What will be the pressure of a sample of 48.0 grams of oxygen gas in a glass container of volume 5.2 L at 25 degrees Celsius?

Molar Mass O2: 32.00, Molar Mass CO2: 44.01, T= 21C+273= 294Kelvins V=5.00 L Grams of O2=20.0 so moles O2= 20.0g/32.00g=.625g/mol

Grams of CO2=4.4 so moles CO2= 4.4g/ 44.01=.1

Total moles: .625+.1= .725 mols (this is n)

Ideal Gas Law: PV=nRT....Rearrange and solve for pressure.

P=nRT/V.....R is the constant which is equal to .08206L*atm*mol

P=(.725)(.08206)(294K)/(5.00L)

P= 3.5atm