J.R. S. answered 03/25/20
Ph.D. University Professor with 10+ years Tutoring Experience
Concentration of PCl5 initially = 0.2215 mol/2.80 L = 0.0791 M
PCl5(g) ===> PCl3(g) + Cl2(g)
0.0791.............0..............0..........Initial
-x.....................+x............+x.......Change
0.0791-x...........x.............x.........Equilibrium
Kc = 1.80 = [PCl3][Cl2]/[PCl5]
1.80 = (x)(x)/0.0791 - x
0.142 - 1.80x = x2
x2 + 1.80x -0.142 = 0
x = 0.076 M
At equilibrium...
[PCl5] = 0.0791 - 0.076 = 0.0031 M
[PCl3] = 0.076 M