📚 IB & CIE Chemistry: Stoichiometry Essentials | IB CIE 化学：化学计量 考点精讲
Stoichiometry is the backbone of quantitative chemistry, bridging the gap between the sub-microscopic world of atoms and molecules and the macroscopic world of measurable quantities. In both IB and CIE curricula, mastering stoichiometry is non-negotiable—it underpins calculations in energetics, kinetics, equilibrium, and even organic synthesis. This guide systematically breaks down every essential subtopic, equipping you with the conceptual clarity and computational fluency needed to secure full marks on stoichiometric problems.
化学计量是定量化学的核心支柱，它连接了原子分子的微观世界与可测量量的宏观世界。在 IB 和 CIE 课程体系中，掌握化学计量是不可或缺的——它支撑着能量学、动力学、平衡乃至有机合成中的各类计算。本指南系统梳理每一个关键子主题，帮助你建立概念清晰度与计算熟练度，在化学计量相关题目中稳拿满分。
1. The Mole Concept & Avogadro’s Number | 摩尔概念与阿伏伽德罗常数
The mole is the SI base unit for amount of substance. One mole of any substance contains exactly 6.02 × 10²³ elementary entities—atoms, molecules, ions, or formula units. This number is Avogadro’s constant (Nₐ). The mole allows chemists to count particles by weighing them, because the mass of one mole of a substance in grams is numerically equal to its relative atomic or molecular mass.
摩尔是国际单位制中物质的基本单位。一摩尔任何物质恰好包含 6.02 × 10²³ 个基本单元——原子、分子、离子或式单元。这个数字就是阿伏伽德罗常数 (Nₐ)。摩尔让化学家能够通过称重来“计数”粒子，因为一摩尔物质的质量（以克计）在数值上等于其相对原子质量或相对分子质量。
The defining equation is: n = N / Nₐ, where n is the amount in moles, N is the number of particles, and Nₐ = 6.02 × 10²³ mol⁻¹. For example, 3.01 × 10²³ water molecules correspond to 0.500 mol H₂O. Both IB and CIE exam questions frequently ask you to interconvert between mass, moles, and number of particles—often as the first step in a multi-stage stoichiometry problem.
定义方程为：n = N / Nₐ，其中 n 为摩尔数，N 为粒子数，Nₐ = 6.02 × 10²³ mol⁻¹。例如，3.01 × 10²³ 个水分子对应 0.500 mol H₂O。IB 和 CIE 考试题目经常要求你在质量、摩尔数和粒子数之间进行换算——这往往是多步化学计量计算题的第一步。
A critical nuance: the term ‘formula unit’ applies to ionic compounds. One mole of NaCl contains 6.02 × 10²³ Na⁺ ions and 6.02 × 10²³ Cl⁻ ions—a total of 1.204 × 10²⁴ ions. Students often overlook this doubling effect, so pay close attention to dissociation stoichiometry in ionic substances.
一个关键细节：“式单元”这一术语适用于离子化合物。一摩尔 NaCl 含有 6.02 × 10²³ 个 Na⁺ 离子和 6.02 × 10²³ 个 Cl⁻ 离子——总共 1.204 × 10²⁴ 个离子。学生常常忽略这种翻倍效应，因此务必仔细关注离子化合物的解离计量关系。
2. Relative Atomic Mass (Ar) and Relative Molecular Mass (Mr) | 相对原子质量与相对分子质量
Relative atomic mass (Ar) is the weighted average mass of an atom of an element relative to 1/12th the mass of a carbon-12 atom. It is dimensionless and accounts for the natural isotopic abundance of each element. For chlorine, Ar = 35.5, reflecting the 3:1 ratio of ³⁵Cl to ³⁷Cl isotopes. Relative molecular mass (Mr) is simply the sum of Ar values for all atoms in a molecule.
相对原子质量 (Ar) 是某元素一个原子的加权平均质量相对于碳-12 原子质量的 1/12 的比值。它是无量纲的，并考虑了每种元素天然同位素的丰度。氯的 Ar = 35.5，这反映了 ³⁵Cl 和 ³⁷Cl 同位素 3:1 的比例。相对分子质量 (Mr) 就是分子中所有原子 Ar 值的总和。
For calculation purposes, use these relationships: n = m / Mr, where m is mass in grams and Mr is relative molecular mass. For instance, to find the moles in 8.80 g of CO₂ (Mr = 44.0): n = 8.80 / 44.0 = 0.200 mol. IB data booklets provide Ar values; CIE candidates should memorise common ones like H = 1.0, C = 12.0, N = 14.0, O = 16.0, Na = 23.0, Cl = 35.5, S = 32.1.
