Partial pressure question?

[KoonS]

A container has O2 gas added to it to a partial pressure of .480 atm and NH3 gas added until its partial pressure is .350 atm. The reaction given below was then sparked. What is the final partial pressure of gaseous water and final total pressure given that all the reaction heat escaped, leaving the container at its original temperature.
4NH3(g) + 5 O2(g) ---> 4NO(g) + 6H2O(g)

anyone have any ideas to tackle this problem?
thanks in advance.

---

Members don't see this ad.

 

[clutchfans]

Alright I will give it shot.

The total P on the reactant side is 0.48 + o.35 = 0.83 atm
P and n are directly proportional (PV=nRT).
The number of moles of gas goes from 9 (4+5) to 10 (4+6).
So the total pressure should also increase.
By using this formula P1/n1 = P2/n2 you get 0.83atm/9mol = P2/10mol.
P2 = 0.922 2 repeating atm.

Also you know that partial pressure is related to mol fraction.
PH20 = XH2O time Ptotal
PH2O = 6/10 times 0.922 = 0.553 atm H2O.

Let me know if this is right answer and anyone else, let me know if my thought process is correct.

 

[KoonS]

I dont have the answer, but conceptual wise your thought process makes a lot sense to me. thank you