Limiting Reactant and Excess Reactant Calculations | 限制反应物与过量反应物计算

📚 Limiting Reactant and Excess Reactant Calculations | 限制反应物与过量反应物计算

In chemical reactions, reactants are rarely present in the exact stoichiometric ratio. The limiting reactant is the substance that is completely consumed first, determining the maximum amount of product formed. The other reactants are called excess reactants, and a portion of them remains unreacted after the reaction stops.

在化学反应中，反应物极少按精确的化学计量比存在。限制反应物是最先被完全消耗的物质，它决定了产物生成的最大量。其他反应物称为过量反应物，反应停止后会有部分剩余。

1. The Core Concept | 核心概念

Every balanced chemical equation gives the mole ratio in which substances react. For example, in the reaction 2H₂ + O₂ → 2H₂O, two moles of hydrogen react with one mole of oxygen. If we mix 2 mol H₂ with 2 mol O₂, hydrogen is the limiting reactant because only 1 mol O₂ is needed for 2 mol H₂; the remaining 1 mol O₂ is excess.

每个配平的化学方程式都给出了物质反应的摩尔比。例如，在反应 2H₂ + O₂ → 2H₂O 中，2 摩尔氢气与 1 摩尔氧气反应。如果我们混合 2 摩尔 H₂ 和 2 摩尔 O₂，氢气是限制反应物，因为 2 摩尔 H₂ 只需要 1 摩尔 O₂；剩余 1 摩尔 O₂ 是过量的。

Limiting reactant = determines the maximum product
限制反应物 = 决定产物的最大生成量

2. Balanced Equation and Mole Ratio | 配平方程式与摩尔比

Before any calculation, you must write a correctly balanced equation. The coefficients in the balanced equation are the mole ratios, not mass ratios. For the Haber process: N₂ + 3H₂ → 2NH₃, the ratio N₂ : H₂ : NH₃ = 1 : 3 : 2.

在任何计算之前，必须写出正确配平的方程式。配平方程式中的系数是摩尔比，而不是质量比。对于哈伯法制氨：N₂ + 3H₂ → 2NH₃，比值为 N₂ : H₂ : NH₃ = 1 : 3 : 2。

If you start with 1 mol N₂ and 3 mol H₂, both react completely. If you start with 1 mol N₂ and only 2 mol H₂, then H₂ is limiting because 1 mol N₂ requires 3 mol H₂, but only 2 mol H₂ is available.

如果以 1 摩尔 N₂ 和 3 摩尔 H₂ 开始，两者完全反应。如果以 1 摩尔 N₂ 和仅 2 摩尔 H₂ 开始，则 H₂ 是限制反应物，因为 1 摩尔 N₂ 需要 3 摩尔 H₂，但只有 2 摩尔 H₂ 可用。

3. Step 1: Convert Masses to Moles | 第一步：将质量转化为摩尔

The first step in any limiting reactant problem is to convert the given mass of each reactant to moles using the molar mass. Moles = mass / molar mass. For example, 4.0 g of H₂ (molar mass 2.0 g/mol) gives 2.0 mol H₂; 32.0 g of O₂ (molar mass 32.0 g/mol) gives 1.0 mol O₂.

在任何限制反应物问题的第一步，都是使用摩尔质量将每种反应物的给定质量转化为摩尔。摩尔 = 质量 / 摩尔质量。例如，4.0 克 H₂（摩尔质量 2.0 克/摩尔）得到 2.0 摩尔 H₂；32.0 克 O₂（摩尔质量 32.0 克/摩尔）得到 1.0 摩尔 O₂。

n = m / M

Use the periodic table to obtain molar masses to the appropriate precision. In exams, always show your working because method marks are awarded even if the final answer is slightly wrong.

使用元素周期表获得适当精度的摩尔质量。在考试中，始终写出计算过程，因为即使最终答案略有错误，也会给方法分。

4. Step 2: Compare the Mole Ratio | 第二步：比较摩尔比

To identify the limiting reactant, compare the available mole ratio with the required mole ratio. Divide the moles of each reactant by its coefficient in the balanced equation. The reactant with the smallest value is the limiting reactant.

要识别限制反应物，需要将可用摩尔比与所需摩尔比进行比较。用每种反应物的摩尔数除以其在配平方程式中的系数。值最小的反应物就是限制反应物。

For the reaction aA + bB → products, calculate n(A)/a and n(B)/b. The smaller value indicates the limiting reactant.

