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!
To find the volume of 19.7g of oxygen at STP (Standard Temperature and Pressure), you can use the ideal gas law equation, PV = nRT.
First, let's convert the mass (grams) of oxygen to moles:
19.7g of oxygen x (1 mol/ 32.00g) = 0.615625 moles of oxygen
Now, we can plug the values into the ideal gas law equation:
PV = nRT
V = (nRT) / P
V = (0.615625 mol x 0.0821 L·atm/(K·mol) x 273K) / 1.00 atm
V ≈ 13.77 L
So, the volume of 19.7g of oxygen at STP is approximately 13.77 liters.
It appears that the correct answer is indeed 13.8L which is very close to our approximated value of 13.77L. It's possible that the difference in the values could be due to rounding errors or using slightly different values for the gas constant (0.0821 vs. 0.08206).