What is the volume (in litres) of oxygen at STP required for complete combustion of $32$ g of $C{H}_4$?

[Molecular weight of $C{H}_4$ is $16$]

The correct option is B $89.6$
We have

$C{H}_4+2O_2\to C{O}_2+2H_{2}O$

Moles of $C{H}_4$ = $\frac{32}{16}=2mol$

For 2 mole of $C{H}_4$ we require 4 moles of $O_2$,

Volume of 1 mole of $O_2$ at $STP=22.4L$

Volume of 4 mole of $O_2$ at STP = $4\times 22.4=89.6L$

Option B is correct.