chemistry hw?

Suppose that the samples are placed in a 1.00 L reaction vessel instead.

_           PCl₃(g)    +    Cl₂(g)    ⇌     PCl₅(g)        Kc = 1.80
Initial:     1.5 M            1 M                0 M
Change:   -y M           -y M               +y M
Eqm:   (1.5 - y) M    (1 - y) M             y M

At equilibrium:
Kc = [PCl₅] / ([PCl₃] [Cl₂])
1.8 = y / [(1.5 - y)(1 - y)]
1.8 = y / (y² - 2.5y + 1.5)
1.8(y² - 2.5y + 1.5) = y
1.8y² - 5.5y + 2.7 = 0
y = [5.5 ± √(5.5² - 4×1.8×2.7)] / (2×1.8)
y = 2.44 (rejected) or y = 0.614

At equilibrium:
[PCl₃] = (1.5 - 0.614) M = 0.886 M
[Cl₂] = (1 - 0.614) M = 0.386 M
[PCl₅] = 0.614 M

I'm going to assume that you meant that the vessel was 1.00 L in volume. 

Kc = [PCl5]/[PCl3][Cl2] = 1.80

Let equilibrium concentration of PCl5 = x. Then, [PCl3] = 1.50-x and [Cl2] = 1.00-x

Then,
Kc = 1.80 = x / (1.50-x)(1.00-x)

Now, I'll leave the math up to you. You will need to rearrange this equation in the a quadratic with the form ax^2 + bx + c = 0 and use the quadratic formula to find x. Once you have x, you can easily calculate the concentrations.