`40%` of a mixture of `2.0` mol of `N_(2)` and `0.6` mol of `H_(2)` reacts to give `NH_(3)` according to the equation:
`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)`
at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are

Options

A. `4:5`
B. `5:4`
C. `7:10`
D. `8:5`

Answer

A