A 2:1 mixture by volume of SO2 and O2 at an initial pressure of 15atm is allowed to reach equilibirum at 250°C.2SO2(g) + O2(g) <-> 2SO3(g)The equilibirum pressure was found to be 11atm.(i) Calculate the partial pressure of each gas at equilibrium.How to do this question???

SO₂ and O₂ are in 2:1 ratio by volume and total pressure is 15 atm. So, Pressure of SO₂ is 15 x (2/3) = 10 atm; Pressure of O₂ is 15 x (1/3) = 5 atm

2 X moles of SO₂ combines with X moles of O₂ to form 2X moles of SO₃. So, the total decrease in number of moles is X.

From the given data, X is (15-11) = 4

At equilibrium, partial pressure of SO₂ is 10-2X= 10 – 8 = 2 atm

At equilibrium, partial pressure of O₂ is 5-X= 5 – 4 = 1 atm

At equilibrium, partial pressure of SO₃ is 2X= 8 atm