we need 400.6 mol Cu but 2 × 400.6 = 801.2 mol H₂SO₄.
根据配平的方程式，1 mol CuO 生成 1 mol CuSO₄（无水盐），结晶后变成 1 mol 水合盐。对于方法一，我们需要 400.6 mol CuO 和 400.6 mol H₂SO₄。对于方法二，1 mol Cu 生成 1 mol CuSO₄，因此我们需要 400.6 mol Cu，但需要 2 × 400.6 = 801.2 mol H₂SO₄。
6. Converting Moles to Masses of Reactants | 将物质的量转换为反应物的质量
. Mᵣ of H₂SO₄ = (2×1.0) + 32.1 + (4×16.0) = 98.1. For Method 1, H₂SO₄ mass = 400.6 × 98.1 ≈ 39,299 g ≈ 39.3 kg. For Method 2, mass of Cu needed = 400.6 × 63.5 ≈ 25,438 g ≈ 25.4 kg; mass of H₂SO₄ = 801.2 × 98.1 ≈ 78,598 g ≈ 78.6 kg.
现在我们将物质的量转换为质量，以便公司订购原料。CuO 的 Mᵣ = 63.5 + 16.0 = 79.5。所需 CuO 质量 = 400.6 × 79.5 ≈ 31,847 g ≈ 31.8 kg。H₂SO₄ 的 Mᵣ = 2×1.0 + 32.1 + 4×16.0 = 98.1。方法一所需 H₂SO₄ 质量 = 400.6 × 98.1 ≈ 39,299 g ≈ 39.3 kg。方法二所需 Cu 的质量 = 400.6 × 63.5 ≈ 25,438 g ≈ 25.4 kg；所需 H₂SO₄ 质量 = 801.2 × 98.1 ≈ 78,598 g ≈ 78.6 kg。
Although Method 2 uses a smaller mass of copper than the mass of copper oxide in Method 1, it consumes almost twice the mass of sulfuric acid. This will influence both cost and waste management.
尽管方法二使用的铜质量比方法一的氧化铜质量小，但它消耗的硫酸质量几乎是方法一的两倍。这将影响成本和废物管理。
7. Theoretical Yield and Percentage Yield Comparison | 理论产量与百分产率比较
The theoretical yield of CuSO₄·5H₂O is 100 kg in both cases, as that is our target. However, real reactions never give 100% yield. Factors such as incomplete reaction, loss during filtration, and product left in solution during crystallisation reduce the actual mass obtained. Suppose laboratory tests show that Method 1 typically has a percentage yield of 85%, while Method 2 achieves 75% due to side reactions and the difficulty of handling hot concentrated acid.
两种情况的理论产量都是 100 kg，因为这是我们的目标。然而，真实反应永远不可能达到 100% 产率。反应不完全、过滤时的损失以及结晶时留在溶液中的产物等因素都会减少实际获得的质量。假设实验室测试表明，方法一的典型百分产率为 85%，而方法二由于副反应和处理热浓酸的难度，只能达到 75%。
- Method 1 actual yield = 100 kg × 0.85 = 85 kg
- 方法一实际产量 = 100 kg × 0.85 = 85 kg
- Method 2 actual yield = 100 kg × 0.75 = 75 kg
- 方法二实际产量 = 100 kg × 0.75 = 75 kg
To obtain the same 100 kg of product, more raw materials would be needed, further increasing costs. Percentage yield is a critical indicator of process efficiency.
要获得同样的 100 kg 产品，需要更多原料，从而进一步增加成本。百分产率是过程效率的一个关键指标。
8. Atom Economy Analysis | 原子经济性分析
d product. It is calculated as: (Mᵣ of desired product ÷ sum of Mᵣ of all reactants) × 100% hydrous Mᵣ 159.6) but we can use the hydrated product Mᵣ if we include water as a reactant? llisation water is not counted. So we should consider the anhydrous CuSO₄ as the immediate product. Atom economy = 159.6 ÷ (79.5 + 98.1) × 100% = 159.6 ÷ 177.6 × 100% ≈ 89.9%. Very high.
原子经济性衡量有多少反应物最终进入了目标产物。其计算公式为：（目标产物的 Mᵣ ÷ 所有反应物的 Mᵣ 之和）× 100%。对于方法一，目标产物为 CuSO₄（无水盐，Mᵣ 159.6）。原子经济性 = 159.6 ÷ (79.5 + 98.1) × 100% ≈ 89.9%。非常高。
ctants Mᵣ = 63.5 + (2 × 98.1) = 63.5 + 196.2 = 259.7. Atom economy = (159.6 ÷ 259.7) × 100% ≈ 61.5%. Nearly 40% of the reactant mass is wasted as SO₂ and extra water.
