At 1.00 atm and 0 ∘ C, a 5.04 L mixture of methane ( CH 4 ) and propane ( C 3 H 8 ) was burned, producing 15.0 g CO 2 . What was the mole fraction of each gas in the mixture? Assume complete combustion.

Answer : The mole fraction of methane and propane is, 0.742 and 0.26

Explanation :

First we have to calculate the moles of mixture by using ideal gas equation.

PV = nRT

where,

P = pressure of the mixture = 1.00 atm

V = Volume of the mixture = 5.04 L

T = Temperature of the mixture =

R = Gas constant =

n = number of moles of mixture = ?

Putting values in above equation, we get:

Let the number of moles of methane be 'x' moles and that of propane be 'y' moles

So,      .....(1)

The chemical equation for the combustion of methane follows:

By Stoichiometry of the reaction:

1 mole of methane produces 1 mole of carbon dioxide

So, 'x' moles of methane will produce = moles of carbon dioxide

The chemical equation for the combustion of propane follows:

By Stoichiometry of the reaction:

1 mole of propane produces 3 mole of carbon dioxide

So, 'y' moles of propane will produce = moles of carbon dioxide

Now we have to calculate the mass of carbon dioxide.

Total moles of carbon dioxide = (x + 3y)

Mass of carbon dioxide = (Total moles) × (Molar mass of carbon dioxide)

Molar mass of carbon dioxide = 44 g/mol

Mass of carbon dioxide =

As we are given:

Mass of carbon dioxide = 15.0 g

So,     .....(2)

Putting value of 'x' from equation 1, in equation 2, we get:

Evaluating value of 'x' from equation 1, we get:

Mole fraction of a substance is given by:

For Methane:

Moles of methane = 0.167 moles

Total moles = 0.225

Putting values in above equation, we get:

For Propane:

Moles of propane = 0.058 moles

Total moles = 0.225

Putting values in above equation, we get:

Hence, the mole fraction of methane and propane is, 0.742 and 0.26

the answer is 75.0%.

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