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.

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

1. o2(g)as a product

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