What is the molar enthalpy using the following reactions? And how much energy is released per gram of hydrazine burned?

Rewrite the four given thermochemical equation as :
3N₂H₄(l)+ 3H₂O(l) → 3N₂O(g) + 9H₂(g) ...... ΔH = -3×(-317 kJ)
N₂H₄(l) + H₂O(l) → 2NH₃(g) + (1/2)O₂(g) ...... ΔH = -(-143 kJ)
2NH₃(g) + 3N₂O(g) → 4N₂(g) + 3H₂O(l) ...... ΔH = -1013 kJ
9H₂(g) + (9/2)O₂(g) → 9H₂O(l) ..... ΔH = 9(-286 kJ)

Add the above four thermochemical equations, and cancel 4H₂O(l), 2NH₃(g), 3N₂O(g), (1/2)O₂(g) and 9H₂(g) on the both side.
4N₂H₄(l) + 4O₂(g) → 4N₂(g) + 8H₂O ...... ΔH = 3(317) + 143 - 1013 - 9(286) kJ = -2493 kJ

Divided the above thermochemical equations by 4.
N₂H₄(l) + O₂(g) → N₂(g) + 2H₂O ...... ΔH = -2493/4 kJ = -623 kJ

Molar enthalpy of the above reaction = -623 kJ/mol of N₂H₄

Molar mass of N₂H₄ = (14.01×2 + 1.008×4) = 32.05 g/mol
When 1 g of N₂H₄ is burned, energy released = (2493/4 kJ/mol) / (32.05 g/mol) = 19.4 kJ