Intro to chem
posted by pooja .
What is the volume of 19.7g of oxygen at STP?
Heres what i got but did not get the same answer..pls tell me where i am wrong!
(grams to moles)
19.7g x 1 mol/ 16.00 = 1.23 mol Oxygen
(Pv=nRT) looking for V
1.23 mol x 0.0821 x 273K / 1.00 atm = 27.57
CORRECT ANSWER is 13.8L!
You used the atomic weight of Oxygen and not the molar mass (molecular weight).
19.7/32 = 0.6156 moles
V = 0.6156*0.08206*273/ = 13.79L which rounds to 13.8L.
There is another way to do this which is a little easier if you remember that a mole of any gas occupies 22.4L at STP.
(19.7/32)*22.4 = 13.79 = 13.8L
Thank you SO MUCH!