`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.

Answer

4.879 gm

Explanation

(i) Monobromination of ethane
`underset("Ethane")(C_(2)H_(6)overset(Br_(2))to C_(2)H_(5)Br`
(ii) Wurtz reaction:
`C_(2)H_(5)Br + 2Na+BrC_(2)H_(5) underset("ether")overset("Dry")to C_(4)H_(10)`
Molecular weight of
`C_(4)H_(10)=(12xx4)+(10+1)=58`
`therefore` Amount of n -butane to be produced `=(55)/(58)` mol=0.948 mol
`therefore` Amount of `C_(2)H_(5)`Br required to obtain 0.948 mol
But the conversion is only 85%
Hence, the amount of `C_(2)H_(5)Br` required `=(1.896)/(85)xx100=2.23` mol
To obtain `C_(2)H_(5)Br` from `C_(2)H_(6)` , the same amount of `C_(2)H_(6)` would be required . But the percent conversion of `C_(2)H_(6)` to required `=(2.23)/(90)xx100=2.478` mol
Thus, required volume of ethane of NTP ` =22400xx2.478 =55507.2 ml=55.50 L`