4 NH3 + 70 O2 ---> 4NO2 + 6 H2O. If 23.7 g of NH3 react with 18.5g of oxygen, which is the limiting reagent?

First, the equation must be balanced correctly:

4 NH3 + 7 O2 → 4 NO2 + 6 H2O

(23.7 g NH3) / (17.03056 g NH3/mol) = 1.3916 mol NH3

(18.5 g O2) / (31.99886 g O2/mol) = 0.57815 mol O2

0.57815 mole of O2 would react completely with 0.57815 x (4/7) = 0.33037 mole of NH3, but there is more NH3 present than that, so NH3 is in excess and O2 is the limiting reagent.

your equation does not balance. 

Ammonia + Oxygen ➜ Nitrogen dioxide + water

4NH₃ + 7O₂ ➜ 4NO₂ + 6H₂O

NO₂ is 46 g/mol

NH₃ is 17 g/mol

H₂O is 18 g/mol

O₂ is 32 g/mol

4NH₃ + 7O₂ ➜ 4NO₂ + 6H₂O

4•17 g of NH₃ + 7•32 g of O₂ ➜ 4•46 g of NO₂ + 6•18 g of H₂O

68 g of NH₃ + 224 g of O₂ ➜ 184 g of NO₂ + 108 g of H₂O

check 68+224 = 184+108 = 292

23.7 g of NH₃ react with 18.5g of O₂

amount of O₂ required to react with 23.7 g of NH₃ is

224/68 = x/23.7

x = 78.1 g

you have only 18, so it is the limiting factor.