# Question bfc24

Dec 8, 2016

${N}_{2} \left(g\right) + 3 {H}_{2} \left(g\right) \to N {H}_{3} \left(l\right)$

Stochiometric ratio of the above raction suggests that 1mole (28g) ${N}_{2} \left(g\right)$ reacts with 3moles $\left(3 \times 2 = 6 g\right) \text{ } {H}_{2} \left(g\right)$ to produce 2moles $\left(2 \times 17 = 34 g\right) \text{ } N {H}_{3} \left(l\right)$

50g of each reactant is taken . So it is obvious that ${N}_{2}$ will be limilting reactant , hence will be consumed completely to give the product.

By the balanced equation,28g ${N}_{2} \left(g\right)$ reacts with 6g ${H}_{2} \left(g\right)$ to produce 34g $N {H}_{3} \left(l\right)$

So 50 g ${N}_{2} \left(g\right)$ reacts completely with 6/28xx50g~~ 10.71g" "H_2(g)# leaving $\left(50 - 10.71\right) = 39.29 g \text{ } {H}_{2} \left(g\right)$

Now we are to calculate the volume (V) of the remaining $w = 39.29 g \text{ } {H}_{2} \left(g\right)$ at pressure
$P = 760 m m = 1 a t m$ and temperature $T = 200 K$

Now by equation of ideal gas we have

$P V = \frac{w}{M} R T$

here $M = 2 \frac{g}{\text{mol}}$
$R = 0.082 L a t m {J}^{-} 1 m o {l}^{-} 1$

So
$V = \frac{w}{M} \times \frac{R T}{P}$

$\implies V = \frac{39.29}{2} \times \frac{0.082 \times 200}{1} L = 322.178 L$