Lesson 3.12 · AQA Combined Science

Higher Tier: limiting reactants

Higher Tier only. This is an optional extension after the core lessons. Go to exam practice.

The limiting reactant runs out

Higher Tier only. The limiting reactant is used up first and limits how much product can form. Another reactant may be in excess: some of it remains once the limiting reactant is used up.

Compare what the equation needs

2H₂ + O₂ → 2H₂O.
You have 3 mol H₂ and 1 mol O₂. That oxygen needs only 2 mol H₂.
O₂ is limiting. At most 2 mol H₂O forms, leaving 1 mol H₂.

A reliable comparison

Divide the available moles of each reactant by its coefficient. The smaller result gives the number of complete reaction “batches”.
H₂: 3 ÷ 2 = 1.5.
O₂: 1 ÷ 1 = 1.
Oxygen permits fewer batches, so it limits the reaction. The smaller mass or smaller raw mole number is not always the limiting reactant.

Worked example with masses

2Mg + O₂ → 2MgO. Start with 12 g Mg and 4 g O₂. Molar masses: Mg 24, O₂ 32, MgO 40 g/mol.
Mg: 12 ÷ 24 = 0.50 mol; O₂: 4 ÷ 32 = 0.125 mol.
Compare: 0.50 ÷ 2 = 0.25; 0.125 ÷ 1 = 0.125. O₂ limits.
MgO formed = 2 × 0.125 = 0.25 mol.
Maximum mass = 0.25 × 40 = 10 g MgO.
Mg used = 0.25 × 24 = 6 g, leaving 6 g Mg.

Predict what happens if you add more

Once the limiting reactant runs out, adding more of the reactant already in excess does not make more product. Adding more of the limiting reactant can make more product while the other reactant is still available. Real collected product can also be reduced by incomplete reaction or losses; this calculation gives a maximum from the stated reaction.

Remember

Use the equation ratio to find what runs out. Calculate product from the limiting reactant, not from an excess amount.