A0.210 mol sample of pcl5(g) is injected into an empty 2.45 l reaction vessel held at 250 â°c. calculate the concentrations of pcl5(g) and pcl3(g) at equilibrium.

When the balanced equation for this reaction is:
                        PCl5(g)       ↔    PCl3(g) + Cl2(g)
initial  C           (0.21/2.45)              0                 0
change                -X                      +X              +X
equilibruimC  (0.21/2.45)-X           X                 X

So by substitution in Ka formula: when we have Ka at 250°C = 1.8 (must be given- missing in your question)
Ka = [PCl3][Cl2]/[PCl5]
1.8 = (X)(X) / (0.21/2.45)-X
1.8*0.086- 1.8X = X^2
X^2 + 1.8X - 0.1548 = 0 by solving this equation
X= 0.082 mol
∴[PCl5] = (0.21/2.45) - X
             = 0.0857 - 0.082 = 0.0037 mol
∴[PCl3] = X = 0.082 mol