A sample of oxygen gas was collected via water displacement. Since the oxygen was collected via water displacement, the sample is saturated with water vapor. If the total pressure of the mixture at 26.4 °C is 725 torr, what is the partial pressure of oxygen? The vapor pressure of water at 26.4 °C is 25.81 mm Hg.

You're right about Pgas = X*Ptotal but I don't think you understand the equation I wrote.

Dalton's Law of partial pressures tells us that the total pressure inside a container is the sum of the individual gas pressures. So I wrote the equation
Ptotal = pO2 + pH2O.
That means that the pressure of the oxygen inside the container + the pressure of the H2O inside the container is equal to the total pressure. And the total pressure is given in the problem as 725 torr.
So 725 = pO2 + pH2O
The problem also tells you that the pressure of the H2O at the temperature of the experiment is 25.81 mm (which is the same as 25.81 torr). So you solve this equation for pO2.
725 = pO2 + 25.81
725-25.81 = pO2
pO2 = 699.19 mm Hg or 699.19 Torr. That answer probably will not be counted correct by your prof because it contains too many significant figures. Based on your post of 725 for total P I would round that to 699 mm Hg.

Ptotal = pO2 + pH2O

I got .988atm, but what do I do with the temp?

The temperature is used to look in a table of T vs vapor pressure. It tells you the vapor pressure of the water (in this case 25.81 mm Hg) at 26.4 C.

I don't agree with your answer. And why in the world would you change mm and torr to atm if the problem didn't ask for the answer in atm?

To find the partial pressure of oxygen in the mixture, we need to subtract the vapor pressure of water from the total pressure of the mixture.

Given:
Total pressure of the mixture = 725 torr
Vapor pressure of water at 26.4 °C = 25.81 mm Hg

Since the units of the vapor pressure are different from the units of the total pressure, we need to convert the vapor pressure to torr before subtracting.

1 mm Hg = 1 torr

So, the vapor pressure of water can be converted to torr:
Vapor pressure of water at 26.4 °C = 25.81 mm Hg = 25.81 torr

Now, we can calculate the partial pressure of oxygen:
Partial pressure of oxygen = Total pressure of the mixture - Vapor pressure of water
Partial pressure of oxygen = 725 torr - 25.81 torr
Partial pressure of oxygen = 699.19 torr

Therefore, the partial pressure of oxygen in the mixture is 699.19 torr.

Ok then, I got 750.81 torr.

I know that Partial pressure = total pressure x the mole fraction

Does this mean that I have to find moles of O2 and H2?

Wow, I feel stupid...I need some sleep. Thanks so much!!