Chem Help?

_____________ X₂ ___ + ___ Y₂ ___ ⇌  ___ 2XY___ Kc = 64.0
Initial (M): __ 0.250 ______ 0.250 _________ 0
Change (M): __ -y ________ -y ___________ +2y
Eqm (M): __ 0.250 - y ___ 0.250 - y ________ 2y

At equilibrium:
Kc = [XY]² / ([X₂] [Y₂])
64.0 = (2y)² / (0.250 - y)²
2y / (0.250 - y) = √64.0
2y / (0.250 - y) = 8
2y = 8(0.250 - y)
2y = 2 - 8y
10y = 2
y = 0.200

At equilibrium:
[X₂] = (0.250 - 0.200) M = 0.050 M
[Y₂] = (0.250 - 0.200) M = 0.050 M
[XY] = 2 * 200 M = 0.400 M

Write the expression for Kc:

Kc = [XY]^2 / [X2][Y2] = 64.0

Here it will help to set up an ICE table:

..............[X2]............[Y2]...............[XY]
Initial......0.250 ......0.250...............0
Change...-x..............-x.................+2x
Equil........0.25-x......0.25-x...........2x

Plug the equilibrium concentrations into the expression for Kc:

Kc = 64.0 = (2x)^2 / (0.250-x)^2

In this case, since both the numerator and denominator are squares, the easiest first step in solving this is to take the square root of both sides of the equation. This gives:

8.0 = 2x / (0.25-x)

I'll let you do the arithmetic to find x. Once you do, calculate the equilibrium concentrations of the reactants and product as shown in the ICE table.