Ask a New Question
Menu

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?

2 answers

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
see other post.
Similar Questions
  1. Consider the reaction below:4NH3(g) + 3O2(g) -> 2N2 + 6H20, K=10^80 @ certain temp initially, all the reactants and products
    1. answers icon 2 answers
  2. 4NH3(g) + 3O2(g) -> 2N2 + 6H20, K=10^80 @ certain tempinitially, all the reactants and products have concentrations equal to
    1. answers icon 2 answers
  3. 4NH3(g) + 3O2(g) -> 2N2 + 6H20, K=10^80 @ certain tempinitially, all the reactants and products have concentrations equal to
    1. answers icon 1 answer
  4. 4NH3(g) + 3O2(g) -> 2N2 + 6H20, K=10^80 @ certain tempinitially, all the reactants and products have concentrations equal to
    1. answers icon 0 answers
more similar questions