What is the ∆g for the following reaction under standard conditions (t = 298 k) for the formation of nh4no3(s)? 2nh3(g) + 2o2(g) nh4no3(s) + h2o(l)

given:
nh4no3(s): ∆hf = -365.56 kj ∆sf = 151.08 j/k.
nh3(g): ∆hf = -46.11 kj ∆sf = 192.45 j/k.
h2o(l): ∆hf = -285.830 kj ∆sf = 69.91 j/k.
o2(g): ∆hf = 0.00 kj ∆sf = 205 j/k.

∆G =  ∆H - T∆S

NH3: ∆G = -46.11x10^3 - (298)(192.45) = -103460.1 J
O2: ∆G = 0 - (298)(205) = -61090 J
NH4NO3: ∆G = -365.56x10^3 - (298)(151.08) = -410581.84 J
H2O: ∆G = -285.830x10^3 - (298)69.91) = -306663.18 J

∆Grex = ∆Gproducts - ∆Greactants
∆Grex = (-410581.84 +  -306663.18) - (-103460.1/2 + -61090/2)
∆Grex =-634969.97 J/mol = -634.97 kJ/mol

The answer is 4 i don't know if its right let me see.