📚 Stoichiometry & Reaction Calculations for A-Level Chemistry | A-Level化学考点：反应计算与化学计量关系
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. Mastering this topic is essential for solving nearly every numerical problem in A-Level Chemistry, from mole conversions to titration calculations.
化学计量学研究化学反应中反应物与产物之间的定量关系。掌握这一主题对于解决A-Level化学中几乎所有计算题至关重要——从摩尔换算到滴定计算，无一例外。
1. The Mole & Avogadro’s Constant | 物质的量与阿伏伽德罗常数
The mole is the SI unit for the amount of substance. One mole contains exactly 6.022 × 10²³ elementary entities (atoms, molecules, ions, or electrons), a value known as the Avogadro constant (L or Nₐ).
摩尔是物质的量的国际单位制（SI）基本单位。1摩尔恰好包含6.022 × 10²³个基本实体（原子、分子、离子或电子），这个数值称为阿伏伽德罗常数（L 或 Nₐ）。
To convert between number of particles and moles, use:
在粒子数与摩尔之间进行换算时，使用以下公式：
n = N / Nₐ
- n = amount in moles (mol) | 物质的量（单位：mol）
- N = number of particles | 粒子数目
- Nₐ = 6.022 × 10²³ mol⁻¹ | 阿伏伽德罗常数
Example: How many moles are present in 1.204 × 10²⁴ molecules of water?
例题：1.204 × 10²⁴个水分子中含有多少摩尔？
n = (1.204 × 10²⁴) / (6.022 × 10²³) = 2.00 mol
2. Molar Mass & Mass–Mole Conversions | 摩尔质量与质量–摩尔换算
Molar mass (M) is the mass of one mole of a substance, expressed in g mol⁻¹. Numerically, it equals the relative atomic mass (Aᵣ) or relative molecular mass (Mᵣ) from the periodic table.
摩尔质量（M）是1摩尔物质的质量，单位为g mol⁻¹。数值上，它等于元素周期表中的相对原子质量（Aᵣ）或相对分子质量（Mᵣ）。
The core conversion formula is:
核心换算公式为：
n = m / M
where m is the mass in grams and M is the molar mass in g mol⁻¹.
其中 m 为质量（克），M 为摩尔质量（g mol⁻¹）。
Worked example: Calculate the number of moles in 10.0 g of calcium carbonate (CaCO₃).
例题演示：计算10.0 g碳酸钙（CaCO₃）中含有的物质的量。
M(CaCO₃) = 40.1 + 12.0 + (3 × 16.0) = 100.1 g mol⁻¹
n = 10.0 / 100.1 = 0.0999 ≈ 0.100 mol
3. Balancing Equations & Mole Ratios | 配平方程式与摩尔比
For the balanced equation aA + bB → cC + dD, the coefficients a, b, c, d give the mole ratios in which reactants combine and products form.
对于配平方程式 aA + bB → cC + dD，系数 a、b、c、d 给出了反应物结合和产物生成的摩尔比。
Mole ratios allow conversion from moles of one substance to moles of another:
摩尔比使我们能够从一种物质的物质的量换算为另一种物质的物质的量：
n(B) = n(A) × (coefficient of B / coefficient of A)
Consider the reaction: N₂ + 3H₂ → 2NH₃. The mole ratio N₂ : H₂ : NH₃ is 1 : 3 : 2. Thus 4 moles of hydrogen produce (4 × 2/3) ≈ 2.67 moles of ammonia.
以反应 N₂ + 3H₂ → 2NH₃ 为例，N₂ : H₂ : NH₃ 的摩尔比为1 : 3 : 2。因此4摩尔氢气生成（4 × 2/3）≈ 2.67摩尔氨气。
Always begin stoichiometric calculations by writing a balanced equation — it provides the quantitative “map” for the entire reaction.
进行化学计量计算时，务必先写出配平的化学方程式——它为整个反应提供了定量的”地图”。
4. Limiting Reagent & Excess | 限制试剂与过量试剂
The limiting reagent is the reactant that is completely consumed first, determining the maximum amount of product that can form. Any reactant present in more than the stoichiometric amount is in excess.