对于方法二，Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O。目标产物 CuSO₄ (159.6)。反应物 Mᵣ 之和 = 63.5 + (2 × 98.1) = 259.7。原子经济性 = (159.6 ÷ 259.7) × 100% ≈ 61.5%。近 40% 的反应物质量被浪费为 SO₂ 和多余的水。
9. Environmental and Safety Assessment | 环境与安全评估
Method 1 produces only water as a by-product, no toxic gases. The dilute acid is less corrosive, and the reaction is easily controlled. Waste solution contains some unreacted acid but can be neutralised with lime before disposal. Method 2 generates sulfur dioxide, which causes acid rain and respiratory problems. Extra equipment is needed for gas scrubbing, and hot concentrated acid poses severe burn risks. Spill containment and fume extraction increase capital and operational costs.
方法一仅产生水作为副产物，没有有毒气体。稀酸腐蚀性较小，反应易于控制。废液中含有少量未反应的酸，但可在排放前用石灰中和。方法二会产生二氧化硫，导致酸雨和呼吸问题。需要额外设备进行气体洗涤，而且热浓酸会带来严重的灼伤风险。泄漏处理和排风系统会增加资本投入和运营成本。
| Factor | Method 1 | Method 2 |
| By-products | Water only | SO₂ and water |
| Acid concentration | Dilute | Concentrated |
| Gas scrubbing required | No | Yes |
| Corrosion risk | Low | High |
These factors make Method 1 significantly safer and more environmentally friendly.
这些因素使得方法一明显更安全、更环保。
10. Economic Considerations and Raw Material Costs | 经济考量和原材料成本
Copper oxide is generally cheaper than pure copper metal because it is often produced from recycled copper or as a by-product of metallurgy. Sulfuric acid cost is proportional to the quantity used. Method 2 uses nearly double the acid mass and requires more expensive safety infrastructure. Energy costs are also higher because concentrated acid must be heated strongly. Moreover, lower percentage yield in Method 2 means that for every kilogram of product, the effective raw material cost rises.
氧化铜通常比纯金属铜便宜，因为它通常由回收铜或冶金副产品制得。硫酸成本与使用量成正比。方法二使用的硫酸质量接近方法一的两倍，并且需要更昂贵的安全基础设施。由于浓酸需要强烈加热，能源成本也更高。此外，方法二较低的百分产率意味着，每生产一公斤产品，有效原料成本都会上升。
A simple cost simulation would show that Method 1 is the more economical route, even if copper oxide were slightly more expensive per kilogram than copper. The combination of lower acid usage, higher atom economy, and higher yield reduces total production cost.
一个简单的成本模拟会显示，即使氧化铜每千克价格略高于铜，方法一仍是更经济的路线。较低的酸用量、较高的原子经济性和较高的产率共同降低了总生产成本。
11. Improving the Chosen Method | 改进所选方法
Engineers can further optimise Method 1. Unreacted copper oxide can be filtered and recycled back into the process. The acidic filtrate can be neutralised and used to precipitate copper hydroxide, which is then converted back to CuO by heating. Crystallisation can be carried out in stages to maximise crystal purity and minimise loss. Using waste heat from other plant operations reduces energy input for warming the acid.
工程师可以进一步优化方法一。未反应的氧化铜可以过滤并回用到工艺中。酸性滤液可以中和，用于沉淀氢氧化铜，然后通过加热将其转化回 CuO。结晶可以分阶段进行，以最大化晶体纯度并最小化损失。利用工厂其他操作产生的废热可减少加热硫酸所需的能量输入。
These circular principles align with green chemistry goals and improve long-term profitability.
这些循环原则符合绿色化学的目标，并能提高长期盈利能力。
12. Conclusion and Final Recommendation | 结论与最终建议
Based on higher atom economy, greater percentage yield, lower toxic emissions, and reduced handling hazards, Method 1 — reacting copper oxide with dilute sulfuric acid — is the superior choice. It demonstrates how careful chemical analysis can steer industry toward sustainable practices. By practising such case studies, you strengthen your ability to link equations, mole calculations, and environmental impact, exactly as required in AQA GCSE Chemistry.
基于更高的原子经济性、更高的百分产率、更少的毒性排放和更低的操作风险，方法一——将氧化铜与稀硫酸反应——是更优的选择。它展示了细致的化学分析如何引导工业走向可持续实践。通过练习此类案例分析，你加强了将方程式、摩尔计算和环境影响联系起来的能力，这正是 AQA GCSE 化学所要求的。
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