The concentrations of the products at equilibrium are [pcl3] = 0.180 m and [cl2] = 0.250 m. what is the concentration of the reactant, pcl5, at equilibrium?

We have Kc = 4.2 x 10^-2 (given but missing in the question)
and When the balanced equation for this reaction is:
PCl5(g) ↔ PCl3(g) + Cl2(g)
so, according to the Kc formula:
Kc = the concentration of products / the concentration of the reactants
so, to get the concentration of the reactants in equilibrium, the concentration of the products / the concentration of the reactants should equal the Kc value which is given in the question (missing in your question).
So by substitution in Kc formula: 
Kc   = [PCl3]*[Cl2] / [PCl5]
4.2 x 10^-2 =  0.18 * 0.25 /[PCl5]
∴[PCl5] = 0.18*0.25 / 4.2x10^-2 = 1.07
So the concentration of the reactants in equilibrim = 1.07

257 grams h2o i think