What volume of ${\mathrm{H}}_2$ at $\mathrm{NTP}$ is required to convert $2.8 \mathrm{g}$ of ${\mathrm{N}}_2$ in to ${\mathrm{NH}}_3$ ?

$2.8 \mathrm{g}$ of ${\mathrm{N}}_2$ in terms of moles $=\frac{2.8}{28}=0.1 \mathrm{mol}$

          ${\mathrm{N}}_2+3{\mathrm{H}}_2\to 2{\mathrm{NH}}_3$

1 mole of ${\mathrm{N}}_2$ reacts with 3 moles of ${\mathrm{H}}_2$.

$0.1$ mole of ${\mathrm{N}}_2$ will react with $=3\times 0.1=0.3$ mole of ${\mathrm{H}}_2$

The volume of ${\mathrm{H}}_2$ required $=22.4\times 0.3=6.72 \mathrm{L}$