O Not set A 82.6/100 3/16/2017 03:44 PM Gradebook Print m Calculator Periodic Ta

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O Not set A 82.6/100 3/16/2017 03:44 PM Gradebook Print m Calculator Periodic Table x Incorrect Incorrect X Incorrect Incorrect Incorrect Incorrect X Question 23 of 35 Incorrect Map A Learning macmillan learning At 1.00 atm and 0 °C, a 5.04 L mixture of methane (CH4) and propane (C3H8) was burned, producing 15.7 g of CO2. What was the mole fraction of each gas in the mixture? Assume complete combustion. Number 52.83 methane Number 47.17 propane A Previous Give Up & View Solution Check Answer Next Exit Hint

Explanation / Answer

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 = 15.7 /44.01= 0.3567 mol

1mol CH4 will produce 1 mol CO2
1 mol C3H8 will produce 3 mol CO2

X mol CH4 will produce X mol CO2

0.3567 mol C = (0.22 mol)(x) (1 mol C / 1 mol CH4) + (0.22 mol) (1-x) (3 mol C / 1 mol C3H8)

0.3567 = 0.22x + 0.66 - 0.66x

0.3567-0.66 = -0.44 x

0.3033 = 0.44 x

x = 0.69 Mole methane

1- x = 1-0.69 = 0.31 Mole Propane

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