60 ml of a mixture of nitrous oxide and nitric oxide was exploded with excess of hydrogen. If 38 ml of N2 was formed, calculate the volume of nitric oxide in mixture before reaction. All measurements are made at STP conditions. Assume H2O in liquid phase.

NO+   N2O+   2H2→32 N2+2H2O 'a'ml   'b'ml   excess volume  0         0                       38 ml volume after reaction   One mole of  NO gives 12 volume of N2  One mole of N2O gives 1 volume of N2 ∴a2+b=38  (1) Also, a+b = 60   (2) By equations (1) and (2)  a = 44 ml, b = 16 ml