A 10.0L flask contains 3.0atm of ethane and 8.0 atm of oxygen at 28 degrees C. The contents of the gas are reacted and allowed to return to 28 degrees C. What is the pressure of the flask?

Question

Using n=RT/PV I converted atm to moles and then used the balanced equation to find that Oxygen was the LR. I then calculated the moles of Ethane that were unreacted. Using the moles of Oxzygen I calculated the moles of carbon dioxide and water formed. I then used P=nRT/V to find the pressure of each gas and Dalton's law to find the final pressure of 15.1 atm. It doesn't match the book's answer of 12.1 atm. Why?

DrBob222 · Answer

I can get the answer of 12.1 atm but I don't think it is correct and I don't agree with your 15.1 either.
I agree O2 is the limiting reagent.
I calculate moles CO2 formed as 1.851.
moles H2O = 2.776
moles ethane reacted = 0.9254 which leaves 0.2892 remaining.
IF I add CO2 + H2O + excess ethane I get 4.916 moles and plug that into PV = nRT and I get 12.14 which rounds to 12.1 atm.
However, at 28C H2O should not be a gas and I don't think it should be included. I think just CO2 and unused ethane should make up the moles. At best, we could look up the vapor pressure of water (memory tells me this is ABOUT 30 mm---that's mm Hg and not atm) and add that in AND we could correct for the volume occupied by the liquid water but I don't think anything like that is called for, in my opinion, from the intent of the problem. Check my thinking.