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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+ 3H_2 hArr 2NH_3`
t=0 0.2 0.6 0
at t=t 0.2-n 0.6 -3x 2x
40% of `N_2 =0.2xx0.4 =0.08`
40% of `H_2=0.6 xx0.4 =0.24`
`therefore` Number of moles of `N_2` reamaining =0.2 -0.08 =0.12
Number of moles of `H_2` remaining =0.6 -0.24 =0.36
Number of moles of `NH_3` formed =0.16
Total number of moles =0.12+0.36 +0.16 =0.64
`therefore ("Final volume")/("Initial volume")=("Final moles")/("Initial moles")=(0.64)/(0.80)=4/5`
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