Question

If you have 6 moles of H and 2 moles of O what's the partial pressure of O at 11.2 atm?

Answer 1

I believe the partial pressure is proportional to the mole fraction of O, which is just 1/3.

So, \[P_1 = \left( \frac{n_1}{n_t} \right) \cdot P_t = X_1 P_t\]

Or 1/3 times 11.2atm or 11.2/3 atm or 3.73 atm. Let me know if this sounds right.

Answer 2

your idea is right, your numbers are off. The partial pressure of O2 isn't 1/3, it's 1/4. The total # of moles in the container is 8, 6 moles of H2 and 2 moles of O2.

Answer 3

\[P_{O2} = P_{t} \times (\frac{P_{O2}}{P_{H2 + O2}}) = 11.2atm \times (\frac{2mol O_{2}}{8mol total})\]