对于反应 aA + bB → 产物，计算 n(A)/a 和 n(B)/b。较小的值表示限制反应物。

Reactant	Moles available	Coefficient	n / coefficient
N₂	2.0	1	2.0
H₂	4.0	3	1.33

Here 1.33 < 2.0, so H₂ is the limiting reactant even though its actual mole number is larger than that of N₂.

这里 1.33 < 2.0，所以 H₂ 是限制反应物，尽管其实际摩尔数大于 N₂。

5. Step 3: Use the Limiting Reactant to Calculate Product | 第三步：用限制反应物计算产物

Once the limiting reactant is known, all product amounts are based on it. Using the mole ratio from the balanced equation, convert moles of limiting reactant to moles of product, then to mass if required.

一旦确定了限制反应物，所有产物的量都基于它计算。使用配平方程式中的摩尔比，将限制反应物的摩尔数转换为产物的摩尔数，然后在需要时转换为质量。

Example: For 2H₂ + O₂ → 2H₂O, if 2.0 mol H₂ is limiting, then moles of H₂O = 2.0 mol × (2/2) = 2.0 mol. Mass of H₂O = 2.0 mol × 18.0 g/mol = 36.0 g.

示例：对于 2H₂ + O₂ → 2H₂O，如果 2.0 摩尔 H₂ 是限制反应物，则 H₂O 的摩尔数 = 2.0 摩尔 × (2/2) = 2.0 摩尔。H₂O 的质量 = 2.0 摩尔 × 18.0 克/摩尔 = 36.0 克。

6. Calculating the Amount of Excess Reactant Remaining | 计算过量反应物的剩余量

To find how much excess reactant remains, first calculate the amount that actually reacts using the mole ratio with the limiting reactant. Then subtract this from the initial amount.

要找出有多少过量反应物剩余，首先使用与限制反应物的摩尔比计算实际反应的量。然后从初始量中减去这个量。

Using the earlier example: 2.0 mol H₂ reacts with 1.0 mol O₂. If you started with 2.0 mol O₂, the remaining O₂ = 2.0 − 1.0 = 1.0 mol. Convert to mass if needed: 1.0 mol × 32.0 g/mol = 32.0 g.

使用前面的例子：2.0 摩尔 H₂ 与 1.0 摩尔 O₂ 反应。如果你起始有 2.0 摩尔 O₂，则剩余 O₂ = 2.0 − 1.0 = 1.0 摩尔。如果需要，转换为质量：1.0 摩尔 × 32.0 克/摩尔 = 32.0 克。

7. Theoretical Yield and Percent Yield | 理论产率与百分产率

The theoretical yield is the maximum mass of product calculated from the limiting reactant. The percent yield compares the actual yield (from experiment) to the theoretical yield:

理论产率是根据限制反应物计算出的产物最大质量。百分产率将实际产率（实验获得）与理论产率进行比较：

Percent yield = (actual yield / theoretical yield) × 100%

For example, if the theoretical yield is 10.0 g but only 8.5 g is collected, percent yield = (8.5 / 10.0) × 100% = 85%. Percent yield is always ≤ 100% in practice due to losses, side reactions, or incomplete reactions.

例如，如果理论产率为 10.0 克，但实际只收集到 8.5 克，则百分产率 = (8.5 / 10.0) × 100% = 85%。由于损失、副反应或反应不完全，实际百分产率总是 ≤ 100%。

8. Gas-Phase Reactions and Volume Ratios | 气相反应与体积比

For gases under the same conditions of temperature and pressure, the mole ratio equals the volume ratio. This is a direct consequence of Avogadro’s law. Therefore, limiting reactant problems involving gases can be solved using volumes instead of masses.

对于在相同温度和压力条件下的气体，摩尔比等于体积比。这是阿伏伽德罗定律的直接推论。因此，涉及气体的限制反应物问题可以用体积代替质量来解决。

Consider the reaction C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. If 10 dm³ of C₃H₈ reacts with 40 dm³ of O₂, the stoichiometric volume ratio is 1:5, meaning 10 dm³ C₃H₈ requires 50 dm³ O₂. Since only 40 dm³ is available, O₂ is limiting.

考虑反应 C₃H₈ + 5O₂ → 3CO₂ + 4H₂O。如果 10 立方分米 C₃H₈ 与 40 立方分米 O₂ 反应，化学计量体积比为 1:5，意味着 10 立方分米 C₃H₈ 需要 50 立方分米 O₂。由于只有 40 立方分米可用，O₂ 是限制反应物。

9. Solutions and Concentration | 溶液与浓度

When reactants are in solution, the amount of solute in moles is calculated using concentration and volume: moles = concentration × volume. Ensure volume is in dm³ when concentration is in mol/dm³.

