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 179l: 132


Mass percent of Fe = $80.51$% Mass percent of Fe$_{2}$O$_{3}$ = $19.49$%

Work Step by Step

Using the chemical equation: MnO$_{4}$$^{-} + 5$Fe$^{2+} +8$H$_{3}$O$^{+}$ $\rightarrow$ Mn$^{2+} + 5$Fe$^{3+} + 12$H$_{2}$O Given that we used $37.50mL$ of $0.04240M$ of KMnO$_{4}$ $\frac{0.03750L}{1}\times\frac{0.04240 mol\space KMnO_{4}}{1 L}\times\frac{5 mol\space Fe^{2+} }{1mol\space KMnO_{4}}\times\frac{1mol\space Fe}{1 mol\space Fe^{2+}}\times\frac{55.8g\space Fe}{1 mol \space Fe} = 0.44361g\space Fe$ Using this we can calculate the mass percent of Fe in the sample: $\frac{ 0.44361g\space Fe}{0.5510g\space sample}\times100 = 80.51$% of Fe Now we can subtract the mass percent of Fe from 100% to get the mass percent of Fe$^{3+}$: $100$% - $80.51$% = $19.49$% of Fe$^{3+}$ or Fe$_{2}$O$_{3}$
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