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Calculate the delta H for the reaction
N2H4(l)+ O2(g) --> N2(g) + 2H20(l)

Given the following data:

2NH3(g) + 3N20(g) --> 4N2(g) + 3H20(l)
N20(g) + 3H2(g) --> N2H4(l) + H20(l)
2NH3(g) + 1/2O2(g) --> N2H4(l) + H20(l)
H2(g) + 1/2O2(g) --> H20(l)

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    Reverse equation 3.
    Add equation 1.
    Reverse and multiply equation 2 by 3 and add.
    Multiply equation 4 by 9 and add.
    Don't forget to multiply delta H by appropriate multipliers and don't forget to change the sign of those equations reversed. (You didn't show any of the delta H values).
    You will end up with
    4 N2H4 + 4O2 ==> 4N2 + 8H2O so all of the delta H values (multiplied and sign changed as appropriate) will be 4 times the value of delta H for 1 mole (instead of the four in the equation).

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