在计算中使用以下关系：n = m / Mr，其中 m 为质量（克），Mr 为相对分子质量。例如，求 8.80 g CO₂ (Mr = 44.0) 中的摩尔数：n = 8.80 / 44.0 = 0.200 mol。IB 数据手册提供 Ar 值；CIE 考生应熟记常见元素的 Ar，如 H = 1.0、C = 12.0、N = 14.0、O = 16.0、Na = 23.0、Cl = 35.5、S = 32.1。
A common pitfall is confusing Mr calculations for hydrated salts. For CuSO₄·5H₂O, Mr = 63.5 + 32.1 + (4 × 16.0) + 5 × (2 × 1.0 + 16.0) = 249.6. Always include the water of crystallisation in Mr unless the question explicitly refers to the anhydrous form. Both IB and CIE love testing this distinction in empirical formula and titration contexts.
一个常见错误是在计算水合盐的 Mr 时混淆。对于 CuSO₄·5H₂O，Mr = 63.5 + 32.1 + (4 × 16.0) + 5 × (2 × 1.0 + 16.0) = 249.6。务必在计算 Mr 时包含结晶水，除非题目明确指代无水形式。IB 和 CIE 都喜欢在经验式和滴定相关题目中考察这一区别。
3. Balancing Chemical Equations | 化学方程式的配平
A balanced chemical equation obeys the law of conservation of mass: atoms are neither created nor destroyed, so the number of each type of atom must be identical on both sides. Balancing is the indispensable prerequisite to all stoichiometric calculations—an incorrectly balanced equation guarantees a wrong answer downstream.
配平的化学方程式遵循质量守恒定律：原子既不能被创造也不能被消灭，因此每种原子的数量在方程两边必须相同。配平是一切化学计量计算不可或缺的前提——配平错误的方程式必然导致后续答案全错。
The systematic approach: balance metals first, then non-metals (excluding H and O), then hydrogen, and finally oxygen. For combustion reactions, consider using fractional coefficients temporarily and then multiply through to clear fractions. Example: C₃H₈ + O₂ → CO₂ + H₂O. Balance C: C₃H₈ + O₂ → 3CO₂ + H₂O. Balance H: C₃H₈ + O₂ → 3CO₂ + 4H₂O. Balance O: on right, O = 3×2 + 4×1 = 10; so left needs 5 O₂. Final equation: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
系统方法：先配平金属，再配平非金属（不包括 H 和 O），然后配平氢，最后配平氧。对于燃烧反应，可考虑临时使用分数系数，然后将整个方程式乘以分母来消除分数。示例：C₃H₈ + O₂ → CO₂ + H₂O。配平 C：C₃H₈ + O₂ → 3CO₂ + H₂O。配平 H：C₃H₈ + O₂ → 3CO₂ + 4H₂O。配平 O：右边 O = 3×2 + 4×1 = 10；因此左边需要 5 个 O₂。最终方程式：C₃H₈ + 5O₂ → 3CO₂ + 4H₂O。
For ionic equations, balance both atoms and charge. In the redox reaction MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ (acidic medium), the half-equation method is indispensable. MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O and Fe²⁺ → Fe³⁺ + e⁻, multiplied and combined yield: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. IB requires proficiency in both acidic and basic redox balancing; CIE focuses primarily on acidic conditions.
对于离子方程式，需要同时配平原子和电荷。在氧化还原反应 MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺（酸性介质）中，半反应法是必不可少的工具。MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O 和 Fe²⁺ → Fe³⁺ + e⁻，经倍数调整后合并得到：MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O。IB 要求熟练掌握酸性和碱性条件下的氧化还原配平；CIE 则主要关注酸性条件。
4. Molar Ratios and Stoichiometric Calculations | 摩尔比与化学计量计算
The coefficients in a balanced equation establish the molar ratios between reactants and products—the central principle of stoichiometry. For the reaction 2Al + 3Cl₂ → 2AlCl₃, the ratio tells us that 2 mol Al reacts with 3 mol Cl₂ to produce 2 mol AlCl₃. These ratios serve as conversion factors in all quantitative predictions.
