With this problem you can use PV = nRT and solve for the total mol of all the gases combined and then since you hae been given the partial mol of N2 subtract this and you will have the remainder which is CO2 and O2 moles. From this you can use atomic weight to decipher what the moles of the O2 is by itself then re-plug it back into the equation PV = nRT and you can solve for the partial pressure of O2.

PV =nRT

5(3) = n(0.0820573)273

15 = n22.4016429

15/22.4016429 = n

0.6695937 moles

Now you will subract the moles given for N2 so you will have the remainder moles:

0.6695937 - 0.230 = 0.4395937 mol

this left over mole is 100 % between the CO2 and O2 because we already have the partial mole for N2.

So now to figure out what this would mean between the two,

look at their mole and grams by thier atomic weight for the two combined:

CO2 at wt = 44g/mol

O2 atomic weight = 32g/mol

Now add them both up and you get:

44+32 = 76g/mol

44/76 = 0.578947 or 57.8 % is the CO2 between the two

so now take the fraction and multiply this by this new left over mole:

0.578947 x 0.4395937 = 0.2545 moles of CO2

So the remainder must be the subtraction or what is left over and that is what the Oxygen's moles will be, so:

0.4395937- 0.2545 = 0.1850937 moles for O2

and when you add up all the partial moles they will all amount to total moles.

O2 + CO2 + N2 = total moles

0.1850937 + 0.2545 + 0.230 = 0.6695937 moles

now take the partial moles of O2 and use the same conditions to solve for the partial pressure of O2:

PV =nRT

P(3) = (0.1850937)(0.0820573)273

P = ((0.1850937)(0.0820573)273)/3

P = 4.1464/3

P = 1.38213 atm for Oxygen

you can check it to make sure all the partial pressures add up to the total pressure and you get:

CO2 + O2 +N2 = 5.00 atm

0.250 atm + 1.38213 atm + 3.367866 atm = 5.00 atm

So Oxygen's partial pressure is 1.38 atm for 3 sig figs