Answer: The mole fraction of methane is 0.67 and that of propane is 0.33
Explanation:
- The equation given by ideal gas follows:
\(PV=nRT\)
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:
\(1.00atm\times 5.04L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 273K\
_{mix}=\frac{1.00\times 5.04}{0.0821\times 273}=0.225mol\)
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
- 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 = \((x+3y)\times 44\)
We are given:
Mass of carbon dioxide = 16.5 g
So, \(44(x+3y)=16.5\) .....(2)
Putting value of 'x' from equation 1, in equation 2, we get:
\(44(0.225-y+3y)=16.5\\\\0.225+2y=0.375\\\\y=\frac{0.15}{2}=0.075\)
Evaluating value of 'x' from equation 1, we get:
\(x+0.075=0.225\\x=0.15\)
- Mole fraction of a substance is given by:
\(\chi_A=\frac{n_A}{n_A+n_B}\)
For Methane:
Moles of methane = 0.15 moles
Total moles = 0.225
Putting values in above equation, we get:
\(\chi_{(Methane)}=\frac{0.15}{0.225}=0.67\)
For Propane:
Moles of propane = 0.075 moles
Total moles = 0.225
Putting values in above equation, we get:
\(\chi_{(Propane)}=\frac{0.075}{0.225}=0.33\)
Hence, mole fraction of methane is 0.67 and that of propane is 0.33
