We are asked to **calculate the equilibrium concentration of PCl _{5}(g) and PCl_{3}(g)**

We’re given the following equilibrium reaction:

PCl_{5(g) }⇌ PCl_{3(g)} + Cl_{2(g)}; K_{c} = 1.80 at 250°C

We know that the initial amount of PCl_{5} is 0.250 mol. In a 2.50 L container, **the concentration of each is:**

$\overline{){\mathbf{Molarity}}{\mathbf{}}{\mathbf{\left(}}{\mathbf{M}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}}{\mathbf{L}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}}}$

${\mathbf{M}}_{{\mathbf{PCl}}_{\mathbf{5}}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{250}\mathbf{}\mathbf{mol}}{\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{L}}$

**M _{PCl5} = 0.1**

Phosphorus pentachloride decomposes according to the chemical equation.

PCl_{5}(g)_{ }⇌ PCI_{3}(g) + Cl_{2}(g) _{ }K_{c} = 1.80 at 250 °C

A 0.250 mol sample of PCl_{5}(g) is injected into an empty 2.50 L reaction vessel held at 250 °C. Calculate the concentrations of PCl_{5}(g) and PCl_{3}(g) at equilibrium.

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