Question

The partial pressure of CH4 is 0.175 atm and that of O2 is 0.250 atm in a mixture of the two gases. What is the mole fraction of each gas in the mixture? If the mixture occupies a volume of 10.5 L at 35oC, calculate the total number of moles of gas in the mixture. Calculate the number of grams of each gas in the mixture.

Answer 1

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.

Step-by-step 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 =
`n_1+n_2`

Using an ideal gas equation :

`PV=nRT`

`P* 10.0 L=n* 0.0821 atm L/mol K* 308.15 K`

`P=2.509 n`

Partial pressure of the methane=
`p_1=0.175 atm`

Moles of the methane=
`n_1`

Partial pressure of the oxygen gas=
`p_2=0.250 atm`

Moles of the methane=
`n_2`

Mole fraction of the methane=
`\chi_1=(n_1)/(n_1+n_2)`

Mole fraction of the oxygen gas=
`\chi_2=(n_2)/(n_1+n_2)`

`p_1=p* \chi_1` (Dalton's law)

`0.175 atm=P* (n_1)/(n_1+n_2)`

`0.175 atm=(2.509 n)* (n_1)/(n)`

`n_1=0.06975 mol`

Mass of 0.06975 moles of methane gas :

0.06975 mol × 18 g/mol =1.25550 g

`p_2=p* \chi_2` (Dalton's law)

`0.250 atm=P* (n_2)/(n_1+n_2)`

`0.250 atm=(2.509 n)* (n_2)/(n)`

`n_2=0.09964 mol`

Mass of 0.06975 moles of oxygen gas :

0.09964 mol × 32 g/mol =3.18848 g

Total moles of mixture = n =
`n_1+n_2`

[/tex]=0.06975 mol+0.09964 mol=0.16939 mol[/tex]