How do I solve a heat of formation problem like this?

Question

11. 3N2O(g) + 2NH3(g) -----> 4N2(g) + 3H2O(g) Enthalpy change = -879.6 kJ
What is the heat of formation for N2O in kJ/mole (heats of formation: NH3 = -45.9 kJ/mole and
H2O = -241.8 kJ/mole)?
a) +246 b) +82
c) -82 d) -246

Answer 1

This time you DO have a correctly balanced equation.  Good job.

You should recall that ∑∆H_products - ∑∆H_reactants = ∆H_reaction

Thus  [(3 x -241.8) + (4 x 0)] - [(3x) + (2 x -45.9)] = -870.6

Solve for x to get ∆H_formation N2O in kJ/mole