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40% of a mixture of 2.0 mol of N(2) and ...

`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

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40% of a mixture of 0.2 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

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