Question: What mass of water, in grams, would have to be decomposed to produce 17.6 L of molecular oxygen at STP?
Work:
n=PV/RT
x=(1 atm)(17.6 L)/(0.0821)(273 k)
x = 1.27348295454546
1.27348295454546 x 2 = 2.54696590909092
Grams= Moles x 18.0
Grams = 2.54696590909092 x 18.0
Grams = 45.8
I think I'm messing up somewhere, but I don't know what I'm doing wrong.
Your first mistake (I didn't look past that one) is that you didn't plug in the numbers into the calculator. You have the reciprocal of the number of moles.
n = 1*17.6/[(0.09206)(273)]
n should be 0.7857 and that should be rounded to the correct number of significant figures. (I wonder why you didn't just divide 17.6/22.4 to arrive at the number).
It seems that you are on the right track with your calculation. However, there is a slight mistake.
To find the mass of water that needs to be decomposed, we need to use the equation:
grams = moles x molar mass of water
You have correctly found the moles of molecular oxygen using the ideal gas law equation (n = PV/RT).
n = (1 atm)(17.6 L) / (0.0821)(273 K)
n = 0.6779 moles
Now, to find the mass of water, we need to multiply the number of moles by the molar mass of water. The molar mass of water (H2O) is 18.0 g/mol.
grams = 0.6779 moles x 18.0 g/mol
grams = 12.2022 g
Therefore, the mass of water that would have to be decomposed to produce 17.6 L of molecular oxygen at STP is approximately 12.2022 grams.