calculate enthalpy of H for the reaction

N2H4(l) + 2H2O(l) -> N2(g) + 4H2)(l)

Given the reactions
N2H4(l) + O2(g) -> N2(g) + 2H2O(l) Enthalpy of H = -6.22.2 kJ

H2(g) + (1/2)O2(g) -> H2O(l) enthalpy of H = -285.8 kJ/mol

H2(g) + O2(g) -> H2O2(l) enthalpy of H = -187.8 kJ

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1. You need to check the post carefully. The equation you want is not balanced. I think you have made two typos. I think the left H2O should be H2O2 and I think the right H2) should be H2O

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2. You are defiantly right.
it is suppose to be 2H2O2(l) and 4H2O(l)

and the first equation is suppose to be -622.2 kJ/mol

actually they are all suppose to be kJ/mol, but that was a typo on the exercise.

For a final answer I got -818.2 kJ/mol

i used the first equation as is. then i used the second equation and multiplied it by two and then for the last equation i reversed it and also multiplied it by 2.

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3. N2 + 2F2 ---> 2NF3 calculate the standard enthalpy

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