11.2 litres each of  ${\mathrm{N}}_2$ and ${\mathrm{H}}_2$ react  to form Ammonia. Moles of Ammonia formed is

$\begin{array}{l}22.4\mathrm{lit} 3(22.4\mathrm{lit})2\mathrm{moles}\\ {\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NH}}_3\left(\mathrm{g}\right)\\ \frac{11.2}{3}\mathrm{lit}\text{ \hspace{0.17em}1}1.2\mathrm{lit} \text{'}\mathrm{X}\text{'}\mathrm{mole}\\ \end{array}$
Amount of Product formed is decided by limiting reagent
 ${\mathrm{N}}_2$……….Excess reagent;    ${\mathrm{H}}_2$     ………. Limiting reagent
$\mathrm{X}=\frac{11.2\times 2}{3\left(22.4\right)}=\frac{1}{3}$