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A 0.366 mol sample of pcl5(g) is injected into an empty 4.45 l reaction vessel held at 250 °c. calculate the concentrations of pcl5(g) and pcl3(g) at equilibrium.

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Chemical reaction: PCl₅ → PCl₃ + Cl₂.
n(PCl₅) = 0,366 mol.
V(PCl₅) = 4,45 L.
c(PCl₅) = n(PCl₅) ÷ V(PCl₅).
c(PCl₅) = 0,366 mol ÷ 4,45 L.
c(PCl₅) = 0,082 mol/L.
Kc = 1,80.
[PCl₃] = [Cl₂] = x.
Kc = [PCl₃] · [Cl₂] ÷ [PCl₅].
1,80 = x² ÷ (0,082 mol/L - x).
Solve quadratic eqaution: x = [PCl₃] = 0,078 mol/L.
[PCl₅] = 0,082 mol/L - 0,078 mol/L.
[PCl₅] = 0,004 mol/L.
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