GCSE Chemistry revision

Limiting reactants

Use of amount of substance in relation to masses of pure substances

AQA 4.3.2.4 Higher Tier only

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A limiting reactant is the reactant that is completely used up during a chemical reaction, causing the reaction to stop.
When calcium carbonate is added to hydrochloric acid, the fizzing stops once all the calcium carbonate is consumed, making it the limiting reactant.
The limiting reactant determines the maximum amount of product that can be produced in a reaction.

1. Introduction AQA

A limiting reactant is the reactant that is completely used up during a chemical reaction, causing the reaction to stop.
When calcium carbonate is added to hydrochloric acid, the fizzing stops once all the calcium carbonate is consumed, making it the limiting reactant.
The limiting reactant determines the maximum amount of product that can be produced in a reaction.

A reactant described as being in excess is one that is not fully used up — there is more than enough of it for the reaction to proceed.
In the calcium carbonate and hydrochloric acid example, the hydrochloric acid is in excess because some remains unreacted after the reaction stops.
Adding more of the excess reactant (e.g. more hydrochloric acid) will not produce more product, as there is no limiting reactant left to react with.

2. Reactants in Excess AQA

A reactant described as being in excess is one that is not fully used up — there is more than enough of it for the reaction to proceed.
In the calcium carbonate and hydrochloric acid example, the hydrochloric acid is in excess because some remains unreacted after the reaction stops.
Adding more of the excess reactant (e.g. more hydrochloric acid) will not produce more product, as there is no limiting reactant left to react with.

When a question states a reaction occurs 'in air', the oxygen is assumed to be in excess because air contains a large and effectively unlimited supply of oxygen.
If one reactant is given a specific mass and the other is implied to be plentiful, the reactant with the given mass is the limiting reactant.
Identifying the limiting reactant is the essential first step before carrying out any mass calculation.

3. Identifying the Limiting Reactant in Questions AQA

When a question states a reaction occurs 'in air', the oxygen is assumed to be in excess because air contains a large and effectively unlimited supply of oxygen.
If one reactant is given a specific mass and the other is implied to be plentiful, the reactant with the given mass is the limiting reactant.
Identifying the limiting reactant is the essential first step before carrying out any mass calculation.

Always begin a calculation by writing out the word equation and then the balanced symbol equation for the reaction.
For sodium burning in air, the balanced equation is: 4Na + O₂ → 2Na₂O The balanced equation provides the molar ratio between reactants and products, which is essential for the calculation.

4. Step 1 – Writing and Balancing the Equation AQA

Always begin a calculation by writing out the word equation and then the balanced symbol equation for the reaction.
For sodium burning in air, the balanced equation is: 4Na + O₂ → 2Na₂O The balanced equation provides the molar ratio between reactants and products, which is essential for the calculation.

Use the formula moles = mass / Mr to calculate the number of moles of the limiting reactant.
For 115 g of sodium with an Mᵣ of 23: moles = 115 / 23 = 5 moles of Na

The molar ratio from the balanced equation is used to convert moles of the limiting reactant into moles of product.
From 4Na + O₂ → 2Na₂O, the ratio of Na to Na₂O is 4:2, which simplifies to 2:1.
Therefore, 5 moles of sodium produces 5 ÷ 2 = 2.5 moles of sodium oxide.

6. Step 3 – Using the Molar Ratio to Find Moles of Product AQA

The molar ratio from the balanced equation is used to convert moles of the limiting reactant into moles of product.
From 4Na + O₂ → 2Na₂O, the ratio of Na to Na₂O is 4:2, which simplifies to 2:1.
Therefore, 5 moles of sodium produces 5 ÷ 2 = 2.5 moles of sodium oxide.

Rearrange the moles formula to find mass: mass = moles × Mᵣ The Mᵣ of sodium oxide (Na₂O) is calculated as (2 × 23) + 16 = 62 Substituting the values: mass = 2.5 × 62 = 155 g of sodium oxide is produced.