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 345 mL . Assuming that the water form

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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 345 mL . Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: (Assume complete combustion of reactant)

TestHub · Accepted Answer

Volume of N 2 in air = 375 × 0.8 = 300 ml Volume of O 2 in air = 375 × 0.2 = 75 ml C x H y + x + y 4 O 2 ⟶ xCO 2 g + y 2 H 2 O ℓ 1 5 ml 1 5 x + y 4 0 0 1 5 x - After combustion, total volume 345 = V N 2 + V CO 2 345 = 300 + 15 x x = 3 Volume of O 2 used 15 x + y 4 = 75 x + y 4 = 5 y = 8 So, hydrocarbon is C 3 H 8 .

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