Under S.T.P. 1 mol of N2 and 3 mol of H2 will form on complete reaction

Question

Under S.T.P. 1 mol of N2 and 3 mol of H2 will form on complete reaction (A) 4 moles of NH3 (B) 89.6 L of NH3 (C) 22.4 L of NH3 (D) 44.8 L of NH3

Tardigrade Admin · Accepted Answer

underset text1 molN2+ underset text3 mol3H2→ underset text2 mol2NH3 1 mol of N2 reacts with 3 moles of H2 to give 2 moles of NH3 which occupies a volume =2× 22.4=44.8 L at S.T.P.

Q. Under S.T.P. 1 mol of $N_{2}$ and $3$ mol of $H_{2}$ will form on complete reaction

4 moles of $N H_{3}$

89.6 L of $N H_{3}$

22.4 L of $N H_{3}$

44.8 L of $N H_{3}$

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

Q. Under S.T.P. 1 mol of N2​ and 3 mol of H2​ will form on complete reaction

4 moles of NH3​

89.6 L of NH3​

22.4 L of NH3​

44.8 L of NH3​