Finding moles of a specific product at equilibrium constant?

Question

For the reaction:

2H₂O(g) ⇌ 2H₂(g) + O₂(g)

K_C = 2.40 x 10^-3 at a given temperature.

At equilibrium, it is found that [H₂O] = 0.205 M and [H₂] = 1.20 x 10^-2 M.

If the reaction is run in a 7.00 L container, how many moles of O₂ are present at equilibrium?

=_____ moles O₂

Answer 1

2H₂O(g) ===> 2H₂(g) + O₂(g)

0.205...............0.012 ...............

K = [O₂][H₂]² / [H₂O]²

2.40x10^-3 = (x)(0.022)² / (0.205)²

x = 1.00x10^-4 / 4.84x10^-4

x = 0.2066 mol/L

moles O₂ = 0.207 mol/L x 7.00 L = 1.45 moles