Some Basic Concepts of Chemistry 1 Question 32
52. $n$-butane is produced by monobromination of ethane followed by Wurtz’s reaction.Calculate volume of ethane at NTP required to produce $55 \mathrm{~g} n$-butane, if the bromination takes place with $90 %$ yield and the Wurtz’s reaction with $85 %$ yield.
(1989, 3M)
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Solution:
- Reactions involved are
$$ \begin{aligned} \mathrm{C}{2} \mathrm{H}{6}+\mathrm{Br}{2} & \longrightarrow \mathrm{C}{2} \mathrm{H}{5} \mathrm{Br}+\mathrm{HBr} \ 2 \mathrm{C}{2} \mathrm{H}{5} \mathrm{Br}+2 \mathrm{Na} & \longrightarrow \mathrm{C}{4} \mathrm{H}_{10}+2 \mathrm{NaBr} \end{aligned} $$
Actual yield of $\mathrm{C}{4} \mathrm{H}{10}=55 \mathrm{~g}$ which is $85 %$ of theoretical yield.
$\Rightarrow$ Theoretical yield of $\mathrm{C}{4} \mathrm{H}{10}=\frac{55 \times 100}{85}=64.70 \mathrm{~g}$
Also, 2 moles ( $218 \mathrm{~g}$ ) $\mathrm{C}{2} \mathrm{H}{5} \mathrm{Br}$ gives $58 \mathrm{~g}$ of butane.
$\Rightarrow \quad 64.70 \mathrm{~g}$ of butane would be obtained from
$$ \frac{2}{58} \times 64.70=2.23 \text { moles } \mathrm{C}{2} \mathrm{H}{5} \mathrm{Br} $$
Also yield of bromination reaction is only $90 %$, in order to have 2.23 moles of $\mathrm{C}{2} \mathrm{H}{5} \mathrm{Br}$, theoretically
$$ \frac{2.23 \times 100}{90}=2.48 \text { moles of } \mathrm{C}{2} \mathrm{H}{5} \mathrm{Br} \text { required. } $$
Therefore, moles of $\mathrm{C}{2} \mathrm{H}{6}$ required $=2.48$
$\Rightarrow$ Volume of $\mathrm{C}{2} \mathrm{H}{6}$ (NTP) required $=2.48 \times 22.4=55.55 \mathrm{~L}$.
