If `5g C_(2)H_(5)|` reacts with `Na` (metallic) in presence of ether, and the yield is `60%` then how many grams of n-butane will you get

Explanation

`2C_(3)H_(5)|_2NararrC_(4)H_(10)+2NaI`
Molecular weight of `2C_(2)H_(5)I=24+5+127=156`
Molecular weight of `C_(4)H_(10)=48+10=58`
Two molecule of `C_(2)H_(5)I` are taking part in above reaction.
`therefore` We get `58g` of `C_(4)H_(10)` from `2xx156g` of `C_(2)H_(5)I`
`therefore` We get `(58)/(2xx156)g C_(4)H_(10)` from 1g` of `C_(2)H_(5)I`
`therefore` We get `(58xx5)/(2xx156)xxg C_(4)H_(10)` from `5g` of `C_(2)H_(5)`
yield is `60%`
So the quantity of `C_(4)H_(10)` will be `(58xx5)/(2xx156)xx(60)/(100)g=0.55g`