📚 Case Study in Action: Stoichiometry, Yield and Energy | 案例分析实战演练：计量、产率与能量
A student investigates the reaction between a magnesium strip and excess dilute hydrochloric acid. The aim is to determine the purity of the magnesium sample, the percentage yield of hydrogen gas collected, and the enthalpy change of the reaction. This case study will guide you through the step-by-step analysis of the experimental data, linking the key quantitative aspects of the CIE IGCSE Chemistry syllabus.
一名学生探究镁条与过量稀盐酸之间的反应，目的是测定镁样品的纯度、收集的氢气的百分产率以及该反应的焓变。本案例分析将逐步引导你分析实验数据，串联起 CIE IGCSE 化学大纲中的关键定量知识。
1. Case Study Scenario | 案例情境
A 0.30 g piece of magnesium ribbon is placed into a conical flask containing 50 cm³ of 1.0 mol/dm³ hydrochloric acid, which is in excess. The flask is connected to a gas syringe to collect the hydrogen gas evolved. A thermometer is placed in the acid to record the temperature change. The experiment is conducted at room temperature and pressure (rtp), where the molar volume of a gas is 24 000 cm³ mol⁻¹.
将 0.30 g 镁条放入装有 50 cm³ 1.0 mol/dm³ 盐酸（过量）的锥形瓶中。锥形瓶与气体注射器相连，用于收集生成的氢气。酸液中放有温度计，记录温度变化。实验在常温常压（rtp）下进行，气体的摩尔体积为 24 000 cm³ mol⁻¹。
The student records the following data:
学生记录的数据如下：
- Mass of magnesium ribbon = 0.30 g | 镁条质量 = 0.30 g
- Final volume of gas collected = 270 cm³ | 收集气体的最终体积 = 270 cm³
- Initial temperature of acid = 20.0 °C | 盐酸初始温度 = 20.0 °C
- Final temperature of mixture = 35.0 °C | 混合物最终温度 = 35.0 °C
- Density of dilute acid assumed to be 1.00 g/cm³ | 稀酸密度按 1.00 g/cm³ 计
- Specific heat capacity of solution = 4.2 J g⁻¹ °C⁻¹ | 溶液比热容 = 4.2 J g⁻¹ °C⁻¹
2. Writing the Balanced Equation | 书写平衡化学方程式
The reaction between magnesium and hydrochloric acid produces magnesium chloride and hydrogen gas. The balanced symbol equation is:
镁与盐酸反应生成氯化镁和氢气。配平的化学方程式如下：
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
This equation shows that 1 mole of magnesium reacts with 2 moles of HCl to produce 1 mole of hydrogen gas. The stoichiometric ratio Mg : H₂ is 1 : 1.
该方程式表明，1 摩尔镁与 2 摩尔 HCl 反应生成 1 摩尔氢气。镁与氢气的化学计量比为 1 : 1。
3. Calculating Moles of Hydrogen Produced | 计算生成氢气的物质的量
Using the molar volume at rtp, we convert the volume of hydrogen collected into moles.
利用常温常压下的摩尔体积，我们将收集到的氢气体积转换为物质的量。
n(H₂) = Volume (cm³) / 24 000 cm³ mol⁻¹
n(H₂) = 270 cm³ / 24 000 cm³ mol⁻¹ = 0.01125 mol
n(H₂) = 270 cm³ / 24 000 cm³ mol⁻¹ = 0.01125 mol
4. Determining Moles of Magnesium That Reacted | 计算参与反应的镁的物质的量
From the balanced equation, the mole ratio of Mg to H₂ is 1:1. Therefore, the number of moles of magnesium that actually reacted is equal to the moles of hydrogen produced.
根据配平的方程式，Mg 与 H₂ 的摩尔比为 1:1。因此，实际参与反应的镁的物质的量等于生成氢气的物质的量。
n(Mg) reacted = n(H₂) = 0.01125 mol
n(Mg) 已反应 = n(H₂) = 0.01125 mol
5. Purity of the Magnesium Sample | 镁样品的纯度
calculate the mass of pure magnesium that reacted using its molar mass (Aᵣ of Mg = 24.3).
我们可以用镁的摩尔质量（Mg 的相对原子质量 Aᵣ = 24.3）计算出参与反应的纯镁质量。
Mass = n × Mᵣ = 0.01125 mol × 24.3 g mol⁻¹ ≈ 0.2734 g
质量 = n × Mᵣ = 0.01125 mol × 24.3 g mol⁻¹ ≈ 0.2734 g
The original ribbon weighed 0.30 g. Hence, the percentage purity is:
原镁条称重 0.30 g。因此，纯度百分比为：
Purity = (0.2734 g / 0.30 g) × 100% = 91.1%
纯度 = (0.2734 g / 0.30 g) × 100% = 91.1%
The magnesium ribbon is not 100% pure; it may contain oxide layers or other impurities that do not react with the acid.
该镁条并非 100% 纯净，可能含有不与酸反应的氧化层或其他杂质。
6. Theoretical Yield of Hydrogen | 氢气的理论产量
If the entire 0.30 g sample were pure magnesium, the theoretical number of moles of Mg would be:
如果 0.30 g 样品全为纯镁，则镁的理论物质的量为：
n(Mg) theoretical = 0.30 g / 24.3 g mol⁻¹ = 0.01235 mol
n(Mg) 理论 = 0.30 g / 24.3 g mol⁻¹ = 0.01235 mol
Theoretical volume of H₂ at rtp = 0.01235 mol × 24 000 cm³ mol⁻¹ = 296 cm³ (approx.).
