Question #bfc24

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Question #bfc24

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Answer 1 · 2016-12-08T03:17:09

$N_{2} (g) + 3 H_{2} (g) \to N H_{3} (l)$ $N_2(g)+3H_2(g)->NH_3(l)$

Stochiometric ratio of the above raction suggests that 1mole (28g) $N_{2} (g)$ $N_2(g)$ reacts with 3moles $(3 \times 2 = 6 g) H_{2} (g)$ $(3xx2=6g)" "H_2(g)$ to produce 2moles $(2 \times 17 = 34 g) N H_{3} (l)$ $(2xx17=34g)" "NH_3(l)$

50g of each reactant is taken . So it is obvious that $N_{2}$ $N_2$ will be limilting reactant , hence will be consumed completely to give the product.

By the balanced equation,28g $N_{2} (g)$ $N_2(g)$ reacts with 6g $H_{2} (g)$ $H_2(g)$ to produce 34g $N H_{3} (l)$ $NH_3(l)$

So 50 g $N_{2} (g)$ $N_2(g)$ reacts completely with $\frac{6}{28} \times 50 g \approx 10.71 g H_{2} (g)$ $6/28xx50g~~ 10.71g" "H_2(g)$ leaving $(50 - 10.71) = 39.29 g H_{2} (g)$ $(50-10.71)=39.29g" "H_2(g)$

Now we are to calculate the volume (V) of the remaining $w = 39.29 g H_{2} (g)$ $w=39.29g" "H_2(g)$ at pressure
$P = 760 m m = 1 a t m$ $P=760mm=1atm$ and temperature $T = 200 K$ $T=200K$

Now by equation of ideal gas we have

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

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

So
$V = \frac{w}{M} \times \frac{R T}{P}$ $V=w/Mxx(RT)/P$

$\Rightarrow V = \frac{39.29}{2} \times \frac{0.082 \times 200}{1} L = 322.178 L$ $=>V=39.29/2xx(0.082xx200)/1L=322.178L$

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