2 NH3 → N2 + 3 H2 ΔHrxn = 92.2 kJ/mol How much energy, in kJ, is required to decompose 83.5 grams of Ammonia (NH3)?

Answer 1

NOTE: ∆H_rxn isn't usually given in kJ/mole, but rather in kJ. I'll assume that to be the case and assume the 92.2 kJ/mole should be 92.2 kJ.

This being the case, then per mole, it will be 92.2 kJ/2 = 46.1 kJ/mole since the balance equation has 2 moles

83.5 g NH₃ x 1 mol/17 g x 46.1 kJ/mol = 226 kJ to decompose 83.5 grams NH₃