Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 4 Stoichiometry: Quantitative Information about Chemical Reactions - Study Questions - Page 179c: 26

Answer

$91.9\%$

Work Step by Step

Number of moles of water removed: $(2.634\ g - 2.125\ g)/18.015\ g/mol=0.0282\ mol$ Number of moles of the hydrated salt: $0.0282\ mol/2=0.0141\ mol$ Mass fraction of the hydrated salt: $0.0141\ mol\times 170.482\ g/mol=2.42\ g$ $2.42/2.634\times100\%=91.9\%$
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