Chemistry II

Calculate the enthalpy for the following reaction: 4 NH3 (g) + 5 O2 (g) into 4 NO (g) + 6 H2O (g).
You may only use the following information:
N2 (g) + O2 (g) into 2 NO (g); Delta Hf = 180.6 kJ
N2 (g) + 3 H2 (g) into 2 NH3 (g); Delta Hf = -91.8 kJ
2 H2 (g) + O2 (g) into 2 H2O (g); Delta Hf = -483.7 kJ

My answer was -910.6 KJ but the real answer was -918.3 KJ. I need to know what I am doing wrong.

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  1. I didn't get either answer.
    Look at your equation to make sure it adds to the desired equation in the problem. Here is what I did.
    equation 1 x 2
    equation 2 reversed and x 2.
    equation 3 x 3.
    Add it. First look at the equation.
    (2N2) + 2O2 + 4NH3 + (6H2) +3O2 ==>
    (2N2) + + (6H2) + 4NO + 6H2O

    The 2N2 and 6H2 cancel since they are on opposite sides of the equation and we are left with
    4NH3 + 5O2 ==>4NO + 6H2O which is exactly what we want.
    Then (180.6*2)+(91.8*2)+(-483.7*3) = -906.3 kJ.
    Check my work.

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