36 ml of a gaseous mixture consisting of a gaseous organic compound A and just sufficient amount of oxygen required for complete combustion to produce 16 ml ofCO2,24 ml of water vapor and 8 ml of N2. (The volumes are measured at STP conditions), volume of O2required for complete combustion is ____?

Volume of CO2=16ml 22400 ml of CO2 contain 32 gram of Oxygen  ∴16 ml of CO2 contain 16×3222400 gram of O2=0.0228 gram   Volume of H2O=24 ml 22400 ml of H2O contain 16 gram of Oxygen  24 ml of H2O contain 24×1622400 gram of oxygen =0.0171 gram  Total mass of oxygen = 0.0228+0.0171 = 0.0399 gram  32 gram of oxygen has volume 22400 ml at STP  Then 0.0399 gram of oxygen has volume 22400×0.039932=27.9≃28ml