Take following sequential reactions :
$2 H_{2} O (l) ⟶ 2 H_{2} (g) + O_{2} (g)$
$N_{2} (g) + 3 H_{2} (g) ⟶ 2 N H_{3} (g)$
$C a C O_{3} (s) ⟶ C a O (s) + C O_{2} (g)$
$N H_{3} (g) + C O_{2} (g) + H_{2} O (l) ⟶ N H_{4} H C O_{3} (a q)$
We want to produce $0.10$ mol $N H_{4} H C O_{3}$ . $H_{2} O (l)$ required is $\frac{y}{100}$ mol.
What is $y$ ?

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Solution

The reactions are as follows:
(a) $2 H_{2} O (l) ⟶ 2 H_{2} (g) + O_{2} (g)] \times 0.75$
(b) $N_{2} (g) + 3 H_{2} (g) ⟶ 2 N H_{3} (g)] \times 0.50$
(c) $C a C O_{3} (s) ⟶ C a O (s) + C O_{2} (g)]$
(d) $N H_{3} (g) + C O_{2} (g) + H_{2} O (l) ⟶ N H_{4} H C O_{3} (a q)$
$0.75 \times$ (a) $+ 0.50 \times$ (b) gives,
$0.5 N_{2} (g) + 15 H_{2} O (l) ⟶ N H_{3} (g) + 0.75 O_{2} (g)$
Thus, $N_{2} (g) r e q u i r e d = 0.05 m o l f o r 0.1 m o l N H_{4} H C O_{3}$ .
$H_{2} O (l) r e q u i r e d = 0.15 m o l i n (a) a n d (b) + 0.10 m o l i n f i n a l r e a c t i o n = 0.25 m o l$ .
So, the value of $y$ is $25$ .

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