Question

what volume of oxygen gas measured at 25C and 760 torr is required to react with 1.0L of methane measured under the same conditions of temperature and pressure?

Answer 1

CH4 + 2O2 ==> CO2 + 2H2O

You may use L as a shortcut without converting to moles and back to L.
1 mol CH4 will use 2 mols O2.

Answer 2

Would PV=nRT apply to this? if so, how do I find "n"?

Answer 3

Yes, you can use PV = nRT and find n = number of mols. To answer your last question, P = 1 atm (760 torr), V = 1 L, T = 298K (25 C + 273 = 298) and R = 0.08206 L*atm/mol*K.
However, you don't need that.The short cut is that as long as the T and P stay the same (as they do), and you're dealing with gases, you may use L directly and not convert to moles.
CH4 + 2O2 ==> CO2 + 2H2O.
L O2 required = 1L CH4 x (2 moles O2/1 mole CH4) = 1L CH4 x (2/1) = 2L O2 required.
You can prove that to yourself, if you wish, by using PV = nRT as above, solving for n CH4 for the 1.0 L CH4, converting moles CH4 to moles O2, then using PV = nRT to convert moles O2 back to L. The answer is the same by either method but the shortcut is considerably quicker.