📚 Mastering Numerical Questions: Answers and Worked Solutions | 掌握数值计算题：答案与解题步骤
Numerical questions are a core component of the Cambridge IGCSE Chemistry examination. They test your ability to apply concepts such as the mole, stoichiometry, concentration, and energy changes to solve real problems. This article provides a structured approach to answering these questions, with worked examples and explanations that mirror the style of textbook exercises.
数值计算题是剑桥 IGCSE 化学考试的核心组成部分。它们考查你运用摩尔、化学计量、浓度和能量变化等概念解决实际问题的能力。本文提供了一套结构化的答题方法，配合典型例题和详细解析，与课本练习风格一致。
1. The Mole Concept | 摩尔概念
One mole of any substance contains 6.02 × 10²³ particles (Avogadro’s constant). The number of moles (n) is calculated using the formula:
任何物质的 1 摩尔都含有 6.02 × 10²³ 个粒子（阿伏加德罗常数）。物质的量（n）用下列公式计算：
n = mass (g) ÷ molar mass (g mol⁻¹)
Example: Calculate the number of moles in 4.0 g of sodium hydroxide (NaOH). Molar mass = 23 + 16 + 1 = 40 g mol⁻¹. So n = 4.0 ÷ 40 = 0.10 mol.
例题：计算 4.0 g 氢氧化钠（NaOH）的物质的量。摩尔质量 = 23 + 16 + 1 = 40 g mol⁻¹，所以 n = 4.0 ÷ 40 = 0.10 mol。
2. Stoichiometry and Reacting Masses | 化学计量与反应质量
Use balanced equations to relate the moles of reactants and products. The coefficients in the equation give the mole ratio.
利用配平的化学方程式建立反应物与产物之间的物质的量关系。方程式中的系数即为摩尔比。
Example: 2H₂ + O₂ → 2H₂O. What mass of oxygen reacts with 4.0 g of hydrogen? Moles of H₂ = 4.0 ÷ 2 = 2.0 mol. Mole ratio H₂:O₂ = 2:1, so moles of O₂ = 1.0 mol. Mass of O₂ = 1.0 × 32 = 32 g.
例题：2H₂ + O₂ → 2H₂O。4.0 g 氢气需要多少克氧气？氢气的物质的量 = 4.0 ÷ 2 = 2.0 mol。H₂:O₂ 摩尔比 = 2:1，所以氧气物质的量 = 1.0 mol。氧气质量 = 1.0 × 32 = 32 g。
3. Concentration and Titration Calculations | 浓度与滴定计算
Concentration (mol dm⁻³) = moles ÷ volume (dm³). In titrations, use the formula C₁V₁ = C₂V₂ for reactions with a 1:1 mole ratio, or adjust for other ratios.
浓度（mol dm⁻³）= 物质的量 ÷ 体积（dm³）。在滴定中，当反应摩尔比为 1:1 时使用 C₁V₁ = C₂V₂，否则需按比例调整。
Example: 25.0 cm³ of 0.10 mol dm⁻³ HCl neutralises 20.0 cm³ of NaOH. Find the concentration of NaOH.
例题：25.0 cm³ 的 0.10 mol dm⁻³ HCl 恰好中和 20.0 cm³ 的 NaOH。求 NaOH 的浓度。
HCl + NaOH → NaCl + H₂O, so moles HCl = moles NaOH
Moles HCl = 0.10 × 25.0/1000 = 0.0025 mol. Concentration NaOH = 0.0025 ÷ (20.0/1000) = 0.125 mol dm⁻³.
HCl 的物质的量 = 0.10 × 25.0/1000 = 0.0025 mol。NaOH 的浓度 = 0.0025 ÷ (20.0/1000) = 0.125 mol dm⁻³。
4. Gas Volumes and Molar Volume | 气体体积与摩尔体积
At room temperature and pressure (r.t.p., 25 °C, 1 atm), one mole of any gas occupies 24 dm³. Volume (dm³) = moles × 24.
在室温常压（r.t.p.，25 °C，1 atm）下，1 摩尔任何气体占 24 dm³。体积（dm³）= 物质的量 × 24。
Example: What volume does 0.50 mol of CO₂ occupy at r.t.p.? Volume = 0.50 × 24 = 12 dm³.
例题：在室温常压下，0.50 mol CO₂ 占多大体积？体积 = 0.50 × 24 = 12 dm³。
5. Empirical and Molecular Formulas | 经验式与分子式
To find the empirical formula, divide each element’s mass by its atomic mass, then divide by the smallest ratio. Multiply to whole numbers if necessary. For the molecular formula, divide the relative molecular mass by the empirical formula mass.
求经验式时，将各元素质量除以各自的相对原子质量，再除以最小比值。如有必要，乘以整数得到最简整数比。分子式 = 相对分子质量 ÷ 经验式相对质量，然后扩大相应倍数。
Example: A compound contains 3.2 g of sulfur and 3.2 g of oxygen. Moles S = 3.2/32 = 0.10, moles O = 3.2/16 = 0.20. Ratio S:O = 1:2, so empirical formula is SO₂.
例题：某化合物含 3.2 g 硫和 3.2 g 氧。硫的物质的量 = 3.2/32 = 0.10，氧的物质的量 = 3.2/16 = 0.20。S:O = 1:2，经验式为 SO₂。
6. Percentage Yield and Atom Economy | 产率与原子经济性
Percentage yield = (actual yield ÷ theoretical yield) × 100%. Atom economy = (mass of desired product ÷ total mass of reactants) × 100%.
