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
