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