当反应物在溶液中时，溶质的摩尔数使用浓度和体积计算：摩尔 = 浓度 × 体积。当浓度以摩尔/立方分米为单位时，确保体积以立方分米为单位。

For example, 25.0 cm³ of 0.200 mol/dm³ HCl contains 0.0250 dm³ × 0.200 mol/dm³ = 0.00500 mol HCl. In a titration, the known moles of one reactant and the stoichiometric ratio reveal the moles of the other reactant, which can then be used to find its concentration.

例如，25.0 立方厘米的 0.200 摩尔/立方分米 HCl 含有 0.0250 立方分米 × 0.200 摩尔/立方分米 = 0.00500 摩尔 HCl。在滴定中，已知一种反应物的摩尔数和化学计量比可以揭示另一种反应物的摩尔数，进而求其浓度。

10. Multistep and Sequential Reactions | 多步与顺序反应

In some industrial processes, a product from one reaction is used as a reactant in a subsequent reaction. The limiting reactant must be identified for each step separately. The amount of intermediate product that carries forward is the theoretical yield of the first step, but the actual yield may be lower.

在一些工业过程中，一个反应的产物用作后续反应的反应物。每一步都需要分别确定限制反应物。前一步骤的理论产率作为中间产物带入下一步，但实际产率可能较低。

When multiple reactions are involved, always use moles, never masses, for the chain of conversions. A common exam question involves calculating the mass of final product from a given mass of starting material through two steps.

当涉及多个反应时，始终使用摩尔而不是质量进行一系列转换。一个常见的考试问题是通过两步从给定质量的起始物计算最终产物的质量。

11. Common Pitfalls and Exam Tips | 常见错误与应试技巧

Do not compare masses directly. You must use moles. A reactant with a smaller mass is not necessarily limiting. Do not forget to balance the equation. The coefficients are essential for the mole ratio. Do not use the limiting reactant after identifying it incorrectly. Double-check by comparing n/coefficient values.

不要直接比较质量。必须使用摩尔。质量较小的反应物不一定就是限制反应物。不要忘记配平方程式。系数对于摩尔比至关重要。不要错误识别限制反应物。通过比较 n/系数的值来复核。

Watch units. Convert all volumes to dm³ if using mol/dm³. Convert mass in grams when using molar mass in g/mol. Show your working for partial credit. State the limiting reactant clearly in your final answer.

注意单位。如果使用摩尔/立方分米，将所有体积转换为立方分米。当摩尔质量以克/摩尔为单位时，质量使用克。写出计算过程以获得部分分数。在最终答案中明确指出限制反应物。

12. Worked Example | 完整例题

Question: 5.00 g of iron reacts with 3.00 g of sulfur according to Fe + S → FeS. Which is the limiting reactant and how much FeS is formed? (Fe = 55.8, S = 32.1)

问题：5.00 克铁与 3.00 克硫按 Fe + S → FeS 反应。哪种是限制反应物？生成多少 FeS？（Fe = 55.8，S = 32.1）

Moles of Fe = 5.00 / 55.8 = 0.0896 mol. Moles of S = 3.00 / 32.1 = 0.0935 mol. The balanced equation has a 1:1 ratio. Since 0.0896 < 0.0935, Fe is the limiting reactant.

铁的摩尔数 = 5.00 / 55.8 = 0.0896 摩尔。硫的摩尔数 = 3.00 / 32.1 = 0.0935 摩尔。配平方程为 1:1 比例。因为 0.0896 < 0.0935，铁是限制反应物。

Moles of FeS formed = 0.0896 mol. Mass of FeS = 0.0896 × (55.8 + 32.1) = 0.0896 × 87.9 = 7.88 g. The excess sulfur remaining = 0.0935 − 0.0896 = 0.0039 mol = 0.125 g.

生成 FeS 的摩尔数 = 0.0896 摩尔。FeS 的质量 = 0.0896 × (55.8 + 32.1) = 0.0896 × 87.9 = 7.88 克。过量硫剩余 = 0.0935 − 0.0896 = 0.0039 摩尔 = 0.125 克。

Mastering limiting and excess reactant calculations is essential for stoichiometry. Always work through moles, compare ratios, and base product yields on the limiting reactant. With careful unit handling and clear working, these problems become straightforward.

掌握限制反应物与过量反应物的计算对于化学计量学至关重要。始终通过摩尔计算，比较比例，并将产率建立在限制反应物的基础上。只要注意单位并清晰书写计算过程，这类问题就会变得简单明了。

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Limiting Reactant and Excess Reactant Calculations | 限制反应物与过量反应物计算