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At 1.00 atm and 0 °C, a 5.04 L mixture of methane (CH4) and propane (C3H8) was burned, producing 16.5 g of CO2. What was the mole fraction of each gas in the mixture? Assume complete combustion.

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The chemical reactions involved are:

CH4 + 2 O2 → CO2 + 2 H2O
C3H8 + 5 O2 → 3 CO2 + 4 H2O

Moles of gases = 5.04 L ( 1 mol / 22.41 L ) = 0.22 mol gases

Moles C in CO2 produced = 15.0 g CO2 ( 1 mol CO2 / 44.01 g CO2 ) (1 mol C/1 mol CO2)= 0.34 mol C

Let x be the molar fraction of CH4 in the mixture so 1-x is the molar fraction of C3H8 in the mixture.

We set up a C balance:

0.34083 mol C = (0.22 mol)(x) (1 mol C / 1 mol CH4) + (0.22 mol) (1-x) (3 mol C / 1 mol C3H8)  

x = 0.74 methane
1-x = 0.26 propane

Answer: The mole fraction of methane is 0.67 and that of propane is 0.33

Explanation:

  • The equation given by ideal gas follows:

[tex]PV=nRT[/tex]

where,

P = pressure of the mixture = 1.00 atm

V = Volume of the mixture = 5.04 L

T = Temperature of the mixture = [tex]0^oC=[0+273]K=273K[/tex]

R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

n = number of moles of mixture = ?

Putting values in above equation, we get:

[tex]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[/tex]

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

So, [tex]x+y=0.225[/tex]      .....(1)

  • The chemical equation for the combustion of methane follows:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

By Stoichiometry of the reaction:

1 mole of methane produces 1 mole of carbon dioxide

So, 'x' moles of methane will produce = [tex]\frac{1}{1}\times x=x[/tex] moles of carbon dioxide

  • The chemical equation for the combustion of propane follows:

[tex]C_3H_8+5O_2\rightarrow 3CO_2+4H_2O[/tex]

By Stoichiometry of the reaction:

1 mole of propane produces 3 mole of carbon dioxide

So, 'y' moles of propane will produce = [tex]\frac{1}{1}\times y=y[/tex] moles of carbon dioxide

  • Evaluating 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 = [tex](x+3y)\times 44[/tex]

We are given:

Mass of carbon dioxide = 16.5 g

So, [tex]44(x+3y)=16.5[/tex]     .....(2)

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

[tex]44(0.225-y+3y)=16.5\\\\0.225+2y=0.375\\\\y=\frac{0.15}{2}=0.075[/tex]

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

[tex]x+0.075=0.225\\x=0.15[/tex]

  • Mole fraction of a substance is given by:

[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex]

For Methane:

Moles of methane = 0.15 moles

Total moles = 0.225

Putting values in above equation, we get:

[tex]\chi_{(Methane)}=\frac{0.15}{0.225}=0.67[/tex]

For Propane:

Moles of propane = 0.075 moles

Total moles = 0.225

Putting values in above equation, we get:

[tex]\chi_{(Propane)}=\frac{0.075}{0.225}=0.33[/tex]

Hence, mole fraction of methane is 0.67 and that of propane is 0.33

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