Question

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?

Answer 1

i set up an ICE table then got:
10^80 = [(12+2x)^2 (12+6x)^6]/[(12-4x)^4 (12-3x)^3]

But then it looks waaay to complicated. Am I doing it wrong?
Please help. Thanks

Answer 2

see other post.