Question

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 1

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 = $0^oC=[0+273]K=273K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of mixture = ?

Putting values in above equation, we get:

$1.00atm\times 5.04L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 273K\\n_{mix}=\frac{1.00\times 5.04}{0.0821\times 273}=0.225mol$

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

So, $x+y=0.225$      .....(1)

The chemical equation for the combustion of methane follows:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

By Stoichiometry of the reaction:

1 mole of methane produces 1 mole of carbon dioxide

So, 'x' moles of methane will produce = $\frac{1}{1}\times x=x$ moles of carbon dioxide

The chemical equation for the combustion of propane follows:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

By Stoichiometry of the reaction:

1 mole of propane produces 3 mole of carbon dioxide

So, 'y' moles of propane will produce = $\frac{1}{1}\times y=y$ 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 = $(x+3y)\times 44$

As we are given:

Mass of carbon dioxide = 15.0 g

So, $44(x+3y)=15.0$     .....(2)

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

$44(0.225-y+3y)=15.0\\\\0.225+2y=0.341\\\\y=0.058$

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

$x+0.058=0.225\\x=0.167$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

For Methane:

$\chi_A=\frac{n_A}{n_A+n_B}$

Moles of methane = 0.167 moles

Total moles = 0.225

Putting values in above equation, we get:

$\chi_{(Methane)}=\frac{0.167}{0.225}=0.742$

For Propane:

Moles of propane = 0.058 moles

Total moles = 0.225

Putting values in above equation, we get:

$\chi_{(Propane)}=\frac{0.058}{0.225}=0.26$

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