Question

$n$-butane is produced by monobromination of ethane followed by Wurtz's reaction. Calculate volume of ethane at NTP required to produce $55 \,g\, n$-butane, if the bromination takes place with $90\%$ yield and the Wurtz's reaction with $85\%$ yield.

Answer 1

Reactions involved are
$ C_2H_6 +Br_2 \rightarrow C_2H_5Br + HBr$
$ 2C_2H_5Br + 2Na \rightarrow C_4H_{10} + 2NaBr $
Actual yield of $C_4H_{10} = 55\, g$ which is $85\%$ of theoretical yield.
$\Rightarrow $ Theoretical yield of $C_4H_{10} = \frac{ 55 \times 100}{85 } = 64 .70 \,g $
Also, $2$ moles $(218\, g) C_2H_5Br$ gives $58 \,g$ of butane.
$\Rightarrow 64.70 \,g$ of butane would be obtained from
$ \frac{2}{58} \times 64.70 = 2.23 $ mole $ C_2H_5 Br $
Also yield of bromination reaction is only $90\%$, in order to have $2.23$ moles of $C_2H_5Br$ , theoretically
$ \frac{2.23 \times 100}{90} = 2.48 $ moles of $C_2H_5Br$ required.
Therefore, moles of $C_2H_6$ required $= 2.48$
$\Rightarrow $ Volume of $C_2H_6 (NTP)$ required $= 2.48 \times 22.4 = 55.55 \,L$.