配平方程中的系数建立了反应物与生成物之间的摩尔比——这是化学计量的核心原理。对于反应 2Al + 3Cl₂ → 2AlCl₃，该比例告诉我们 2 mol Al 与 3 mol Cl₂ 反应生成 2 mol AlCl₃。这些比例在所有定量预测中都充当着换算因子的角色。
The three-stage stoichiometric calculation pathway is universal: (1) Convert given quantity to moles using n = m/Mr or n = cV or n = V/Vm; (2) Use the balanced equation’s molar ratio to find moles of the desired substance; (3) Convert back to the required quantity (mass, volume, concentration). This sequential logic applies whether you are following IB Paper 2 extended-response questions or CIE structured problems.
三步化学计量计算路径是通用的：(1) 使用 n = m/Mr 或 n = cV 或 n = V/Vm 将已知量转换为摩尔数；(2) 利用配平方程的摩尔比求出目标物质的摩尔数；(3) 转换回所求的量（质量、体积、浓度）。无论你面对的是 IB Paper 2 的扩展答题还是 CIE 的结构题，这套逻辑同样适用。
Example problem: What mass of CO₂ is produced when 5.00 g of CH₄ undergoes complete combustion? Step 1: n(CH₄) = 5.00 / 16.0 = 0.3125 mol. Step 2: CH₄ + 2O₂ → CO₂ + 2H₂O, so n(CO₂) = n(CH₄) = 0.3125 mol. Step 3: m(CO₂) = 0.3125 × 44.0 = 13.8 g. Always carry significant figures through intermediate steps and round only at the final answer—both IB and CIE are strict on sig fig conventions.
示例题目：5.00 g CH₄ 完全燃烧后生成多少质量的 CO₂？步骤 1：n(CH₄) = 5.00 / 16.0 = 0.3125 mol。步骤 2：CH₄ + 2O₂ → CO₂ + 2H₂O，因此 n(CO₂) = n(CH₄) = 0.3125 mol。步骤 3：m(CO₂) = 0.3125 × 44.0 = 13.8 g。始终在中间步骤保留有效数字，仅在最终答案处进行四舍五入——IB 和 CIE 对有效数字规范都有严格要求。
5. Limiting Reactant | 限量反应物
In most real chemical processes, reactants are not present in exact stoichiometric proportions. The limiting reactant is the substance that is completely consumed first, thereby determining the maximum amount of product that can form. The other reactant(s) remain in excess. Identifying the limiting reactant correctly is arguably the most tested skill in stoichiometry problems across both IB and CIE.
在大多数实际的化学过程中，反应物并不会恰好按化学计量比存在。限量反应物是最先被完全消耗的物质，因此它决定了能够生成的产物的最大量。其他反应物则保持过量。正确识别限量反应物可以说是 IB 和 CIE 化学计量考题中考查最频繁的技能。
The reliable method: convert the given mass (or volume, or concentration) of each reactant to moles; divide each by its stoichiometric coefficient from the balanced equation; the smallest resulting value identifies the limiting reactant. Consider 10.0 g Al reacting with 20.0 g Cl₂ to form AlCl₃. n(Al) = 10.0 / 27.0 = 0.370 mol; divided by coefficient 2 gives 0.185. n(Cl₂) = 20.0 / 71.0 = 0.282 mol; divided by coefficient 3 gives 0.0940. Since 0.0940 < 0.185, Cl₂ is limiting. All further calculations—including theoretical yield—must be based on Cl₂.
可靠的方法：将每种反应物的已知质量（或体积、或浓度）转换为摩尔数；将各摩尔数除以配平方程中对应的计量系数；所得的最小值即为限量反应物。考虑 10.0 g Al 与 20.0 g Cl₂ 反应生成 AlCl₃。n(Al) = 10.0 / 27.0 = 0.370 mol；除以系数 2 得 0.185。n(Cl₂) = 20.0 / 71.0 = 0.282 mol；除以系数 3 得 0.0940。因为 0.0940 < 0.185，所以 Cl₂ 是限量反应物。所有后续计算——包括理论产率——都必须基于 Cl₂。
Beware of a subtle trap: some students mistakenly assume the reactant with the smaller mass or smaller mole quantity is always limiting. This is incorrect. The comparison must be made after dividing by stoichiometric coefficients. A reactant present in smaller mass could still be in excess if its coefficient is proportionally much smaller. Always apply the division step rigorously.
