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

In `NH_(3),17.65g` of `H` combine with `N = 82.35g`
`:. 1 g` of `H` combine with `N = (82.35)/(17.65) g = 4.66g`
In `H_(2)O, 11.10g` of `H` combine with `O = 88.90 g`
`:. 1g` of `H` combine with `O = (88.90)/(11.10) g = 8.00g`
`:.` Ratio of the weights of `N` and `O` which combine with fixed
weight `(=1g) "of" H = 4.66 : 8.00 = 1 : 1.7`
In `N_(2)O_(3)`, ratio of weights of N and O which combine with each other `=36.85 : 63.15 = 1 : 1.7`
Thus the two ratio are the same. Hence it illustrates the law of reciprocal proportions .