Partial pressures and PV = nRT

gkangelexa

1. The problem statement, all variables and given/known data

Ammonia burns in air to form nitrogen dioxide and water

4NH₃(g) + 7O₂(g) --> 4NO₂(g) + 6H₂O(l)

If 8 moles of NH₃ are reacted with 14 moles of O₂ in a rigid container with an initial pressure of 11 atm, what is the partial pressure of NO₂ in the container when the reaction runs to completion? (assume constant temperature)


3. The attempt at a solution

This is what i did:
i used PV = nRT.... since T, R, and V are constant here,
I said P_initial/ n_initial = P_final/n_final

I used 11 atm for P initial and I added the moles of the 2 reactant gases for the initial n to get 8+14 = 22.
(is this allowed? Can I add the 2 moles of gases together for this problem even though they are different gases?)

Then I used 8 moles for my final n (since 8 moles of NO₂ were formed) and then calculated 4 atm as my answer for the final partial pressure of NO₂.


Is there a faster/ easier way to calculate the answer?
Did I do it the right way?

Bohrok

You can do that; you actually probably want to consider P_totalV = n_totalRT (total being the two gases you start with) and P_NO2V = n_NO2RT, then combine the two equations to get P_total/n_total = P_NO2/n_NO2 (similar to what you did).

Your initial amounts of NH₃ and O₂ are in the same ratio as their reaction ratios in the equation, and 8 mol NH₃ will yield 8 mol NO₂, so everything you did looks good.