60 mL of a mixture of nitrous oxide and nitric oxide was exploded with excess of hydrogen. If 38 mL of `N_(2)` was formed, calculate the volume of each gas in the mixture.

Explanation

Let the volume of NO in the mixture be x mL.
`{:(,NO,+,N_(2)O" "+,H_(2),rarr,N_(2),+,H_(2)O),(,"x mL",,"(60 - x)mL",,,"38 mL",,),(or,"x moles",,"(60 - x) moles",,,"38 moles",,):}`
Since N atoms are conserved, applying POAC for N atoms, we get,
`1xx` moles of `NO+2xx` moles of `N_(2)O=2xx` moles of `N_(2)`
`x+2(60-x)=2xx38`
`x=44`.
Hence, volume of `NO=44mL`
and volume of `N_(2)O=(60-44)mL=16mL`