谨防一个微妙的陷阱：一些学生错误地认为质量较小或摩尔数较少的反应物总是限量反应物。这是不正确的。必须在除以计量系数之后进行比较。如果某种反应物的质量较小，但其系数比例上要小得多，它仍然可能是过量的。务必严格应用除法步骤。
6. Percentage Yield and Atom Economy | 百分产率与原子经济性
Percentage yield quantifies the efficiency of a chemical process by comparing the actual mass of product obtained experimentally to the theoretical maximum mass predicted by stoichiometry. The formula is: % yield = (actual yield / theoretical yield) × 100%. Values below 100% reflect incomplete reactions, side reactions, or losses during purification—all common in both school laboratories and industrial settings.
百分产率通过将实验中实际获得的产品质量与化学计量预测的理论最大质量进行比较，来量化化学过程的效率。公式为：% 产率 = (实际产量 / 理论产量) × 100%。低于 100% 的数值反映了不完全反应、副反应或提纯过程中的损失——这些在学校的实验室和工业环境中都很常见。
Atom economy, a concept heavily emphasised in IB and increasingly in CIE, evaluates how efficiently reactant atoms are incorporated into the desired product. The formula: % atom economy = (Mr of desired product / sum of Mr of all reactants) × 100%. Unlike percentage yield, atom economy is a theoretical metric determined solely by the balanced equation, independent of experimental conditions. High atom economy indicates a greener, more sustainable process with minimal waste.
原子经济性是 IB 重点强调、CIE 也越来越重视的概念，它评估反应物原子被纳入目标产品的效率。公式为：% 原子经济性 = (目标产物的 Mr / 所有反应物 Mr 的总和) × 100%。与百分产率不同，原子经济性是一个纯粹由配平方程决定的理论指标，与实验条件无关。高原子经济性意味着过程更绿色、更可持续，产生的废料更少。
Example: Compare the atom economy of two methods for producing ethanol. Method A—fermentation: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. Desired Mr = 2 × 46.0 = 92.0; reactant Mr = 180.0; atom economy = (92.0 / 180.0) × 100% = 51.1%. Method B—hydration of ethene: C₂H₄ + H₂O → C₂H₅OH. Desired Mr = 46.0; sum of reactants Mr = 28.0 + 18.0 = 46.0; atom economy = 100%. Method B is far more atom-efficient, though it relies on non-renewable feedstock. IB Paper 1 and Paper 2 frequently feature such comparative analyses.
示例：比较两种制取乙醇方法的原子经济性。方法 A——发酵：C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂。目标产物 Mr = 2 × 46.0 = 92.0；反应物 Mr = 180.0；原子经济性 = (92.0 / 180.0) × 100% = 51.1%。方法 B——乙烯水合：C₂H₄ + H₂O → C₂H₅OH。目标产物 Mr = 46.0；反应物 Mr 总和 = 28.0 + 18.0 = 46.0；原子经济性 = 100%。方法 B 的原子效率远高于方法 A，尽管它依赖于不可再生的原料。IB Paper 1 和 Paper 2 中经常出现此类比较分析。
7. Empirical and Molecular Formulae | 经验式与分子式
The empirical formula represents the simplest whole-number ratio of atoms of each element in a compound. The molecular formula gives the actual number of atoms of each element in one molecule. For glucose, the molecular formula is C₆H₁₂O₆, while the empirical formula is CH₂O. Many different compounds can share the same empirical formula—ethanoic acid (C₂H₄O₂) and glucose have different molecular formulae but share CH₂O as the empirical formula.
