General Chemistry 10th Edition

Published by Cengage Learning
ISBN 10: 1-28505-137-8
ISBN 13: 978-1-28505-137-6

Chapter 14 - Chemical Equilibrium - Questions and Problems - Page 628: 14.65


Please see the work below.

Work Step by Step

We know that $K_c=1.23\times 10^3=\frac{[COCl_2]}{[CO][Cl_2]}$ $[COCl_2]=(1.23\times 10^{23})[CO][Cl_2]$ We plug in the known values to obtain: $[COCl_2]=1.23\times 10^{23}(0.012)(0.025)=0.369M$
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