Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 3 - Stoichiometry - Exercises - Page 129: 71


a. $165.403\text{ grams}$ b. $3.023\text{ mol $C_2H_3Cl_3O_2$}$ c. $5.5*10^{22}\text{ atoms Cl}$ d. $5.5*10^{22}\text{ atoms Cl}$ e. $1.6\text{ grams $C_2H_3Cl_3O_2$}$ f. $1.37*10^{-19}\text{ grams $C_2H_3Cl_3O_2$}$

Work Step by Step

a. $\text{ Molar Mass} = 2*12.0107+3*1.00794+3*35.453+2*15.9994=165.403\text{ grams}$ b. $500.\text{ grams $C_2H_3Cl_3O_2$}*\frac{1\text{ mol $C_2H_3Cl_3O_2$}}{165.403\text{ grams $C_2H_3Cl_3O_2$}}=3.023\text{ mol $C_2H_3Cl_3O_2$}$ c. $2.0*10^{-2}\text{ mol $C_2H_3Cl_3O_2$}*\frac{165.403\text{ grams $C_2H_3Cl_3O_2$}}{1\text{ mol $C_2H_3Cl_3O_2$}}=3.3\text{ grams $C_2H_3Cl_3O_2$}$ d. $5.0\text{ grams $C_2H_3Cl_3O_2$}*\frac{1\text{ mol $C_2H_3Cl_3O_2$}}{165.403\text{ grams $C_2H_3Cl_3O_2$}}*\frac{3\text{ mol Cl}}{1\text{ mol $C_2H_3Cl_3O_2$}}*\frac{6.0221*10^{23}\text{ atoms Cl}}{1\text{ mol Cl}}=5.5*10^{22}\text{ atoms Cl}$ e. $1.0\text{ grams Cl}*\frac{1\text{ mol Cl}}{35.453\text{ grams Cl}}*\frac{1\text{ mol $C_2H_3Cl_3O_2$}}{3\text{ mol Cl}}*\frac{165.403\text{ grams $C_2H_3Cl_3O_2$}}{1\text{ mol $C_2H_3Cl_3O_2$}}=1.6\text{ grams $C_2H_3Cl_3O_2$}$ f. $500.\text{ molecules $C_2H_3Cl_3O_2$}*\frac{1\text{ mol $C_2H_3Cl_3O_2$}}{6.0221*10^{23}\text{ molecules $C_2H_3Cl_3O_2$}}*\frac{165.403\text{ grams $C_2H_3Cl_3O_2$}}{1\text{ mol $C_2H_3Cl_3O_2$}}=1.37*10^{-19}\text{ grams $C_2H_3Cl_3O_2$}$
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.