Reacting Masses and Limiting Reactants — AQA GCSE Chemistry
The balanced equation tells you the ratio in which substances react, allowing you to calculate masses.
Reacting masses (Higher Tier)
The numbers in front of the formulae in a balanced equation give the mole ratio of the substances. To calculate a reacting mass:
- Balance the equation.
- Convert the known mass to moles (moles = mass ÷ Mr).
- Use the mole ratio to find moles of the substance you want.
- Convert that back to mass (mass = moles × Mr).
Worked example
What mass of magnesium oxide forms from 6 g of magnesium? (Ar: Mg = 24, O = 16)
- 2Mg + O₂ → 2MgO
- Moles of Mg = 6 ÷ 24 = 0.25 mol.
- Ratio Mg : MgO = 2 : 2 = 1 : 1, so moles of MgO = 0.25 mol.
- Mr of MgO = 40, so mass = 0.25 × 40 = 10 g.
Limiting reactants
When two reactants are mixed, one is often used up first — the limiting reactant. It limits the amount of product; the other is in excess.
- The amount of product is directly proportional to the amount of the limiting reactant.
- To identify the limiting reactant, work out the moles of each and compare with the equation ratio.
Exam tips
- Always balance the equation first.
- Use the steps: mass → moles → ratio → moles → mass.
- The limiting reactant is used up first and controls the product amount.
- Show every step clearly for method marks.