Ammonia (nh3(g) hf=-45.9 kj/mol) reacts. with oxygen to produce nitrogen and water (h2o(g) hf = -241.8 kj/mol according to the equation below. 4nh3(g) + 3o2(> 2n2(g) +6h2o(g) what is the enthalpy of the reaction

ΔH°_rxn = -195.9 kJ·mol⁻¹

Explanation:

                              4NH₃(g) + 3O₂(g) ⟶ 2N₂(g) +6H₂O(g)

ΔH°_f/(kJ·mol⁻¹):    -45.9          0                 0        -241.8

The formula relating ΔH°_rxn and enthalpies of formation (ΔH°_f) is

ΔH°_rxn = ΣΔH°_f(products) – ΣΔH°_f(reactants)

ΣΔH°_f(products) = -6(241.8) = -1450.8 kJ

ΣΔH°_f(reactants) = -4(45.9) = -183.6 kJ

ΔH°_rxn =  (-1450.8 + 183.6) kJ = -1267.2 kJ

sry but any answer choices

i believe it is c.) 1575 j.