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40% of a mixture of 0.2 mol of N(2) and ...

40% of a mixture of 0.2 mol of `N_(2)` and 0.6 mol of `H_(2)` react to give `NH_(3)` according to the equation : `N_(2)(g) + 3H_(2)(g) rarr 2NH_(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)hArr2NH_(3)(g)`

Also, `0.4=x/0.2 rArr x=0.08`
Ratio`=V_(f)/V_(i)=((n_(Total))_(f))/((n_(Total))_(i))=(0.8-2x)/(0.8)`
`=1-x/0.4`
`=1-0.08/0.40=4/5`
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