At 1.00 atm and 0 °C, a 5.04 L mixture of methane (CH4) and propane (C3H8) was burned, producing 16.5 g of CO2. What was the mole fraction of each gas in the mixture? Assume complete combustion.

Question

Pooch

Chemistry

19 November, 12:39

At 1.00 atm and 0 °C, a 5.04 L mixture of methane (CH4) and propane (C3H8) was burned, producing 16.5 g of CO2. What was the mole fraction of each gas in the mixture? Assume complete combustion.

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Izabelle Perkins · Answer 1

The chemical reactions involved are:

CH4 + 2 O2 → CO2 + 2 H2O

C3H8 + 5 O2 → 3 CO2 + 4 H2O

Moles of gases = 5.04 L (1 mol / 22.41 L) = 0.22 mol gases

Moles C in CO2 produced = 15.0 g CO2 (1 mol CO2 / 44.01 g CO2) (1 mol C/1 mol CO2) = 0.34 mol C

Let x be the molar fraction of CH4 in the mixture so 1-x is the molar fraction of C3H8 in the mixture.

We set up a C balance:

0.34083 mol C = (0.22 mol) (x) (1 mol C / 1 mol CH4) + (0.22 mol) (1-x) (3 mol C / 1 mol C3H8)

x = 0.74 methane

1-x = 0.26 propane