Chemistry
posted by Scott L. .
Consider the reaction below:
4NH3(g) + 3O2(g) > 2N2 + 6H20, K=10^80 @ certain temp
initially, all the reactants and products have concentrations equal to 12M. At equilibrium, what is the approximate concentration of oxygen?

i set up an ICE table then got:
10^80 = [(12+2x)^2 (12+6x)^6]/[(124x)^4 (123x)^3]
But then it looks waaay to complicated. Am I doing it wrong?
Please help. Thanks 
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