When reactants are not supplied in exactly their stoichiometric proportions, one reactant is consumed first. That limiting reactant sets the maximum amount of product that can form. Other reactants present beyond the required amount are excess reactants.

Consider

$$\mathrm{N_2+3H_2\rightarrow2NH_3}.$$

Suppose $2.0,\mathrm{mol}$ of $N_2$ is mixed with $4.0,\mathrm{mol}$ of $H_2$.

Reacting all $2.0,\mathrm{mol}$ of $N_2$ would require

$$2.0,\mathrm{mol\ N_2}\times \frac{3,\mathrm{mol\ H_2}}{1,\mathrm{mol\ N_2}} =6.0,\mathrm{mol\ H_2},$$

but only $4.0,\mathrm{mol}$ is available. Hydrogen is therefore limiting.

The product amount follows from the limiting reactant:

$$4.0,\mathrm{mol\ H_2}\times \frac{2,\mathrm{mol\ NH_3}}{3,\mathrm{mol\ H_2}} =2.67,\mathrm{mol\ NH_3}.$$

The nitrogen consumed is

$$4.0\times\frac{1}{3}=1.33,\mathrm{mol\ N_2},$$

so about

$$2.0-1.33=0.67,\mathrm{mol\ N_2}$$

remains in excess.

An equivalent method is to calculate how much of the same product each reactant could produce independently. The reactant predicting the smaller product amount is limiting.

The limiting reactant is determined by amount relative to stoichiometric demand, not simply by which reactant has fewer grams or fewer moles.