How many moles of ethylene (c2h4) can react with 12.9 liters of oxygen gas at 1.2 atmospheres and 297 kelvin? c2h4(g) + 3o2(g) yields 2co2(g) + 2h2o(g) 0.21 moles 0.32 moles 0.63 moles 0.84 moles

1) Convert 12.9 liters of Oxygen to mol at the given conditions:

PV = nRT ⇒ n = PV/RT

n = [1.2atm*12.9 l] / [0.082 atm l /K mol * 297K]

n = 0.636 mol of O2

2) use the stoichiometry derived from the balanced chemical equation

 1mol C2H4 / 3  mol O2 =  x mol C2H4 / 0.636 mol O2

x = 0.636 / 3 mol O2 = 0.212 mol O2.

0.212 mol O2