Limiting Reactants (Higher Tier)

This lesson covers the concept of limiting reactants and excess reactants as required by the AQA GCSE Combined Science Trilogy specification (8464) at Higher tier. You will learn how to identify which reactant limits the reaction and how to calculate the mass of product formed when one reactant is in excess.

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Key Definitions

Term	Definition
Limiting reactant	The reactant that is completely used up first in a reaction. It determines the maximum amount of product that can be formed.
Excess reactant	The reactant that is not completely used up. Some of it remains after the reaction has finished.

When the limiting reactant is all used up, the reaction stops — even though there is still some excess reactant left.

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Why Does It Matter?

In the real world, you rarely mix reactants in perfect proportions. One reactant usually runs out first. The limiting reactant controls:

• The maximum amount of product formed.
• The amount of excess reactant left over.

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How to Identify the Limiting Reactant

Step-by-Step Method

flowchart TD
    A["Write balanced equation"] --> B["Calculate moles\nof each reactant"]
    B --> C["Divide each by its\ncoefficient in equation"]
    C --> D{"Which value\nis smallest?"}
    D -->|"Smallest"| E["That reactant is the\nLIMITING REACTANT"]
    D -->|"Larger"| F["That reactant is\nin EXCESS"]
    E --> G["Use limiting reactant\nto calculate product"]
    style E fill:#ef4444,color:#fff,stroke:#dc2626
    style F fill:#22c55e,color:#fff,stroke:#16a34a
    style G fill:#f59e0b,color:#000,stroke:#d97706

1. Write the balanced equation.
2. Calculate the moles of each reactant.
3. Divide each by its coefficient in the balanced equation.
4. The reactant with the smallest value is the limiting reactant.

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Worked Examples

Example 1 — Magnesium and Hydrochloric Acid

Question: 2.4 g of magnesium reacts with 100 cm³ of 1.0 mol/dm³ hydrochloric acid. Which is the limiting reactant? What mass of hydrogen is produced?

$\text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)$Mg(s)+2HCl(aq)→MgCl2​(aq)+H2​(g)

Step 1: Calculate moles of each reactant.

Moles of Mg: $\text{moles} = \frac{2.4}{24} = 0.1 \text{ mol}$moles=242.4​=0.1 mol

Moles of HCl: $\text{moles} = 1.0 \times 0.1 = 0.1 \text{ mol}$moles=1.0×0.1=0.1 mol

Step 2: Compare with the equation ratio.

From the equation: 1 mol Mg requires 2 mol HCl.

• Mg available: 0.1 mol (needs 0.2 mol HCl)
• HCl available: 0.1 mol

We have only 0.1 mol HCl, but we need 0.2 mol to react with all the Mg. HCl is the limiting reactant.

Step 3: Calculate the product based on the limiting reactant.

From the equation: 2 mol HCl produces 1 mol $\text{H}_2$H2​

$\text{moles of H}_2 = \frac{0.1}{2} = 0.05 \text{ mol}$moles of H2​=20.1​=0.05 mol

$\text{mass of H}_2 = 0.05 \times 2 = 0.1 \text{ g}$mass of H2​=0.05×2=0.1 g

Example 2 — Iron and Sulfur

Question: 5.6 g of iron reacts with 4.8 g of sulfur. What mass of iron sulfide (FeS) is formed? Which reactant is in excess?

$\text{Fe}(s) + \text{S}(s) \rightarrow \text{FeS}(s)$Fe(s)+S(s)→FeS(s)

Step 1: Calculate moles.

$\text{moles of Fe} = \frac{5.6}{56} = 0.1 \text{ mol}$moles of Fe=565.6​=0.1 mol

$\text{moles of S} = \frac{4.8}{32} = 0.15 \text{ mol}$moles of S=324.8​=0.15 mol

Step 2: The ratio is 1 : 1. We need equal moles.

• 0.1 mol Fe needs 0.1 mol S.
• We have 0.15 mol S — more than enough.
• Fe is the limiting reactant. Sulfur is in excess.

Step 3: Product is based on Fe (limiting).

$\text{moles of FeS} = 0.1 \text{ mol}$moles of FeS=0.1 mol

$M_r \text{ of FeS} = 56 + 32 = 88$Mr​ of FeS=56+32=88

$\text{mass of FeS} = 0.1 \times 88 = 8.8 \text{ g}$mass of FeS=0.1×88=8.8 g

Excess sulfur remaining: $\text{S used} = 0.1 \text{ mol} \quad \text{S remaining} = 0.15 - 0.1 = 0.05 \text{ mol}$S used=0.1 molS remaining=0.15−0.1=0.05 mol $\text{mass of excess S} = 0.05 \times 32 = 1.6 \text{ g}$mass of excess S=0.05×32=1.6 g

Example 3 — Copper Oxide and Sulfuric Acid

Question: 12 g of copper oxide is added to a solution containing 9.8 g of sulfuric acid. Which is the limiting reactant, and what mass of copper sulfate is formed?

$\text{CuO}(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{CuSO}_4(aq) + \text{H}_2\text{O}(l)$CuO(s)+H2​SO4​(aq)→CuSO4​(aq)+H2​O(l)

$M_r$Mr​ of CuO = 63.5 + 16 = 79.5, $M_r$Mr​ of $\text{H}_2\text{SO}_4$H2​SO4​ = 98, $M_r$Mr​ of $\text{CuSO}_4$CuSO4​ = 159.5

$\text{moles of CuO} = \frac{12}{79.5} = 0.151 \text{ mol (3 s.f.)}$moles of CuO=79.512​=0.151 mol (3 s.f.)

$\text{moles of H}_2\text{SO}_4 = \frac{9.8}{98} = 0.1 \text{ mol}$moles of H2​SO4​=989.8​=0.1 mol

The ratio is 1 : 1. We need equal moles, but we have less $\text{H}_2\text{SO}_4$H2​SO4​.

$\text{H}_2\text{SO}_4$H2​SO4​ is the limiting reactant.

$\text{moles of CuSO}_4 = 0.1 \text{ mol}$moles of CuSO4​=0.1 mol

$\text{mass of CuSO}_4 = 0.1 \times 159.5 = 16.0 \text{ g (3 s.f.)}$mass of CuSO4​=0.1×159.5=16.0 g (3 s.f.)

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Visualising Limiting Reactants

Think of it like making sandwiches. If you have 10 slices of bread and 3 slices of cheese, and each sandwich needs 2 slices of bread and 1 slice of cheese:

• Cheese makes 3 sandwiches (uses 6 slices of bread).
• Bread could make 5 sandwiches — but you run out of cheese first.
• Cheese is the limiting reactant. Bread is in excess (4 slices left over).

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Why Use Excess Reactant?

In many industrial and lab reactions, one reactant is deliberately added in excess to ensure the other (more expensive or important) reactant is completely used up, maximising yield.

Exam Tip (AQA 8464): If a question says a reactant is "in excess", this means the OTHER reactant is the limiting one. Base all your calculations on the limiting reactant.

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Common Mistakes to Avoid

Mistake	Correction
Assuming the reactant with less mass is limiting	Mass is irrelevant — compare moles relative to the equation ratio
Calculating product from the excess reactant	Always use the limiting reactant
Forgetting to use the mole ratio	The ratio from the balanced equation is essential
Not identifying which reactant is limiting before calculating	Always determine the limiting reactant first

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Summary