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For the reaction: PCl3(g) + Cl2(g) ⇌ PCl5(g) at 70.5°C, Kp = 1.05. If one starts...

For the reaction: PCl3(g) + Cl2(g) ⇌ PCl5(g) at 70.5°C, Kp = 1.05. If one starts with 1.80 atm pressure of PCl3(g), 1.72 atm pressure of Cl2(g), and no PCl5(g), what is the partial pressure of PCl5(g) at equilibrium?

The answer is 0.856 atm

can you explain me how to solve this question?

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