Respuesta :
Answer:
The total number of moles of gas in the mixture is 0.16939.
1.25550 grams of methane gas and 3.18848 grams of oxygen gas.
Explanation:
Total volume of the mixture = V = 10.5 L
Temperature of the mixture = T = 35°C = 308.15K
Pressure of the mixture = P
Total moles of mixture = n =[tex]n_1+n_2[/tex]
Using an ideal gas equation :
[tex]PV=nRT[/tex]
[tex]P\times 10.0 L=n\times 0.0821 atm L/mol K\times 308.15 K[/tex]
[tex]P=2.509 n[/tex]
Partial pressure of the methane= [tex]p_1=0.175 atm[/tex]
Moles of the methane= [tex]n_1[/tex]
Partial pressure of the oxygen gas= [tex]p_2=0.250 atm[/tex]
Moles of the methane= [tex]n_2[/tex]
Mole fraction of the methane= [tex]\chi_1=\frac{n_1}{n_1+n_2}[/tex]
Mole fraction of the oxygen gas= [tex]\chi_2=\frac{n_2}{n_1+n_2}[/tex]
[tex]p_1=p\times \chi_1[/tex] (Dalton's law)
[tex]0.175 atm=P\times \frac{n_1}{n_1+n_2}[/tex]
[tex]0.175 atm=(2.509 n)\times \frac{n_1}{n}[/tex]
[tex]n_1=0.06975 mol[/tex]
Mass of 0.06975 moles of methane gas :
0.06975 mol × 18 g/mol =1.25550 g
[tex]p_2=p\times \chi_2[/tex] (Dalton's law)
[tex]0.250 atm=P\times \frac{n_2}{n_1+n_2}[/tex]
[tex]0.250 atm=(2.509 n)\times \frac{n_2}{n}[/tex]
[tex]n_2=0.09964 mol[/tex]
Mass of 0.06975 moles of oxygen gas :
0.09964 mol × 32 g/mol =3.18848 g
Total moles of mixture = n =[tex]n_1+n_2[/tex]
[/tex]=0.06975 mol+0.09964 mol=0.16939 mol[/tex]
