Li metal is one of the few substances that reacts directly with molecular nitrogen. The balanced equation for reaction is :

$6\mathrm{Li}\left(\mathrm{s}\right)+{\mathrm{N}}_2( \mathrm{g})⟶2{\mathrm{Li}}_3 \mathrm{N}( \mathrm{s})$

How many grams of the product, lithium nitride, can be prepared from 3.5 g of lithium metal and 8.4 g of molecular nitrogen ?

Moles present in 3.5 g of Lithium is:

$\mathrm{Moles} \mathrm{of} \mathrm{Li}=\frac{3.5}{7}=0.5 \mathrm{mol}$

Moles present in 8.4 g of molecular nitrogen is:

$\mathrm{Moles} \mathrm{of} {\mathrm{N}}_2=\frac{8.4}{28}=0.3 \mathrm{mol}$

According to the given reaction, one mole of N₂ requires 6 moles of Li and 0.3 moles of N₂ requires 1.8 moles of Li but available moles of Li is 0.5.

Hence, Li is the limiting reagent and decides the quantity of the product forms in the reaction.

6 moles of lithium produces 2 moles of Li₃N. Moles of Li₃N produces from 0.5 moles of Li is:

$\mathrm{Moles} \mathrm{of} {\mathrm{Li}}_3\mathrm{N}=\frac{2}{6}\times 0.5=\frac{1}{6} \mathrm{mol}$

Mass of Li₃N is:

$\mathrm{Mass}=\mathrm{Molar} \mathrm{Mass}\times \mathrm{Moles}=\frac{1}{6}\times 35=5.83 \mathrm{g}$