产率 =（实际产量 ÷ 理论产量）× 100%。原子经济性 =（目标产物质量 ÷ 反应物总质量）× 100%。
Example: If the theoretical yield is 6.0 g and the actual yield is 4.5 g, percentage yield = (4.5 ÷ 6.0) × 100% = 75%.
例题：理论产量为 6.0 g，实际产量为 4.5 g，则产率 = (4.5 ÷ 6.0) × 100% = 75%。
7. Energetics Calculations | 能量变化计算
Heat change (q) = mass (g) × specific heat capacity (J g⁻¹ °C⁻¹) × temperature change (ΔT). Then convert to per mole of reactant.
热量变化（q）= 质量（g）× 比热容（J g⁻¹ °C⁻¹）× 温度变化（ΔT）。然后换算为每摩尔反应物的能量变化。
Example: 50.0 cm³ of solution (assume mass 50 g) rises from 20.0 °C to 26.5 °C. q = 50 × 4.18 × 6.5 = 1358.5 J. If 0.050 mol of reactant was used, ΔH = –1358.5 / 0.050 = –27170 J mol⁻¹ ≈ –27.2 kJ mol⁻¹ (negative because exothermic).
例题：50.0 cm³ 溶液（假设质量为 50 g）从 20.0 °C 升高到 26.5 °C。q = 50 × 4.18 × 6.5 = 1358.5 J。若反应物为 0.050 mol，则 ΔH = –1358.5 / 0.050 = –27170 J mol⁻¹ ≈ –27.2 kJ mol⁻¹（放热，故为负值）。
8. Rates of Reaction and Graphs | 反应速率与图形
Rates can be calculated from mass loss, gas volume, or colour change over time. The gradient of a tangent to a concentration–time graph gives the instantaneous rate.
速率可通过质量损失、气体体积或颜色随时间的变化来计算。浓度-时间图中切线的斜率给出瞬时速率。
Example: In 2 minutes, 48 cm³ of gas is collected. Average rate = 48 ÷ (2 × 60) = 0.40 cm³ s⁻¹.
例题：2 分钟内收集到 48 cm³ 气体。平均速率 = 48 ÷ (2 × 60) = 0.40 cm³ s⁻¹。
9. Electrolysis Calculations | 电解计算
The quantity of charge (C) = current (A) × time (s). One mole of electrons carries 96500 C (Faraday’s constant). Use electrode half-equations to relate moles of electrons to moles of substance.
电荷量（C）= 电流（A）× 时间（s）。1 摩尔电子携带 96500 C（法拉第常数）。利用电极半反应将电子物质的量与物质物质的量联系起来。
Example: How many moles of copper are deposited by 2 A for 965 s? Charge = 2 × 965 = 1930 C. Moles of electrons = 1930 ÷ 96500 = 0.020 mol. Cu²⁺ + 2e⁻ → Cu, so moles Cu = 0.020 ÷ 2 = 0.010 mol.
例题：2 A 电流通过电解池 965 秒，能沉积多少摩尔铜？电荷量 = 2 × 965 = 1930 C。电子物质的量 = 1930 ÷ 96500 = 0.020 mol。Cu²⁺ + 2e⁻ → Cu，所以铜的物质的量 = 0.020 ÷ 2 = 0.010 mol。
10. Common Pitfalls and Exam Tips | 常见错误与考试技巧
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Always convert cm³ to dm³ by dividing by 1000 in concentration calculations.
浓度计算中，注意将 cm³ 除以 1000 换算为 dm³。
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Check your units: mass in grams, volume in dm³, energy in joules or kilojoules.
检查单位：质量用克，体积用 dm³，能量用焦耳或千焦耳。
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Write down the equation and molar ratio before starting any stoichiometric problem.
开始任何化学计量问题前，先写出方程式和摩尔比。
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For titrations, read burette values to 0.05 cm³ and use concordant results.
滴定时，读数精确到 0.05 cm³，并采用平行结果。
11. Worked Practice Questions | 练习例题与答案
Question 1: Calculate the mass of 0.250 mol of calcium carbonate (CaCO₃). Molar mass = 40 + 12 + 3×16 = 100 g mol⁻¹. Mass = 0.250 × 100 = 25.0 g.
问题 1：计算 0.250 mol 碳酸钙（CaCO₃）的质量。摩尔质量 = 40 + 12 + 3×16 = 100 g mol⁻¹。质量 = 0.250 × 100 = 25.0 g。
Question 2: What volume of 0.200 mol dm⁻³ HCl contains 0.0500 mol? Volume = 0.0500 ÷ 0.200 = 0.250 dm³ = 250 cm³.
问题 2：0.200 mol dm⁻³ 的 HCl 中，含有 0.0500 mol 所需的体积是多少？体积 = 0.0500 ÷ 0.200 = 0.250 dm³ = 250 cm³。
Question 3: In a reaction, 0.300 mol of propane (C₃H₈) burns completely in excess oxygen. What volume of CO₂ is produced at r.t.p.? Equation: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. Mole ratio C₃H₈:CO₂ = 1:3, so CO₂ moles = 0.900 mol. Volume = 0.900 × 24 = 21.6 dm³.
问题 3：0.300 mol 丙烷（C₃H₈）在过量氧气中完全燃烧，在 r.t.p. 下产生的 CO₂ 体积是多少？方程式：C₃H₈ + 5O₂ → 3CO₂ + 4H₂O。C₃H₈:CO₂ = 1:3，所以 CO₂ 物质的量 = 0.900 mol。体积 = 0.900 × 24 = 21.6 dm³。
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