$2\; mole \;\mathrm{NH}_{3}(\mathrm{g})$ requires $3\; mole \;\mathrm{H}_{2}(\mathrm{g})$
$20 \;mole\; \mathrm{NH}_{3}(\mathrm{g})$ requires
$=\frac{3}{2} \times 20 \text { mole } \mathrm{H}_{2}(\mathrm{g})$
$=30\;mole$