Reactions and moles - Higher

Limiting reactants

A reaction finishes when one of the reactants is all used up. The other reactant has nothing left to react with, so some of it is left over:

  • the reactant that is all used up is called the limiting reactant
  • the reactant that is left over is described as being in excess

The mass of product formed in a reaction depends upon the mass of the limiting reactant. This is because no more product can form when the limiting reactant is all used up.

Reacting mass calculations

The maximum mass of product formed in a reaction can be calculated using:


12 g of magnesium reacts completely with excess hydrochloric acid to form magnesium chloride and hydrogen:

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Calculate the maximum mass of hydrogen that can be produced. (Ar of Mg = 24, Mr of H2 = 2)

Amount of magnesium = \frac{mass}{relative~atomic~mass}

Amount of magnesium = \frac{12}{24}

= 0.5 mol

Looking at the equation, 1 mol of Mg forms 1 mol of H2, so 0.5 mol of Mg forms 0.5 mol of H2

Mass of H2 = Mr × amount

= 2 × 0.5

= 1 g


1.0 g of calcium carbonate decomposes to form calcium oxide and carbon dioxide:

CaCO3(g) → CaO(s) + CO2(g)

Calculate the maximum mass of carbon dioxide that can be produced. (Mr of CaCO3 = 100, Mr of CO2 = 44)

Amount of calcium carbonate = \frac{1.0}{100}

= 0.01 mol

Looking at the equation, 1 mol of CaCO3 forms 1 mol of CO2, so 0.01 mol of CaCO3 forms 0.01 mol of CO2

Mass of CO2 = relative formula mass × amount

= 44 × 0.01

= 0.44 g

Calculating balancing numbers

The balancing numbers in an equation can be worked out using masses found by experiment.


6.0 g of magnesium reacts with 4.0 g oxygen to produce magnesium oxide, MgO. Deduce the balanced equation for the reaction. (Ar of Mg = 24, Mr of O2 = 32)

1Write the formulae of the substancesMgO2
2Calculate the amounts \frac{6.0}{24} = 0.25 mol \frac{4.0}{32} = 0.125 mol
3Divide both by the smaller amount \frac{0.25}{0.125} = 2 \frac{0.125}{0.125} = 1

This means that 2 mol of Mg reacts with 1 mol of O2, so the left-hand side of the equation is:

2Mg + O2

Then balancing in the normal way gives: 2Mg + O2 → 2MgO