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 in butane, if the bromination takes place with $90 %$ yield and the Wurtz reaction with $85 %$ yield is (as nearest integer) :

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Solution

The reactions are as follows:
(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 as:
$C_{2} H_{6} : C_{2} H_{5} B r : C_{4} H_{10} :: 2 : 2 : 1$
Now, moles of butane formed $= \frac{55}{58}$ .
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
= $\frac{2 \times 55}{58} \times \frac{100}{85} \times \frac{100}{90} = 2.479$ mol.
(Since, this time effciency is $90 %$ )
Therefore, volume of $C_{2} H_{6}$ at NTP actually required
$22.4 \times 2.479 = 55.53$ L.
So, the answer is $56$ .

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