Some Basic Concepts of Chemistry - Result Question 53

Answer:

Correct Answer: 52. (55.55L)

Solution:

52. 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 $=2.48 \times 22.4=55.55 L$.