A gas mixture of 3.0 liters of propane and butane on complete combustion at 25°C produced 10 liters of CO2. Find out the composition of the gas mixture.
C3H8 + 5 O2 → 3 CO2 + 4 H2O C4H10 + 6 12O2 → 4 CO2 + 5 H2O Suppose volume of propane (C3H8) = xL ∴ Volume of butane (C4H10) = (3 – x)L 1 L of C3H8 gives 3 L of CO2 and 1 L of C4H10 gives 4 L of CO2 ∴ CO2 produced = 3x + 4(3 – x) = 12 – x (Given) 12 – x = 10 Hence x = 2 L ∴ Volume of propane = 2 L and volume of butane = 3 – 2 = 1 L
