How many moles of O2 contained in 5.25L of a gas at 26 degrees C and 1.2 atm?

4 answers

Since PV=nRT, we have
PV/nT = R is constant.

So, use STP and your given numbers:

(1*1)/(1*273) = (1.2*5.25)/((273+26)n)

Just solve for n.
I think there is an extra 273 (on the left side) in this solution that doesn't belong there.
PV = nRT
n = PV/RT
n = 1.2*5.25/(0.08206*299)
n = 0.26
Oh, yeah: 22.4L at STP, not 1L.

(1*22.4)/(273*1) = (1.2*5.25)/(299n)
n = 0.256

My 273 is included in the .08026 factor you inserted.
right. and I tried to get 0.08206 out of 1/273 but no matter how hard I tried it wouldn't work. But 22.4/273 = 0.08205 or
22.404/273 = 0.08206 which is the number usually quoted. I expect a mathematician will tell me that 22.404 is a lot of s.f. that probably don't belong there.