Limiting reactant calculations

Reactants combine in the mole ratio shown by a balanced equation. If one reactant runs out before the other, the reaction cannot make any more product. The reactant that runs out first is the limiting reactant; any reactant left over is in excess.

To identify the limiting reactant, compare the amounts available with the required mole ratio. The reactant with fewer moles is not necessarily limiting: it may also be needed in a smaller proportion.

Comparing reactant amounts

If reactant masses are given, first convert each mass to moles:

Use the balanced equation to compare the mole amounts. One method is to divide the amount of each reactant by its coefficient in the equation. The smaller result identifies the limiting reactant: it can support fewer complete sets of the reaction ratio.

Once the limiting reactant is known, use its amount and the mole ratio in the equation to calculate the maximum amount of product. If a product mass is required, multiply the product amount in moles by its molar mass.

Limiting reactant calculation: forming water

Hydrogen reacts with oxygen according to the balanced equation:

Suppose $$4,\mathrm{g}$$ of hydrogen reacts with $$16,\mathrm{g}$$ of oxygen. The molar masses of $$\ce{H2}$$ and $$\ce{O2}$$ are $$2,\mathrm{g,mol^{-1}}$$ and $$32,\mathrm{g,mol^{-1}}$$, respectively. The available amounts are therefore $$2,\mathrm{mol}$$ of $$\ce{H2}$$ and $$0.5,\mathrm{mol}$$ of $$\ce{O2}$$.

The equation requires $$2,\mathrm{mol}$$ of $$\ce{H2}$$ for every $$1,\mathrm{mol}$$ of $$\ce{O2}$$. Dividing each amount by its coefficient gives $$2/2 = 1$$ for hydrogen and $$0.5/1 = 0.5$$ for oxygen. Oxygen gives the smaller result, so oxygen is limiting and hydrogen is in excess.

The equation shows that $$1,\mathrm{mol}$$ of $$\ce{O2}$$ makes $$2,\mathrm{mol}$$ of $$\ce{H2O}$$. Therefore, $$0.5,\mathrm{mol}$$ of $$\ce{O2}$$ can make at most $$1.0,\mathrm{mol}$$ of $$\ce{H2O}$$. Using the molar mass of water, $$18,\mathrm{g,mol^{-1}}$$, this is $$18,\mathrm{g}$$ of water. The excess hydrogen cannot produce more water once the oxygen has run out.

Limiting and excess reactants

Reactants combine in the fixed mole ratio shown by a balanced chemical equation. If the amounts mixed do not match that ratio, one reactant is used up before the other.

The reactant that is completely used up first is the limiting reactant. Once it has been consumed, the reaction stops, so it determines the maximum amount of product that can form. The reactant left over is the excess reactant: more was supplied than was needed to react with the limiting reactant.

Using the mole ratio

Hydrogen and oxygen react to form water:

The equation shows that $$2\ \text{mol}$$ of hydrogen react with $$1\ \text{mol}$$ of oxygen. Suppose $$3\ \text{mol}$$ of hydrogen are mixed with $$2\ \text{mol}$$ of oxygen. The hydrogen needs only $$1.5\ \text{mol}$$ of oxygen to react completely.

Hydrogen is therefore the limiting reactant. Oxygen is in excess: $$0.5\ \text{mol}$$ remains after all the hydrogen has reacted. The equation also shows that $$3\ \text{mol}$$ of hydrogen can form $$3\ \text{mol}$$ of water; no more water can form through this reaction once the hydrogen is used up.

A reactant can be deliberately supplied in excess to ensure that the other reactant is fully used up. If reactants are supplied in exactly the ratio in the balanced equation and react completely, neither is left in excess.