限制试剂是首先被完全消耗的反应物，它决定了产物生成的最大量。任何超过化学计量比所需的反应物即为过量试剂。
To identify the limiting reagent:
判断限制试剂的步骤：
- Convert all reactant masses/volumes to moles | 将所有反应物的质量/体积换算为物质的量
- Divide each mole amount by its stoichiometric coefficient | 将每种物质的量除以其化学计量系数
- The smallest value indicates the limiting reagent | 商值最小的物质为限制试剂
Example: 5.0 g of Mg reacts with 3.65 g of HCl. Which is the limiting reagent? (Mg + 2HCl → MgCl₂ + H₂)
例题：5.0 g 镁与3.65 g 氯化氢反应，哪个是限制试剂？（Mg + 2HCl → MgCl₂ + H₂）
n(Mg) = 5.0 / 24.3 = 0.206 mol
n(HCl) = 3.65 / 36.5 = 0.100 mol
Mg: 0.206 / 1 = 0.206; HCl: 0.100 / 2 = 0.050 → HCl is limiting
5. Calculations Involving Gas Volumes | 涉及气体体积的计算
At room temperature and pressure (r.t.p., 25 °C and 1 atm), one mole of any ideal gas occupies approximately 24.0 dm³. This molar gas volume (Vₘ) enables direct conversion between gas volume and moles:
在室温常压（r.t.p.，25 °C和1 atm）下，1摩尔任何理想气体约占据24.0 dm³的体积。利用这个摩尔气体体积（Vₘ），可在气体体积与物质的量之间直接换算：
n = V / Vₘ
where V is volume in dm³. At standard temperature and pressure (s.t.p., 0 °C, 1 atm), Vₘ = 22.4 dm³ mol⁻¹. Alternatively, the ideal gas equation can be used:
其中 V 为体积（dm³）。在标准温压条件下（s.t.p.，0 °C，1 atm），Vₘ = 22.4 dm³ mol⁻¹。或者也可使用理想气体状态方程：
PV = nRT
Worked example: What volume does 0.60 mol of CO₂ occupy at r.t.p.?
例题演示：0.60 mol CO₂在室温常压下占据多大体积？
V = 0.60 × 24.0 = 14.4 dm³
6. Concentration & Solution Calculations | 浓度与溶液计算
Concentration expresses the amount of solute dissolved in a given volume of solution, commonly in mol dm⁻³ (molarity).
浓度表示单位体积溶液中所含溶质的量，常用单位是mol dm⁻³（摩尔浓度）。
c = n / V (in dm³)
- c = concentration (mol dm⁻³) | 浓度（mol dm⁻³）
- n = moles of solute (mol) | 溶质的物质的量（mol）
- V = volume of solution (dm³) | 溶液体积（dm³）
Remember: 1 dm³ = 1000 cm³. When given volume in cm³, divide by 1000 first.
注意：1 dm³ = 1000 cm³。若体积单位为cm³，需先除以1000进行换算。
Example: Calculate the concentration when 5.85 g of NaCl is dissolved in 250 cm³ of water.
例题：将5.85 g NaCl溶解在250 cm³水中，计算所得溶液的浓度。
n(NaCl) = 5.85 / 58.5 = 0.100 mol
c = 0.100 / 0.250 = 0.400 mol dm⁻³
7. Titration Calculations | 滴定计算
Titration is a quantitative technique used to determine the unknown concentration of a solution by reacting it with a solution of known concentration. The key formula for titration calculations is:
滴定是一种定量分析技术，通过用已知浓度的溶液与未知浓度的溶液反应，从而确定后者的浓度。滴定计算的关键公式为：
(Cₐ × Vₐ) / a = (Cᵦ × Vᵦ) / b
where a and b are the stoichiometric coefficients from the balanced equation, C is concentration, and V is volume.
其中 a 和 b 为配平方程式中的化学计量系数，C 为浓度，V 为体积。
Typical titration problem: 25.0 cm³ of 0.100 mol dm⁻³ NaOH requires 20.0 cm³ of H₂SO₄ for neutralisation. Find the concentration of H₂SO₄.
典型滴定问题：25.0 cm³的0.100 mol dm⁻³ NaOH恰好需要20.0 cm³ H₂SO₄中和。求H₂SO₄的浓度。
Balanced equation: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
配平方程式：2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
Moles NaOH = 0.100 × (25.0/1000) = 0.00250 mol
Moles H₂SO₄ = 0.00250 / 2 = 0.00125 mol
C(H₂SO₄) = 0.00125 / (20.0/1000) = 0.0625 mol dm⁻³
8. Percentage Yield & Atom Economy | 百分产率与原子经济性
Percentage yield measures the efficiency of a reaction in terms of actual product obtained versus theoretical maximum:
百分产率衡量反应的实际获得产物相对于理论最大产物的效率：
Percentage yield = (actual yield / theoretical yield) × 100%
Low yields may result from side reactions, incomplete reactions, or product loss during purification.
产率偏低可能是由于副反应、反应不完全或纯化过程中的产物损失。
Atom economy reflects how much of the starting materials ends up in the desired product:
原子经济性反映起始原料中有多少最终进入了目标产物：
Atom economy = (molar mass of desired product / sum of molar masses of all reactants) × 100%
For the reaction CaCO₃ → CaO + CO₂, if the desired product is CaO:
对于反应 CaCO₃ → CaO + CO₂，若目标产物为CaO：
Atom economy = 56.1 / 100.1 × 100% = 56.0%
High atom economy means fewer by-products and is a key principle of green chemistry.
高原子经济性意味着更少的副产物，是绿色化学的重要原则。
9. Ideal Gas Equation & Stoichiometry | 理想气体方程与化学计量
When reacting gases are involved, the ideal gas equation PV = nRT can determine moles from pressure, volume, and temperature data, which can then be used in stoichiometric ratios.
