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

Answer

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: $[COCl_2]=1.23\times 10^{23}(0.012)(0.025)=0.369M$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.

GradeSaver will pay $15 for your literature essays
GradeSaver will pay $25 for your college application essays
GradeSaver will pay $50 for your graduate school essays – Law, Business, or Medical
GradeSaver will pay $10 for your Community Note contributions
GradeSaver will pay $500 for your Textbook Answer contributions