Limiting Reactants [H]

This lesson covers limiting reactants and excess reactants as required by the AQA GCSE Chemistry specification (4.3.2). This is Higher Tier only content. You need to understand the difference between a limiting and an excess reactant, be able to identify which reactant is limiting, and calculate the mass of product formed based on the limiting reactant.

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What Is a Limiting Reactant?

In most chemical reactions, the reactants are not present in the exact amounts needed for them to react completely with each other. One reactant will be completely used up before the other — this is the limiting reactant. The other reactant is said to be in excess because some of it remains unreacted.

Term	Definition
Limiting reactant	The reactant that is completely used up in a reaction. It determines the maximum amount of product that can be formed.
Excess reactant	The reactant that is left over after the reaction is complete. There is more of it than is needed.

The limiting reactant controls:

• The amount of product formed
• When the reaction stops
• The extent of the reaction

Exam Tip: The limiting reactant is always the one that runs out first. It "limits" the amount of product that can be made. All reacting mass calculations should be based on the limiting reactant, not the excess reactant.

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Identifying the Limiting Reactant

To determine which reactant is limiting, you need to:

1. Calculate the moles of each reactant present.
2. Use the mole ratio from the balanced equation to see which one would run out first.
3. The one that runs out first is the limiting reactant.

Worked Example 1

3.0 g of magnesium reacts with 100 cm3 of 1.0 mol/dm3 hydrochloric acid. Which is the limiting reactant?

Balanced equation: Mg + 2HCl --> MgCl2 + H2

Step 1: Calculate moles of each reactant.

• moles of Mg = mass / Mr = 3.0 / 24 = 0.125 mol
• moles of HCl = concentration x volume = 1.0 x (100/1000) = 0.1 mol

Step 2: From the equation, 1 mol Mg reacts with 2 mol HCl.

• 0.125 mol Mg would need 0.125 x 2 = 0.25 mol HCl
• But only 0.1 mol HCl is available

Step 3: There is not enough HCl to react with all the Mg. Therefore, HCl is the limiting reactant and Mg is in excess.

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Visualising Limiting Reactants

graph TD
    A["Calculate moles of Reactant A"] --> C["Compare using mole ratio"]
    B["Calculate moles of Reactant B"] --> C
    C --> D{"Which runs out first?"}
    D -->|"A runs out"| E["A is the limiting reactant"]
    D -->|"B runs out"| F["B is the limiting reactant"]
    E --> G["Calculate product based on A"]
    F --> H["Calculate product based on B"]

    style A fill:#3498db,color:#fff
    style B fill:#e67e22,color:#fff
    style D fill:#9b59b6,color:#fff
    style E fill:#27ae60,color:#fff
    style F fill:#27ae60,color:#fff

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Worked Example 2: Calculating Product Mass

5.6 g of iron reacts with 3.2 g of sulfur. Calculate the maximum mass of iron sulfide (FeS) that can be formed. Identify the limiting reactant.

Balanced equation: Fe + S --> FeS

Step 1: Calculate moles of each reactant.

• Mr of Fe = 56. moles of Fe = 5.6 / 56 = 0.1 mol
• Mr of S = 32. moles of S = 3.2 / 32 = 0.1 mol

Step 2: From the equation, ratio Fe : S = 1 : 1

• 0.1 mol Fe needs 0.1 mol S
• 0.1 mol S needs 0.1 mol Fe
• Both are present in exactly the right amounts — neither is in excess (this is a special case).

Step 3: moles of FeS = 0.1 mol

• Mr of FeS = 56 + 32 = 88
• mass of FeS = 0.1 x 88 = 8.8 g

Worked Example 3: Unequal Mole Ratio

4.8 g of magnesium reacts with 7.3 g of hydrochloric acid. Which is the limiting reactant and what mass of magnesium chloride is produced?

Balanced equation: Mg + 2HCl --> MgCl2 + H2

Step 1: Calculate moles of each reactant.

• Mr of Mg = 24. moles of Mg = 4.8 / 24 = 0.2 mol
• Mr of HCl = 36.5. moles of HCl = 7.3 / 36.5 = 0.2 mol

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

• 0.2 mol Mg would need 0.2 x 2 = 0.4 mol HCl
• Only 0.2 mol HCl is available

HCl is the limiting reactant.

Step 3: 0.2 mol HCl produces 0.2 / 2 = 0.1 mol MgCl2

• Mr of MgCl2 = 24 + (2 x 35.5) = 95
• mass of MgCl2 = 0.1 x 95 = 9.5 g

Exam Tip: When identifying the limiting reactant, calculate the moles of each reactant and then use the mole ratio to check if there is enough of one to react with all of the other. The one there is "not enough" of is the limiting reactant.

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Effect of Limiting Reactant on Reaction

The amount of product formed is directly proportional to the amount of limiting reactant used. If you double the amount of limiting reactant (keeping the other in excess), you double the amount of product.

Amount of Limiting Reactant	Effect on Product	Amount of Excess Reactant Left
Increased	More product formed	Less excess remains
Decreased	Less product formed	More excess remains
Doubled	Product doubles	Excess used up faster

Exam Tip: Graphs showing the effect of a limiting reactant often appear in exams. If you add more limiting reactant, the graph goes higher (more product). If you add more of the excess reactant, the graph does NOT change — the same amount of product is formed.

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Calculating Excess Reactant Remaining

You can also calculate how much of the excess reactant remains unreacted.

Worked Example 4

6.0 g of magnesium reacts with 100 g of hydrochloric acid (HCl). Calculate the mass of HCl that remains unreacted.

Balanced equation: Mg + 2HCl --> MgCl2 + H2

Step 1: moles of Mg = 6.0 / 24 = 0.25 mol moles of HCl = 100 / 36.5 = 2.74 mol