Respuesta :
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 = [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
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 = [tex](x+3y)\times 44 [/tex]
As we are given:
Mass of carbon dioxide = 15.0 g
So, [tex]44(x+3y)=15.0[/tex] .....(2)
Putting value of 'x' from equation 1, in equation 2, we get:
[tex]44(0.225-y+3y)=15.0\\\\0.225+2y=0.341\\\\y=0.058[/tex]
Evaluating value of 'x' from equation 1, we get:
[tex]x+0.058=0.225\\x=0.167[/tex]
Mole fraction of a substance is given by:
[tex]\chi_A=\frac{n_A}{n_A+n_B} [/tex]
For Methane:
[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex]
Moles of methane = 0.167 moles
Total moles = 0.225
Putting values in above equation, we get:
[tex]\chi_{(Methane)}=\frac{0.167}{0.225}=0.742[/tex]
For Propane:
Moles of propane = 0.058 moles
Total moles = 0.225
Putting values in above equation, we get:
[tex]\chi_{(Propane)}=\frac{0.058}{0.225}=0.26[/tex]
Hence, the mole fraction of methane and propane is, 0.742 and 0.26
