Correct Answer - A
Let `n` mol of `(C_2H_4+H_2)` and `x` mol of `C_2H_4`
`H_2=(n-x)mol`
`underset(x)(C_2H_4)+underset(x)(H_2)rarrunderset(x mol)(C_2H_4)`
After reaction `(C_2H_6+H_2` left)
`x+n-x-x=n-x`
[Total `H_2=(n-x),H_2` reacted `=x]`
`H_2` left `=(n-x-x)`
`n=600,n-x=400`
`(n)/(n-x)=(600)/(400)`
`x=(n)/(3)` volume of `C_2H_4`
`=(1)/(3)` rd of total volume