`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

`(55.55 L)`

Explanation

Reaction involved are
`C_(2)H_(6)+Br_(2) rarr C_(2)H_(5)Br+HBr`
`2C_(2)H_(5)Br+2Na rarr C_(4)H_(10)+2NaBr`
Actual yield of `C_(4)H_(10)=55 g` which is `85%` of throretical yield.
`implies` Theoretical yield of `C_(4)H_(10)=(55xx100)/(85)=64.70 g`
Also, `2` moles `(218 g) C_(2)H_(5)Br` gives `58 g` of butane.
`implies 64.70 g` of butane would be obtained from
`2/58 xx64.70=2.23` 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
`(2.23xx100)/90=2.48` moles of `C_(2)H_(5)Br` required.
Therefore, moles of `C_(2)H_(6)` required `=2.48`
`implies` Volume of `C_(2)H_(6)` (NTP) required `=2.48xx22.4=55.55 L`