Limiting reactants (HT only) — AQA GCSE Combined Science
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Limiting reactants (HT only) explained
When two reactants are mixed, the reaction stops as soon as one of them is used up.
Read the full explanation
That reactant is the limiting reactant; the other is in excess. Using an excess of one reactant is a practical way to make sure the more valuable or harder-to-obtain reactant is fully consumed, so no product is lost. For example, in 2Mg + O₂ → 2MgO, if 0.10 mol Mg reacts with 0.20 mol O₂, Mg is limiting because it runs out first, even though less oxygen was needed. The amount of product is fixed by the limiting reactant, not by the excess. To identify the limiting reactant, compare the mole ratio from the balanced equation with the amounts actually present; the reactant that would produce the smaller amount of product is limiting. Note that limiting reactant concepts are assessed at Higher Tier only.
Students should be able to explain the effect of a limiting quantity of a reactant on the amount of products it is possible to obtain in terms of amounts in moles or masses in grams.
The limiting reactant controls the maximum product because the reaction stops when it is used up. To explain the effect, compare the actual mole amounts with the balanced equation ratio. For example, in N₂ + 3H₂ → 2NH₃, if 1 mol N₂ reacts with 2 mol H₂, hydrogen is limiting because 1 mol N₂ needs 3 mol H₂. The hydrogen can form (2 ÷ 3) × 2 = 1.33 mol NH₃, while nitrogen could form 2 mol NH₃, so the actual maximum is 1.33 mol. In masses, convert the limiting moles to grams using mass = moles × relative formula mass. Extra excess reactant cannot increase the product once the limiting reactant is gone. This topic is assessed at Higher Tier only.
Your focus
- Define limiting reactant and excess reactant using the idea of complete consumption.
- Use a balanced equation to identify which reactant is limiting from given mole amounts.
- Explain why the limiting reactant determines the maximum amount of product formed.
Show all 6 objectives
- Calculate the maximum product in moles from a limiting reactant using the balanced equation ratio.
- Convert the maximum product from moles to mass in grams using relative formula mass.
- Explain why the limiting reactant fixes the amount of product and why excess reactant does not increase it.
Limiting reactants (HT only) exam tips
Marking Points
- The limiting reactant is the reactant that is completely used up when the reaction stops.
- The reactant in excess is not completely used up and remains after the reaction.
- The amount of product formed is determined by the limiting reactant, not by the excess reactant.
- Using an excess of one reactant can ensure that the other reactant is fully used, which is useful when that other reactant is expensive or difficult to obtain.
- To identify the limiting reactant, compare the actual mole amounts with the mole ratio in the balanced equation.
- The limiting reactant is used up first and therefore determines the maximum amount of product.
- The balanced equation gives the mole ratio needed to compare the amounts of reactants actually present.
- Calculate the product that each reactant could form, then choose the smaller value as the maximum possible amount.
- Convert between moles and mass using mass = moles × relative formula mass when the question asks for masses in grams.
Examiner Tips
- 💡Write the balanced equation first and label the mole ratio before doing any comparison.
- 💡Convert masses to moles using moles = mass ÷ relative formula mass, then compare with the equation ratio.
- 💡Remember that limiting reactant calculations and explanations are assessed at Higher Tier only.
- 💡Show the balanced equation and the mole ratio before calculating product amounts.
- 💡Calculate the product from each reactant separately, then compare to identify the limiting reactant and maximum yield.
- 💡Be aware that explaining the effect of a limiting quantity of a reactant is a Higher Tier only requirement.
Common Mistakes
- Assuming the reactant with the smaller mass is always limiting. Correction: compare amounts in moles using the balanced equation, because molar masses differ.
- Thinking the excess reactant is the one that runs out first. Correction: the excess reactant remains after reaction; the limiting reactant is used up.
- Ignoring the coefficients in the balanced equation when comparing mole amounts. Correction: use the ratio from the equation, for example 2 mol Mg to 1 mol O₂.
- Using the larger reactant amount to calculate product. Correction: use the limiting reactant, which gives the smaller possible product amount.
- Forgetting to multiply by the coefficient ratio from the balanced equation. Correction: apply the mole ratio, for example 3 mol H₂ forms 2 mol NH₃.
- Mixing up moles and grams in the final answer. Correction: convert carefully using relative formula mass and state the unit.