How many moles of ethylene can react with 12.9 l of oxygen gas at 1.2 atm in 297 k

The balanced chemical reaction is:

C2H4+3O2→2CO2+2H2O

We are given the volume of the oxygen gas for the reaction to be utilized. In order to use this value in obtaining the amount of the other reactant in moles, it should be converted to units of moles. We must assume that the oxygen gas is an ideal gas.

PV = nRT
n = PV / RT = 1.2 atm (12.9 L) / (0.08206 atm L / mol K ) 297 K
n= 0.6352 mol O2

0.6352 mol O2 ( 1 mol C2H4 / 3 mol O2 ) = 0.2117 mol C2H4