Limiting reactants - (higher tier)

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 - it sets a limit on how much product can form.
  • 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.


How much aluminium chloride is formed when 2.7 g of aluminium are reacted with 9.125 g of HCl?

What mass of aluminium chloride is formed when 2.7g of aluminium are reacted with 9.125g of HCl?

2Al + 6HCl → 2AlCl3 + 3H2

Convert mass into moles for both reactants:

\[{moles~of~Al} = \frac{mass~(g)}{M_r} = \frac{2.7}{27} = 0.1\]

\[{moles~of~HCl} = \frac{mass~(g)}{M_r} = \frac{9.125}{36.5} = 0.25\]

Now look at the ratio of the reactants:

2 mol Al : 6 mol HCl (1 mol Al : 3 mol HCl)

0.1 mol Al : 0.3 mole HCl

There are only 0.25 moles of HCl (instead of 0.3 moles), so the HCl will run out first. It is the limiting reactant.

Now use the moles of the limiting reactant to calculate the mass of the product. Remember to use the molar ratio between the limiting reactant and the product.

Moles of HCl = 0.25

6 mol HCl : 2 mol AlCl3 which simplifies to

3 mol HCl : 1 mol AlCl3


0.25 mol HCl will make 0.0833 mol AlCl3.

Mass of AlCl3 = moles x Mr = 0.0833 x 133.5 = 11.12 g