Calculate the enthalpy change that takes place when 1 mole of hydrazine burns in a lunar lander?

a)
Refer to: https://www.drjez.com/uco/ChemTools/Standard%20Thermodynamic%20Values.pdf
ΔHf°[N₂H₄(ℓ)] = 50.63 kJ/mol
ΔHf°[N₂O₄(g)] = 9.16 kJ/mol
ΔHf°[H₂O(g)] = -241.82 kJ/mol
(The ΔHf° values are to 2 decimal places.)
(ΔHf° values from different sources may be slightly different.)

N₂H₄(ℓ) + (1/2)N₂O₄(g) → (3/2)N₂(g) + 2H₂O(g)     ΔH₁°

Enthalpy change when 1 mole of C₂H₄(ℓ) burns in a lunar lander, ΔH₁°
= (3/2) ΔHf°[N₂(g)] + 2 ΔHf°[H₂O(g)] - ΔHf°[N₂H₄(ℓ)] - (1/2) ΔHf°[N₂O₄(g)]
= (3/2) (0) + 2 (-241.82) - (50.63) - (1/2) (9.16) kJ
= -538.85 kJ

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b)
N₂H₄(ℓ) + O₂(g) → 2N₂(g) + 2H₂O(g)     ΔH₂°

Enthalpy change when O₂ was used instead of N₂O₄, ΔH₂°
= 2 ΔHf°[N₂(g)] + 2 ΔHf°[H₂O(g)] - ΔHf°[N₂H₄(ℓ)] - ΔHf°[O₂(g)]
= 2 (0) + 2 (-241.82) - (50.63) - 2 (0) kJ
= -534.27 kJ