At STP, the volume of ammonia formed using 1 mole of nitrogen and 4 moles of hydrogen on complete reaction is:

The correct option is B 44.8 L
The balanced reaction is:
$N_2+3H_2\to 2N{H}_3$

Thus, one mole of nitrogen reacts with 3 moles of hydrogen.
We are provided with 1 mole of nitrogen and 4 moles of hydrogen.
Hence, nitrogen is the limiting agent.

Again, 1 mole of nitrogen gives 2 moles of ammonia.
Volume of ammonia formed at STP = $2\times 22.4$ L= 44.8 L