Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 15 - Chemical Equilibrium - Exercises - Page 665: 15.49

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
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