[kgor93]

Br2(g) <--> 2 Br(g)

Kc = 1.04 * 10^-3. If a 0.200L container has 4.53 * 10^-2 moles of Br2 at equilibrium, how many moles of Br are present?

The way I set it up is to multiply the 4.53*10^-2 by 5 to get it in terms of moles/L.
Then,
(x^2)/.2265=1.04 * 10^-3
x is the amount of moles present. But, it's not the right answer. The correct answer should be something like 3.04 * 10^-4. What am I doing wrong?

[Tdha]

I suppose, the "correct answer" isn't the right answer.
Your "x" is the concentration of the dissociated Br, not the amount of dissociated bromine.
K_c=(Br)²/(Br₂).
c(Br)=n(Br)/V. We know V and c(Br), so you can calculate n(Br).