Question

n-Butane, $\left(C_4{H}_{10}\right)$ is produced by monobromination of $C_2{H}_6$ followed by Wurtz reaction. Calculate the volume of ethane at STP required to produce $55$ g of n-butane.The bromination takes place with $90\%$ yield and the Wurtz reaction with $85\%$ yield.

• A. $27.75$ L
• B. $55.5$ L
• C. $111$ L
• D. $5.55$ L

Answer 1

The correct option is B $55.5$ L

The bromination followed by Wurtz reaction is given as:

$2C_2{H}_6+B{r}_2\to 2C_2{H}_{5}Br\underset{ether}{\overset{2Na}{\to }}{C}_4{H}_{10}$

12 × 4 + 10 = 58gm

58 gm of n-butane $\Rightarrow$ 2 × 22.4 litre of $C_2{H}_6$ at STP .

that is $2\times 22.4\times \frac{100}{85}\times \frac{100}{90}$ = 58.56 litre at STP

Volume of ethane required to produce 58 gm of n-butane = 58.56 litres

Therefore, the volume of ethane required to produce 55 gm of n-butane = 55.5 litres.

Hence, option B is correct.