常温常压下 H₂ 的理论体积 = 0.01235 mol × 24 000 cm³ mol⁻¹ = 296 cm³（约）。
7. Percentage Yield of Hydrogen | 氢气的百分产率
The actual volume collected was 270 cm³, which is less than the theoretical 296 cm³. The percentage yield is calculated as:
实际收集到的体积为 270 cm³，低于理论值 296 cm³。百分产率计算如下：
% Yield = (Actual volume / Theoretical volume) × 100%
% Yield = (270 cm³ / 296 cm³) × 100% = 91.2%
The yield is below 100% mainly because of the impurities in the magnesium and possibly some gas leaking before the syringe was fully sealed.
产率低于 100% 主要是因为镁中含有杂质，也可能是在注射器完全密封前有少量气体泄漏。
8. Measuring the Energy Change | 测量能量变化
m³ (the volume of HCl used), and its mass is taken as 50 g (since density = 1.00 g/cm³). The heat energy released (Q) is calculated using:
反应混合物的温度从 20.0 °C 上升到 35.0 °C，温升 ΔT = 15.0 °C。溶液总体积约为 50 cm³（所用 HCl 的体积），因其密度为 1.00 g/cm³，溶液质量取为 50 g。放出的热量（Q）按下式计算：
Q = m × c × ΔT
Q = 50 g × 4.2 J g⁻¹ °C⁻¹ × 15.0 °C = 3150 J = 3.15 kJ
This is the energy released to the solution. Because the reaction is exothermic, the value is often recorded with a negative sign in enthalpy terms.
这是释放到溶液中的能量。由于反应放热，焓值通常记为负数。
9. Calculating Enthalpy Change per Mole of Magnesium | 计算每摩尔镁的焓变
We base the calculation on the number of moles of magnesium that actually reacted (0.01125 mol).
我们以实际参与反应的镁的物质的量（0.01125 mol）为基础进行计算。
ΔH = −Q / n(Mg) reacted
ΔH = −3.15 kJ / 0.01125 mol = −280 kJ mol⁻¹
The experimentally determined enthalpy change for the reaction under these conditions is −280 kJ mol⁻¹. The literature value for Mg + 2HCl is more exothermic (around −467 kJ mol⁻¹). The difference highlights significant heat loss to the surroundings and the limited insulation of the apparatus.
在此条件下实验测定的反应焓变为 −280 kJ mol⁻¹。文献中 Mg + 2HCl 的反应焓变更为放热（约 −467 kJ mol⁻¹）。这一差异表明实验装置隔热有限，有大量热量散失到环境中。
10. Error Analysis and Improvements | 误差分析与改进
Several sources of error can be identified from the data.
从数据中可以识别出几种误差来源。
Gas collection: Some hydrogen may have escaped before the syringe was connected or dissolved slightly in the water if a trough was used. Using a well-sealed gas syringe from the start minimises loss.
气体收集： 在连接注射器前，可能有部分氢气逸出，或者如果使用水槽排水集气，氢气可能微量溶解于水。从实验开始就使用密封良好的气体注射器可以减少损失。
Heat loss: The polystyrene cup was not used; a glass flask allowed rapid heat transfer. Using an insulated calorimeter (e.g., a polystyrene cup with a lid) would give a more accurate temperature increase.
热量损失： 实验没有使用聚苯乙烯杯，玻璃瓶散热快。使用隔热量热器（如加盖的聚苯乙烯杯）可以获得更准确的温度上升数据。
Purity assumption: The calculation assumes all impurities are inert. If the impurity also reacts with acid, the analysis becomes more complex.
纯度假设： 计算假设所有杂质均不参与反应。如果杂质也与酸反应，分析将更加复杂。
11. Key Takeaways for CIE Chemistry | CIE 化学核心要点
- Always write a balanced equation first; it gives the mole ratios for all subsequent calculations. | 首先写出配平的方程式，它提供了后续所有计算所需的摩尔比。
- Molar volume at rtp (24 000 cm³ mol⁻¹) links gas volume to moles directly. | 常温常压下的摩尔体积（24 000 cm³ mol⁻¹）可将气体体积与物质的量直接联系起来。
- Purity and percentage yield are distinct concepts: purity concerns the sample composition; yield compares actual vs. theoretical product. | 纯度与百分产率是两个不同的概念：纯度关乎样品的组成；产率则是实际产量与理论产量的比值。
- always identify the total mass of the solution and use the correct specific heat capacity. | Q = mcΔT 用于测量溶液中能量的变化；始终要确定溶液总质量并使用正确的比热容。
- ΔH calculations should be expressed per mole of the limiting reactant, in kJ mol⁻¹. Remember the negative sign for exothermic reactions. | ΔH 计算应以 kJ mol⁻¹ 为单位，表示为每摩尔限量反应物的焓变。放热反应记得使用负号。
12. Further Practice Questions | 拓展练习
1. If the same experiment were repeated with 0.50 g of the same magnesium ribbon and collected 450 cm³ of H₂, what would be the new percentage purity and yield? (Assume molar volume 24 000 cm³ mol⁻¹.)
1. 如果用同种镁条 0.50 g 重复该实验，收集到 450 cm³ H₂，新的纯度与产率各为多少？（假设摩尔体积 24 000 cm³ mol⁻¹。）
2. Why is it important to use excess hydrochloric acid in this investigation? What would happen to the calculated purity if the acid was the limiting reactant?
2. 该探究中为何必须使用过量盐酸？如果酸成为限量反应物，对计算出的纯度有何影响？
3. The student repeats the experiment using a lid and a polystyrene cup, obtaining a temperature rise of 23.5 °C for the same mass of ribbon and acid. Calculate the new ΔH value and suggest why it is closer to the literature value.
3. 学生改用加盖的聚苯乙烯杯重复实验，相同质量镁条和酸得到 23.5 °C 温升。计算新的 ΔH 值，并解释为何更接近文献值。
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