当涉及气体反应时，理想气体方程 PV = nRT 可根据压强、体积和温度数据求出物质的量，进而用于化学计量比计算。
R = 8.314 J K⁻¹ mol⁻¹. Pressure must be in Pa, volume in m³ (1 m³ = 1000 dm³), and temperature in kelvin (K = °C + 273.15).
R = 8.314 J K⁻¹ mol⁻¹。压强单位必须为Pa，体积为m³（1 m³ = 1000 dm³），温度为开尔文（K = °C + 273.15）。
Example: Find the volume of 0.500 mol of oxygen at 300 K and 1.00 × 10⁵ Pa.
例题：求0.500 mol氧气在300 K和1.00 × 10⁵ Pa下的体积。
V = nRT / P = (0.500 × 8.314 × 300) / (1.00 × 10⁵)
V = 1247 / 1.00 × 10⁵ = 0.01247 m³ = 12.5 dm³
10. Water of Crystallisation & Hydrated Salts | 结晶水与水合盐
Hydrated salts contain water molecules within their crystal lattice, e.g., CuSO₄·5H₂O. Stoichiometric calculations frequently involve determining the value of x in a formula such as MSO₄·xH₂O.
水合盐在晶格中含有水分子，例如 CuSO₄·5H₂O。化学计量计算中常需要确定类似于 MSO₄·xH₂O 的化学式中的 x 值。
The general approach is:
一般解题步骤为：
- Measure the mass of the hydrated salt | 称量水合盐的质量
- Heat to remove water and reweigh the anhydrous salt | 加热去除水分并称量无水盐
- Find moles of anhydrous salt and moles of water | 计算无水盐和水的物质的量
- Determine the simplest ratio to find x | 求最简整数比以确定 x
Example: 2.50 g of hydrated Na₂CO₃·xH₂O is heated, leaving 0.93 g of anhydrous Na₂CO₃. Determine x.
例题：将2.50 g水合碳酸钠 Na₂CO₃·xH₂O 加热后剩余0.93 g无水 Na₂CO₃。求 x 值。
Mass of water = 2.50 − 0.93 = 1.57 g
n(Na₂CO₃) = 0.93 / 106.0 = 0.00877 mol
n(H₂O) = 1.57 / 18.0 = 0.0872 mol
Ratio H₂O : Na₂CO₃ = 0.0872 / 0.00877 ≈ 10 → x = 10
11. Combustion & Empirical Formula Analysis | 燃烧分析与实验式确定
Combustion analysis is a classical method for determining empirical formulas of organic compounds. Masses of CO₂ and H₂O produced allow calculation of carbon and hydrogen content.
燃烧分析是确定有机化合物实验式的经典方法。通过测定生成的 CO₂ 和 H₂O 的质量，可计算碳和氢的含量。
Consider a hydrocarbon that produces 0.880 g CO₂ and 0.360 g H₂O upon combustion:
考虑某烃完全燃烧生成0.880 g CO₂和0.360 g H₂O：
n(C) = 0.880 / 44.0 = 0.0200 mol
n(H) = 2 × (0.360 / 18.0) = 0.0400 mol
C : H = 0.0200 : 0.0400 = 1 : 2 → empirical formula CH₂
To determine the molecular formula, the empirical formula mass must be multiplied by the ratio (molar mass / empirical formula mass).
要确定分子式，需用摩尔质量除以实验式质量的比值乘以实验式。
12. Back Titration & Multi-Step Calculations | 返滴定与多步综合计算
Back titration is used when the analyte is insoluble, volatile, or reacts slowly. An excess of a known reagent is added, and the leftover is then titrated against another standard solution.
当待测物不溶、易挥发或反应缓慢时，采用返滴定法。先加入过量的已知试剂，然后用另一种标准溶液返滴剩余的试剂。
Multi-step stoichiometric problems often chain several conversions: mass → moles → mole ratio → moles of product → volume/concentration or mass. The conversion pathway is:
多步化学计量问题往往串联多个换算：质量 → 物质的量 → 摩尔比 → 产物物质的量 → 体积/浓度或质量。换算路径为：
m → n (via M) → ratio → n (product) → V or m (via Vₘ or M)
Example: What mass of CO₂ is produced when 5.0 g of CaCO₃ reacts with excess HCl? (CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂)
例题：5.0 g CaCO₃与过量HCl反应生成多少克CO₂？(CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂)
n(CaCO₃) = 5.0 / 100.1 = 0.0500 mol
n(CO₂) = 0.0500 mol (1:1 ratio)
Mass CO₂ = 0.0500 × 44.0 = 2.2 g
In exams, always clearly show units in every step and check that the final result is sensible — this greatly reduces careless errors.
考试中，每一步都应清晰标注单位，并检查最终结果是否合理——这样可以大大减少粗心错误。
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