Chemistry - Ideal Gas Question with Solution | TestHub

$n$($N_2$) = 5.6/28 = 0.20 mol. $n$($H_2$) = 1.5/2 = 0.75 mol.

$N_2$ + $3H_2$ $\to$ $2N{H}_3$. $N_2$ is limiting: 0.20 mol $N_2$ uses 0.60 mol $H_2$ and forms 0.40 mol $N{H}_3$. $H_2$ left = 0.15 mol.

Total moles after reaction = 0.40 + 0.15 = 0.55.

$P$($N{H}_3$) = 1100 x 0.40/0.55 = 800 mm.

Answer: (B).