`n`-butane is produced by the monobromination of ethane followed by Wurtz reaction. Calculate the volume of ethane at `NTP` to produce `55 g` n-butane if the bromination takes place with `90%` yield and the Wurtz reaction with `85%` yield.

Explanation

The reaction involved are as follows
`underset(2xx30=60g)(2CH_(3)CH_(3))+Br_(2)tounderset(2xx(29+80)=218g)(2CH_(3)CH_(2)Br)`
`2CH_(3)CH_(2)Br+2Natounderset(58g)(CH_(3)CH_(2)CH_(2)CH_(3))+2NaBr`
Amount of bromoethane actually formed in the reaction `=((218g)xx90)/(100)=196*2g`
calculation of the amount of n-butane actually formed
from the available data
218 g of bromoethane are expected to form n-butane =58g
`196*2g` g of bromoethane are expected to form n-butane `=((58g))/((218g))xx(196*2g)=52*2g`
But since the yield of n-butane is only 85 percent therefore, the amount of n-butane actually formed in the wurtz reaction `=((52*2g)xx85)/(100)=44*37g`
Calculation of volume of ethane required at NTP
58g of n-butane are produced from ethane `=((60g))/((58g))xx(44*37g)=45*9g`
Now, 30 g of ethane at N.T.P. correspond to `22*4L`
`45*9g` of ethane at N.T.P. correspond to `=(22*4Lxx(45.9g))/((30g))=34*27L`