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

Answer: (B)

$55.5$ L

The molar mass of n butane is $58.12g/mol.$
$55$ g of n butane corresponds to $\frac{55}{58.12}=0.946mol$ of n butane.
They will be produced by $2\times 0.946=1.89$ mole of ethane which will occupy a volume of $1.89\times 22.4=42.4L$ at STP.
These calculatios are assuming $100$% yield.
However, actual yields are $90$% and $85$%.
Hence, the volume of ethane required at $STP$ will be $\frac{42.4\times 100\times 100}{90\times 85}=55.5L$.