Reacting masses, limiting reactants and balancing from masses

Higher tier: use moles to work out how much product a reaction makes, which reactant runs out first, and even the balanced equation itself.

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Moles and reacting masses

moles = mass ÷ Mr (or Ar for atoms). One mole contains 6.02 × 1023 particles (the Avogadro constant).
To find a mass from an equation:
1. Find the moles of the substance you know.
2. Use the ratio in the balanced equation.
3. Convert moles back to mass.
Example: 2Mg + O2 → 2MgO. 6 g of Mg is 6 ÷ 24 = 0.25 mol, so 0.25 mol of MgO forms: 0.25 × 40 = 10 g.

The limiting reactant

The reactant that is used up first is the limiting reactant. It decides how much product forms; the other reactant is in excess.
Example: 2H2 + O2 → 2H2O with 4 g H2 (2 mol) and 16 g O2 (0.5 mol).
2 mol H2 would need 1 mol O2, but only 0.5 mol is there, so O2 is limiting.
0.5 mol O2 makes 1 mol H2O = 18 g.
If you double the amount of the limiting reactant, you double the amount of product.

Balancing an equation from masses

1. Change each mass into moles.
2. Divide by the smallest number of moles to get a whole-number ratio.
3. Use the ratio as the numbers in front of each formula.
Example: 4.8 g Mg reacts with 3.2 g O2 to make 8.0 g MgO.
Mg: 4.8 ÷ 24 = 0.2; O2: 3.2 ÷ 32 = 0.1; MgO: 8.0 ÷ 40 = 0.2. Ratio 2 : 1 : 2, so 2Mg + O2 → 2MgO.
Mass is conserved: 4.8 + 3.2 = 8.0 g.
Worked example

What mass of calcium oxide forms when 50 g of calcium carbonate is heated? CaCO3 → CaO + CO2 (Mr: CaCO3 = 100, CaO = 56)

  1. Moles of CaCO3 = 50 ÷ 100 = 0.5 mol
  2. Ratio 1 : 1, so 0.5 mol CaO
  3. Mass = 0.5 × 56 = 28 g

Answer: 28 g

Key idea

moles = mass ÷ Mr. Use the equation ratio to go from one substance to another. The limiting reactant runs out first and fixes the amount of product. Turn masses into moles and simplify the ratio to balance an equation.

Key words
Reacting masses
1. Find moles of the known substance. 2. Use the balanced equation ratio to find moles of the other. 3. mass = moles × M_{r}.
Mole
The amount of substance containing 6.02 × 10^{23} particles. Its mass in grams equals the A_{r} or M_{r}.
Avogadro constant
The number of particles in one mole of a substance: 6.02 × 10^{23}.

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  1. What mass of magnesium oxide forms from 6 g of magnesium? 2Mg + O2 → 2MgO (Ar: Mg 24, O 16)

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    Answer: 10 g

    Moles of magnesium = 6 ÷ 24 = 0.25. Ratio 2 : 2, so moles of MgO = 0.25. Mass = 0.25 × 40 = 10 g

  2. How many moles of carbon dioxide (Mr 44) are in 66 g?

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    Answer: 1.5 mol

    moles = 66 ÷ 44 = 1.5 mol

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Sources and references

Checked by the GCSE-ready editorial team, 16 October 2026. Last updated 2 October 2026. Spotted a mistake? Tell us and we will fix it.

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