Answer
3.62 x 10^-2g
Work Step by Step
Kc = 3.1 x 10^-5
Mass of I at equilibrium = 2.67 x 10^-2g
Moles of I = 2.67 x 10-2g/126.9(g/mol) = 2.104 x 10^-4mol
Concentration of I = 2.104 x 10^-4mol/10.0L = 2.104 x 10^-5M
Concentration of I2
KC = [I]^2/[I2]
3.1 x 10^-5 = [2.104 x 10^-5M]/I2
[I2] = 1.428 x 10^-5
Mass of I2
Moles of I2 = (1.428 x 10^-5M)(10.0L) = 1.428 x 10^-4
(1.428 x 10^-4)(253.8g/mol) = 3.62 x 10^-2g