Fifty millilitre of a mixture of `NH_3` and `H_2` was completely decomposed into `N_2` and `H_2` by sparking. Forty millilitre of `O_2` was then added and the mixture was sparked again. After cooling, the mixture was shaken with alkaline pyrogallol and a contraction of 6 ml was observed. Calculate the percentage of `NH_3` in the original mixture.

Explanation

`underset(x)(NH_3)+underset(y)(H_2)to(1)/(2)underset((x)/(2))(N_2)+(3)/(2)underset((3)/(2)x)(H_2)+underset(ymL)(H_2)`
`x+y=50mL`
`underset((3)/(2)x)((3)/(2)H_2)+underset((3)/(4)x)((3)/(4)O_2)to(3)/(2)H_2O`
`underset(ymL)(H_2)+underset((y)/(2)mL)(1)/(2)O_2tounderset(ymL)(H_2O)`.
[After shaki ng with pyrogallol solution, unreacted
`O_2` was adsobed and 6 mL contraction was observed
Initial volume of `O_2=40ml`
unreacted `O_2=6mL`, Reacted `O_2=40-6=34mL`]
From equation (ii) and (iii), `O_2` used
`(3)/(4)x+(y)/(2)=34implies3x+2y=34xx4`
Soving equation (i) and (iv) we get
`{:(3x+2y=136),(x+y=50):}]impliesy=14,x=36`
Percentage of `NH_3=(36)/(50)=100=72%`
Total volume of organic compound `+O_2=95mL`
Volume of compound `=20mL`
Volume of `O_2` used `=95-20=75mL`