## Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall

# Chapter 3 - Sections 3.1-3.12 - Exercises - Problems by Topic: 84b

#### Answer

$19kg\ Cl$

#### Work Step by Step

Divide the compound by its molar mass to get the number of moles of the compound. Multiply by the ratio of moles of the compound to moles of chlorine to get the number of moles of sodium. Multiply by the molar mass of chlorine ($\frac{35.45g}{1mol}$) to get the mass of sodium in grams.Divide by 1000 to get the mass in kilograms. $25kg=25,000g$ $25,000g\ CFCl_{3} \times\frac{1mol\ CFCl_{3}}{137.37g\ CFCl_{3}}\times\frac{3}{1}\times\frac{35.45g\ Cl}{1mol\ Cl}\times\frac{1kg}{1000g}= 19kg\ Cl$

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