3 N2H4(g)+N2O4(g)�¨ 4 N2(g) + 4H2O(g) + 2 H2(g)

∆H= -1310 kJ (-312 kcal)
How much energy is produced per mole of hydrazine burned?
a. 13.6 kJ (3.25kcal)
b. 40.9 kJ (9.75kcal)
c. 437 kJ (104kcal)
d. 1310 kJ (312 kcal)

My assumption is that this reaction, as written, produces 1310 kJ heat. That's for 3 mols N2H4; therefore, 1 mol would be just 1/3 of that. Right?

To determine how much energy is produced per mole of hydrazine burned, we need to compare the stoichiometric coefficients in the balanced chemical equation with the given enthalpy change (∆H).

From the balanced equation:
3 moles of N2H4 produce 4 moles of N2, 4 moles of H2O, and 2 moles of H2

We are given that the enthalpy change (∆H) for the reaction is -1310 kJ.

To find the energy produced per mole of hydrazine burned, we divide ∆H by the stoichiometric coefficient of N2H4 in the balanced equation (3 moles):

Energy produced per mole of hydrazine burned = ∆H / stoichiometric coefficient of N2H4

Energy produced = -1310 kJ / 3

Calculating the result:
Energy produced = -436.7 kJ

Since the given options are in kJ, we round the result to the nearest kJ:
Energy produced = -437 kJ

Therefore, the correct option is c. 437 kJ (104 kcal).