At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 ml air containing 20% O_2 by volume for complete combustion. After combustion the gases occupy 330 ml. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formulaof the hydrocarbon is :

At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 ml air

1.	At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 ml air containing 20% O_2 by volume for complete combustion. After combustion the gases occupy 330 ml. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formulaof the hydrocarbon is :
Answer» `C_4H_10` `C_3H_6` `C_3H_8` `C_4H_8` SOLUTION :`C_x H_(y(g)) + ((4x+y)/4)O_(2(g)) to CO_(2(g)) + Y/2 H_2O_((l))` Vol. of `O_2` used =`20/100xx375` =75 mL From the REACTION of combustion , 1 mL `C_x H_y` requires =`(4x+y)/4` mL `O_2` `because` 15 mL `=15((4x+y))/4`=75 mL So, 4x+y=20 So, x=3 , y=8hence formula of hydrocarbon is `C_3H_8`.