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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

A

`4:5`

B

`5:4`

C

`7:10`

D

`8:5`

Text Solution

Verified by Experts

The correct Answer is:
A
`N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g)`
Also : `0.4 = x/0.2 implies x = 0.08`
Ratio = `V_(f)/V_(i) = (n_(t otal))_(f)/(n_(t otal))_(i) = 0.8-2x/0.8 = 1 - x/0.4 = 1 - 0.80/0.40 = 4/5`
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