How many liters of oxygen can react with 84.0 grams of lithium metal at standard temperature and pressure? show all of the work used to find the answer. equation: 4li + o2--> 2li2o

  60. 4Li(s) + O2(g) >>>>>>>>>>> 2Li2O(s) 

moles O2 required = (84 grams Li) * (1 mol Li / 6.941 grams Li) * (1 mol O2 / 4 mol Li) = 3.026 moles O2 required. 

PV = nRT, P = 1 atm, V = ?, n = 3.026 moles, R = 0.0821 L*atm/mol*K, T = 273 K 
(1 atm)*(V) = (3.026 moles O2) * (0.0821 L*atm/mol*K) * (273 K) 
V = 67.82 Liters is the answer. 

hope this helps

The third choice .. (:

It is called a neutralization reaction