n-butane is produced by the monobromination of ethane by Wurtz reaction. The amount of volume of ethane required at NTP to produce 55 g of butane, if the bromination takes place with 90% yield and the Wurtz reaction with 85% yield, is __________(as the nearest integer).

Open in App

Solution

Given reactions are
(i) $2 C_{2} H_{6} + 2 {B r}_{2} \to 2 C_{2} H_{5} B r + 2 H B r$
(ii) $2 C_{2} H_{5} B r + 2 N a \to C_{4} H_{10} + 2 N a B r$
Reactions (i) and (ii) reveal mole ratio
$C_{2} H_{6} : C_{2} H_{5} B r : C_{4} H_{10} :: 2 : 2 : 1$
Now mole of butane formed $=$ 55/58
Mole of $C_{2} H_{5} B r$ actually required $=$ $\frac{2 \times 55}{58} \times \frac{100}{85}$
( $∵$ reaction efficiency is 85%)
Now similarly mole of $C_{2} H_{6}$ actually required
$=$ $\frac{2 \times 55}{58} \times \frac{100}{85} \times \frac{100}{90} = 2.479 m o l$
(Since, this time effciency is 90%)
Therefore, volume of $C_{2} H_{6}$ at NTP actually required
$22.4 \times 2.479 l i t r e = 55.53 l i t r e$
So, answer is 56.

Suggest Corrections

Similar questions

55

90 %

85 %

55 g

90

85

(C_{4} H_{10})

C_{2} H_{6}

55

90 %

85 %

(C_{4} H_{10})

C_{2} H_{6}

55

90 %

85 %

Join BYJU'S Learning Program