Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 3 - Section 3.9 - Composition of Compounds - For Practice - Page 111: 3.15


83.9 g of iron(III) oxide.

Work Step by Step

Since iron (III) oxide is 69.94% iron by mass, each 100% of this compound has that value of iron. So, we can use this as a conversion factor to calculate the mass of iron (III) oxide: $58.7g (Iron)\times \frac{100\% (Iron(III)-Oxide)}{69.94\% (Iron)} = 83.9$ $g (Iron(III)-Oxide)$
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