Some Basic Concepts of Chemistry - Result Question 53
52. $n$-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.
(1989, 3M)
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Answer:
Correct Answer: 52. (55.55L)
Solution:
- Reactions involved are
$\begin{aligned} C _2 H _6+Br _2 & \longrightarrow C _2 H _5 Br+HBr \\ 2 C _2 H _5 Br+2 Na & \longrightarrow C _4 H _{10}+2 NaBr \end{aligned}$
Actual yield of $C _4 H _{10}=55 g$ which is $85 %$ of theoretical yield.
$\Rightarrow$ Theoretical yield of $C _4 H _{10}=\frac{55 \times 100}{85}=64.70 g$
Also, 2 moles ( $218 g$ ) $C _2 H _5 Br$ gives $58 g$ of butane.
$\Rightarrow \quad 64.70 g$ of butane would be obtained from
$\frac{2}{58} \times 64.70=2.23 \text { moles } C _2 H _5 Br$
Also yield of bromination reaction is only $90 %$, in order to have 2.23 moles of $C _2 H _5 Br$, theoretically
$\frac{2.23 \times 100}{90}=2.48 \text { moles of } C _2 H _5 Br \text { required. }$
Therefore, moles of $C _2 H _6$ required $=2.48$
$\Rightarrow$ Volume of $C _2 H _6$ (NTP) required $=2.48 \times 22.4=55.55 L$.
BETA