经验式表示化合物中各元素原子的最简整数比。分子式给出了一个分子中每种元素的实际原子数。对于葡萄糖，分子式为 C₆H₁₂O₆，而经验式为 CH₂O。许多不同的化合物可以共享相同的经验式——乙酸 (C₂H₄O₂) 和葡萄糖的分子式不同，但都以 CH₂O 作为经验式。
To determine empirical formula from experimental data: (1) Convert the mass (or percentage by mass) of each element to moles by dividing by Ar; (2) Divide all mole values by the smallest to obtain the simplest ratio; (3) If ratios are not close to integers, multiply by an appropriate factor (e.g., 1.5 × 2 = 3). A compound containing 40.0% C, 6.67% H, and 53.3% O by mass yields n(C) = 40.0/12.0 = 3.33, n(H) = 6.67/1.0 = 6.67, n(O) = 53.3/16.0 = 3.33. Dividing by 3.33 gives ratio C:H:O = 1:2:1, so empirical formula = CH₂O.
通过实验数据确定经验式：(1) 将各元素的质量（或质量百分比）除以 Ar 转换为摩尔数；(2) 将所有摩尔数除以其中的最小值以获得最简比例；(3) 如果比例不接近整数，则乘以一个适当的因子（例如 1.5 × 2 = 3）。某化合物按质量计含 40.0% C、6.67% H 和 53.3% O，则 n(C) = 40.0/12.0 = 3.33，n(H) = 6.67/1.0 = 6.67，n(O) = 53.3/16.0 = 3.33。除以 3.33 得比例 C:H:O = 1:2:1，因此经验式 = CH₂O。
| Step 步骤 | Action 操作 |
| 1 | Mass or % mass to moles 质量或质量百分比换算为摩尔数 |
| 2 | Divide by smallest mole value 除以最小摩尔数 |
| 3 | Multiply to clear fractions if needed 必要时乘以因子清除分数 |
| 4 | Molecular formula = (empirical formula)n where n = Mr / empirical formula mass 分子式 = (经验式)n，其中 n = Mr / 经验式质量 |
To derive molecular formula from empirical formula, you need the Mr. If the empirical formula is CH₂O (mass = 30.0) and Mr = 180.0, then n = 180.0 / 30.0 = 6, so molecular formula = C₆H₁₂O₆. Combustion analysis is a classic experimental context for these calculations, appearing regularly in both IB and CIE examination papers.
要从经验式推导分子式，需要知道 Mr。如果经验式为 CH₂O（质量 = 30.0），Mr = 180.0，则 n = 180.0 / 30.0 = 6，故分子式 = C₆H₁₂O₆。燃烧分析是此类计算的经典实验背景，经常出现在 IB 和 CIE 的考试卷中。
8. Molar Volume of Gases | 气体摩尔体积
Avogadro’s law states that equal volumes of all gases, under the same conditions of temperature and pressure, contain the same number of molecules. The molar volume (Vm) is the volume occupied by one mole of any ideal gas. The numerical value depends on the chosen standard conditions, and this is where IB and CIE diverge slightly—a crucial distinction for exam candidates.
阿伏伽德罗定律指出，在相同温度和压力条件下，等体积的所有气体都含有相同数量的分子。摩尔体积 (Vm) 是一摩尔任何理想气体所占的体积。其数值取决于所选的标准条件，而 IB 和 CIE 在此处略有分歧——这对考生来说是一个至关重要的区别。
| Curriculum 课程 | Conditions 条件 | Molar Volume (Vm) 摩尔体积 |
| IB (STP) | 273 K, 100 kPa | 22.7 dm³ mol⁻¹ |
| CIE (RTP) | 25 °C (298 K), 1 atm (101 kPa) | 24.0 dm³ mol⁻¹ |
The relationship n = V / Vm is used to interconvert between moles and gas volume. For example, under IB STP conditions, 0.500 mol O₂ occupies 0.500 × 22.7 = 11.35 dm³. Under CIE RTP conditions, the same 0.500 mol O₂ occupies 0.500 × 24.0 = 12.0 dm³. Always check which set of conditions your syllabus specifies and use the corresponding Vm value—using the wrong value is a common and costly mistake.
