Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 3 - Sections 3.1-3.12 - Exercises - Cumulative Problems - Page 135: 124

Answer

$59.0kg\ \frac{Cl}{year}$

Work Step by Step

If the car emits $12kg\ CHF_{2}Cl$ a month then it will emit $12×12=144kg\ CHF_{2}Cl$ or $144,000g\ CHF_{2}Cl$ in a year. Divide this number by the molar mass of $CHF_{2}Cl$ to get the number of moles of $CHF_{2}Cl$. The multiply by the molar mass of Cl to convert to grams. Divide by 1000 to convert back to kilograms. $144,000g\ CHF_{2}Cl\times\frac{1 mol\ CHF_{2}Cl}{86.47g\ CHF_{2}Cl}\times\frac{35.45g\ Cl}{1mol\ Cl}\times\frac{1}{1000}=59.0kg\ Cl$
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