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

First, we will use the ideal gas law to determine the number of moles of the mixture that was present.

n = PV/RT

n = 1 * 5.04 / 0.082 * (273.15 + 0)
n = 0.225 moles


Let the moles of propane be P and the moles of methane be M.


P + M = 0.225

Next, we find the moles of carbon dioxide formed:

Moles = mass / Molecular weight
Moles = 19.4 / 44 = 0.441

Each mole of propane produces 3 moles of carbon dioxide and each mole of methane produces one mole of carbon dioxide. So:

3P + M = 0.441

Subtracting the first equation from the second, we get:

2P = 0.216
P = 0.108 moles

M = 0.225 - 0.108
M = 0.117

Fraction of propane = 0.108/0.225 = 0.48
Fraction of methane = 0.52