Limiting reactants (HT only) — AQA GCSE Chemistry
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Limiting reactants (HT only) explained
Limiting reactants (HT only) covers 3 specification points, including In a chemical reaction involving two… and The reactant that is completely used up is….
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Open a point below for its explanation and revision guidance.
In a chemical reaction involving two reactants, it is common to use an excess of one of the reactants to ensure that all of the other reactant is used.
When two reactants could react, adding extra of one ensures the other is fully consumed. The reactant in excess is left over at the end; the other is the limiting reactant. For example, in burning magnesium in air, oxygen is in excess so all the magnesium reacts. In a school preparation, an acid may be added in excess to dissolve all of a metal carbonate, ensuring complete reaction. Using excess is common because it drives the reaction to completion, maximises yield based on the valuable reactant, and avoids unreacted starting material. In calculations, you identify which reactant is limiting by comparing mole ratios, then use that reactant to find the amount of product formed.
The reactant that is completely used up is called the limiting reactant because it limits the amount of products.
The limiting reactant is the substance that runs out first and therefore controls how much product can form. Once it is used up, the reaction stops, even if other reactants remain. For example, if 0.10 mol of magnesium reacts with 0.20 mol of oxygen, the equation 2Mg + O₂ → 2MgO shows magnesium is limiting because 0.10 mol Mg needs only 0.05 mol O₂, leaving oxygen in excess. The amount of product is calculated from the limiting reactant using the mole ratio. Identifying the limiting reactant is essential in theoretical yield calculations and in predicting whether a reaction goes to completion, but it does not affect atom economy.
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.
When reactants are not in the exact ratio of the balanced equation, the reactant that runs out first is the limiting reactant; it fixes the maximum amount of product. The other reactant is in excess and some remains unreacted. Use the balanced equation to convert the limiting amount into product amount. For example, 2 mol H₂ reacts with 1 mol O₂ to give 2 mol H₂O. If only 1 mol O₂ is available with 3 mol H₂, oxygen is limiting, so at most 2 mol H₂O forms and 1 mol H₂ is left over. The same logic works with masses: convert masses to moles using n = m ÷ Mᵣ, apply the mole ratio, then convert back to mass if required.
Your focus
- Define excess reactant and explain why it is used.
- Determine which reactant is limiting from mole ratios.
- Describe the fate of the excess reactant after reaction.
Show all 9 objectives
- Define limiting reactant and explain its effect on product amount.
- Identify the limiting reactant by comparing mole ratios.
- Calculate the amount of product from the limiting reactant.
- Identify the limiting reactant from given amounts and a balanced equation.
- Calculate the maximum moles or mass of product from the limiting reactant.
- Explain why the excess reactant remains after reaction and does not increase product yield.
Limiting reactants (HT only) exam tips
Marking Points
- Explain that excess means more of a reactant is present than is needed to react with the other reactant.
- State that the purpose of excess is to ensure the other reactant is completely used up.
- Identify that the reactant not in excess is the limiting reactant and determines the amount of product.
- Recognise that the excess reactant remains after the reaction and may need to be separated or recovered.
- Apply the idea to a named reaction, such as an acid in excess reacting with a metal or carbonate.
- Define the limiting reactant as the reactant that is completely used up in a reaction.
- Explain that the limiting reactant determines the maximum amount of product that can form.
- Compare amounts in moles with the balanced equation ratio to identify which reactant is limiting.
- Use the limiting reactant's amount and the equation ratio to calculate the amount of product.
- Recognise that reaction stops when the limiting reactant is exhausted, leaving excess reactants unreacted.
- Identify the limiting reactant by comparing the amounts of each reactant with the ratio required by the balanced equation.
- State that the limiting reactant determines the maximum amount of product that can be formed.
- Use the balanced equation to convert the amount of limiting reactant into the amount of product in moles.
- Convert between mass and moles using n = m ÷ Mᵣ when the question gives masses in grams.
- Explain that the reactant in excess is not fully used up and remains after the reaction.
Examiner Tips
- 💡Define excess and limiting reactant clearly before applying them to a calculation.
- 💡Use the balanced equation to compare the mole ratio of the two reactants when deciding which is limiting.
- 💡Link the purpose of excess to a practical context, such as ensuring complete reaction or maximising use of an expensive reactant.
- 💡Write the balanced equation and note the mole ratio before deciding which reactant is limiting.
- 💡Show the comparison of required versus available moles for each reactant to justify your choice.
- 💡Use the limiting reactant in subsequent product calculations and state this explicitly.
- 💡Write the balanced equation and label the mole ratio before doing any calculation.
- 💡Show clearly which reactant is limiting and state that it controls the product amount.
- 💡Check units and significant figures at the end, and include the unit with any mass answer.
Common Mistakes
- Assuming the reactant with the larger mass is always in excess; correct this by comparing moles and the equation ratio, not masses alone.
- Thinking excess reactant is used up first; correct this by stating that the limiting reactant is consumed completely while excess remains.
- Believing excess guarantees a higher percentage yield of product from the limiting reactant; correct this by explaining it ensures complete reaction of the limiting reactant, not extra product beyond the stoichiometric amount.
- Choosing the reactant with the smaller mass as limiting without converting to moles; correct this by comparing moles using the equation ratio.
- Assuming the reactant present in the smallest mole amount is always limiting; correct this by checking the stoichiometric ratio, since coefficients matter.
- Calculating product from the excess reactant; correct this by always using the limiting reactant for product calculations.
- Assuming the reactant with the smaller mass is always limiting; the correct approach is to compare mole amounts with the equation ratio.
- Using the amount of the excess reactant to calculate product; the correct method uses the limiting reactant only.
- Forgetting to convert grams to moles before applying the ratio; the corrected method converts mass to moles first, then uses the ratio.