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.

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 =

Using an ideal gas equation :

Partial pressure of the methane=

Moles of the methane=

Partial pressure of the oxygen gas=

Moles of the methane=

Mole fraction  of the methane=

Mole fraction  of the oxygen gas=

 (Dalton's law)

Mass of 0.06975 moles of methane gas :

0.06975 mol × 18 g/mol =1.25550 g

 (Dalton's law)

Mass of 0.06975 moles of oxygen gas :

0.09964 mol × 32 g/mol =3.18848 g

Total moles of mixture = n =

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

more info ? ? i don't understand what the question is

explanation: