mole fraction of gas in mixture

Question

At 1.00 atm and 0 °C, a 5.04 L mixture of methane (CH4) and propane (C3H8) was burned, producing 19.4 g of CO2. What was the mole fraction of each gas in the mixture? Assume complete combustion.

Answer 1

Hello Ma

CH4 + 2O2 --> CO2 + 2H2O

C3H8 + 5O2 --> 3CO2 + 4H2O

Convert g of CO2 to moles of CO2 using its molar mass (44 g/mol).

19.4 g CO2 x (1 mol CO2 / 44 g CO2) = 0.441 mol CO2

1.00 atm and 0 oC indicates conditions at STP. At STP, 1 mole of a gas occupies 22.4 L. Therefore, 5.04 L of the mixture will contain

5.04 L x (1 mol / 22.4 L) = 0.225 mol CO2

0.225 mol of CO2 was formed from the combustion of CH4 and C3H8.

1 mol CH4 forms 1 mol CO2

1 mol C3H8 forms 3 mol CO2

Let x = mol of CH4 burned. The amount of C3H8 burned = 0.225 - x

x mol CH4 will form x mol CO2

(0.225 -x) mol C3H8 will form 3(0.225 -x ) mol of CO2 or 0.675 - 3x

mol CO2 from CH4 + mol CO2 from C3H8 = 0.441 mol CO2 in total

x + 0.675 - 3x = 0.441

-2x = 0.441 - 0.675

-2x = -0.234

x = 0.234/2 = 0.117

x = 0.117 mol CH4

mol C3H8 = 0.225 - x = 0.225 - 0.117 = 0.108

mol fraction of CH4 = 0.117 / 0.225 = 0.520

mol fraction of C3H8 = 0.108 / 0.225 = 0.480