The `%` composition of `NH_(3), H_(2)O` and `N_(2)O_(3)` is as given below:
`NH_(3) rarr 82.35% N` and `17.65%H`
`H_(2)O rarr 88.90% ` and `11.10% H`
`N_(2)O_(3) rarr 63.15%O` and `36.85%N`
On the basis of above data prove law of reciprocal proportions.

Explanation

(i) For `NH_(3)`:
`1` part hydrogen reacts with `=(82.35)/(17.65)= 4/67` Part `N`
(ii) For `H_(2)O`
`1` part hydrogen reacts with `=(88.9)/(11.10)= 8.01` part `O`
Thus the ratio `N:O:4.67:8.01implies0.58`
(iii) For `N_(2)O_(3)`: In which `N` and `O` reacts with each other and ratio `N:O::36.85:63.15~~0.58`
beacuse the two ratio are saem, thus law of reciprocal proportions is correct.