5.9 - Limiting Reactants
Excess and the limiting reactant
When two reactants react, the reaction stops once one of them is completely used up.
Adding calcium carbonate to dilute hydrochloric acid makes the mixture fizz, because carbon dioxide is given off:
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
The fizzing stops when one of the two reactants has been used up completely, however much of the other is left in the beaker.
Key terms
- Limiting reactant - The reactant completely used up. It limits how much product can be made.
- In excess - Any reactant still left over when the reaction stops.
Product depends on the limiting reactant
The amount of product formed is directly proportional to the amount of the limiting reactant.
- Double the limiting reactant → twice as much product.
- Halve it → half as much product.
Why a reactant is often added in excess
Using a surplus of the cheaper or more easily supplied reactant makes sure the other reactant is used up completely, so none of the expensive one is left unreacted.
Finding which reactant is limiting
When two reactants are mixed, one runs out first unless their amounts match the ratio in the balanced equation.
The four steps
- Work out the amount in moles of each reactant:
- Read the balancing numbers as the ratio in which the two reactants react.
- Work out how much of one reactant is needed to react with all of the other.
- Compare that with the amount actually present:
- If there is more present than needed, it's in excess.
- If there is less present than needed, it runs out first and is the limiting reactant.
Calculating product formed
Once the limiting reactant is known, only its amount is used to calculate the mass or amount of product formed.
Worked example - Which reactant limits
A mixture contains 135 g of aluminium and 480 g of iron(III) oxide, which react together:
2Al + Fe2O3 → 2Fe + Al2O3
Which reactant is the limiting reactant? (Relative atomic masses: O = 16, Al = 27, Fe = 56.)
Step 1 - Work out the amount in moles of each reactant
Aluminium is an element, so its relative formula mass is its relative atomic mass, 27. For iron(III) oxide:
Dividing each mass by its relative formula mass:
Step 2 - Read the balancing numbers as the ratio
The equation consumes 2 moles of aluminium for every 1 mole of iron(III) oxide.
Step 3 - Work out how much aluminium would be needed
Using up all 3.00 mol of the oxide needs:
2 × 3.00 = 6.00 mol of aluminium
Step 4 - Compare with the amount actually there
There's 5.00 mol of aluminium against the 6.00 mol needed, so there isn't enough aluminium to use up all the iron(III) oxide. The aluminium runs out first.
The aluminium is the limiting reactant, and the iron(III) oxide is in excess.
Step 5 - Interpretation
There was more aluminium than iron(III) oxide in moles (5.00 mol against 3.00 mol), yet the aluminium still ran out first - because the equation takes aluminium two moles at a time and iron(III) oxide one. Neither the masses nor the amounts settle the question on their own.