Question

Under S.T.P. 1 mol of $N_2$ and 3 mol of $H_2$ will form on complete reaction

• A. 4 moles of $NH_3$
• B. 89.6 L of $NH_3$
• C. 22.4 L of $NH_3$
• D. 44.8 L of $NH_3$

Answer 1

Answer: (D)

44.8 L of $NH_3$

Answer 2

$\underset{\text{1 mol}}{ {N_{2}}}+\underset{\text{3 mol}}{ {3H_{2}}}\to \underset{\text{2 mol}}{ {2NH_{3}}}$ $1$ mol of $N_{2}$ reacts with $3$ moles of $H_{2}$ to give $2$ moles of $NH_{3}$ which occupies a volume $=2\times 22.4=44.8\,L$ at $S.T.P.$