It means its OK
Yes, looks right to me.
As usual, could mean I am missing something.
The Whole Question is...
The partial pressure of CH4 is 0.175atm and partial pressure of O2 is 0.250 atm in a mixture of gases..
a) what is the mole fraction of each gas in the mixture?
b) If the volume of teh mixture is 10.5 L At 65 deg C, calculate total number of moles of gas in the mixture
c) Calculate grams of each gas in the mixture.
What I did:
partial presure / total pressure
a) mole fraction of CH4 := 17.7kPa/43.02kPa
of O2 := 25.3 kPa/ 43.02 KPa
b) V- 10.5 L
T = 338 K
P = 43.02 kPa (total)
so using PV=nRt
n= PV/RT
n= 0.160 mol
c) using the mole fraction
moles of CH4 present := 17.7/43.02 * 0.160 = 0.065 mol
moles of CH4 present := 25.3/43.02 * 0.160 = 0.094 mol
mass of CH4 = 0.065 mol * 16g/mol (molar mass) = 1.04 g
mass of O2 = 0.094mol * 32g/mol = 3.0 g