关系式 n = V / Vm 用于在摩尔数和气体体积之间进行换算。例如，在 IB 的 STP 条件下，0.500 mol O₂ 占据 0.500 × 22.7 = 11.35 dm³。在 CIE 的 RTP 条件下，相同的 0.500 mol O₂ 占据 0.500 × 24.0 = 12.0 dm³。务必确认你的课程大纲指定了哪一套条件，并使用相应的 Vm 值——使用错误的数值是一种常见且代价高昂的错误。
Stoichiometric problems involving gases often combine mass, volume, and concentration calculations. Example: What volume of CO₂ (at RTP) is produced when 10.0 g CaCO₃ decomposes? Equation: CaCO₃ → CaO + CO₂. n(CaCO₃) = 10.0 / 100.1 = 0.0999 mol. n(CO₂) = 0.0999 mol. V(CO₂) = 0.0999 × 24.0 = 2.40 dm³ (CIE) or 0.0999 × 22.7 = 2.27 dm³ (IB). Note how the same chemical scenario yields different numerical answers depending on the Vm used.
涉及气体的化学计量题往往将质量、体积和浓度计算结合在一起。示例：10.0 g CaCO₃ 分解时产生多少体积的 CO₂（在 RTP 下）？方程式：CaCO₃ → CaO + CO₂。n(CaCO₃) = 10.0 / 100.1 = 0.0999 mol。n(CO₂) = 0.0999 mol。V(CO₂) = 0.0999 × 24.0 = 2.40 dm³ (CIE) 或 0.0999 × 22.7 = 2.27 dm³ (IB)。请注意，相同的化学情景因所用 Vm 不同而得出不同的数值答案。
9. Solution Concentration and Titration Calculations | 溶液浓度与滴定计算
Concentration (c) is defined as the amount of solute per unit volume of solution, typically expressed in mol dm⁻³. The fundamental equation is c = n / V, where V is the volume of solution in dm³. Rearranged forms n = cV and V = n / c are equally essential for stoichiometric problem-solving. In titrations, this relationship connects the measured volume of a standard solution to the unknown concentration of an analyte.
浓度 (c) 定义为单位体积溶液中溶质的量，通常以 mol dm⁻³ 表示。基本方程为 c = n / V，其中 V 是溶液的体积，单位为 dm³。其变形形式 n = cV 和 V = n / c 在化学计量解题中同样至关重要。在滴定中，这一关系将标准溶液的测量体积与分析物未知浓度联系起来。
Titration calculations often involve a dilution step. If a solution is diluted, the amount of solute (n) remains constant: n₁ = n₂, so c₁V₁ = c₂V₂. This dilution formula is ubiquitous in both IB and CIE. For back titrations, where an excess of one reagent is added and the unreacted portion is titrated, careful accounting of moles is required. The key principle: n(reacted with analyte) = n(initial total) − n(reacted in back-titration).
滴定计算常常涉及稀释步骤。如果溶液被稀释，溶质的量 (n) 保持不变：n₁ = n₂，因此 c₁V₁ = c₂V₂。这个稀释公式在 IB 和 CIE 中都无处不在。对于返滴定，即先加入过量的一种试剂，然后对未反应的部分进行滴定，需要仔细核算摩尔数。关键原则：n(与分析物反应的) = n(初始总量) − n(返滴定中反应的)。
Example of a complete titration calculation: 25.0 cm³ of H₂SO₄ solution requires 18.5 cm³ of 0.100 mol dm⁻³ NaOH for neutralisation. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. n(NaOH) = 0.100 × (18.5/1000) = 1.85 × 10⁻³ mol. From the equation, n(H₂SO₄) = n(NaOH) / 2 = 9.25 × 10⁻⁴ mol. c(H₂SO₄) = 9.25 × 10⁻⁴ / (25.0/1000) = 0.0370 mol dm⁻³. Notice the conversion of cm³ to dm³ by dividing by 1000—forgetting this conversion is among the most frequent student errors.
完整的滴定计算示例：25.0 cm³ H₂SO₄ 溶液需要 18.5 cm³ 0.100 mol dm⁻³ NaOH 进行中和。H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O。n(NaOH) = 0.100 × (18.5/1000) = 1.85 × 10⁻³ mol。根据方程式，n(H₂SO₄) = n(NaOH) / 2 = 9.25 × 10⁻⁴ mol。c(H₂SO₄) = 9.25 × 10⁻⁴ / (25.0/1000) = 0.0370 mol dm⁻³。请注意将 cm³ 转换为 dm³ 时需要除以 1000——忘记这一转换是学生最常犯的错误之一。
10. Common Pitfalls and Exam Tips | 常见错误与应试技巧
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