Equilibrium constant problem... Please help, kinda difficult?

Question

You're given the eqn 3O2 + 2N2O <--> 4NO2
You're also given .016 mol O2 and .082 mol N2O in a 1 Liter vessel. They react to form NO2. At equilibirum, the concetration for NO2 was .015 M.

Find the equlibirum constant?

I did the ICE thing..
3O2 + 2N2O <--> 4NO2
initial: .016 M .082 M 0 M
change: -3x -2x +4x
@equilibrum: ___ ___ .015 M
x = .00375 right?
So what the heck do you do after?...

Because the answer apparently is K = 85.

Answer 1

True, x = 0.00375, so the concentrations at equilibrium are:

[O2] = 0.00475 M
[N2O] = 0.0725 M
[NO2] = 0.015 M

K = [NO2]^4/([O2]^3*[N2O]^2
K = (0.015)^4/(0.00475)^3(0.0725)^2
K = 5.062 * 10^-8/5.633 * 10^-10
K = 89.8

It's in the ballpark.