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