Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 3 - Stoichiometry - Exercises: 105

Answer

$21.4\text{ grams $Fe_2O_3$}$ $7.25\text{ grams Al}$ $13.7\text{ grams $Al_2O_3$}$

Work Step by Step

$15.0\text{ grams Fe}*\frac{1\text{ mol Fe}}{55.845\text{ grams Fe}}*\frac{1\text{ mol $Fe_2O_3$}}{2\text{ mol Fe}}*\frac{159.6882\text{ grams $Fe_2O_3$}}{1\text{ mol $Fe_2O_3$}}=21.4\text{ grams $Fe_2O_3$}$ $15.0\text{ grams Fe}*\frac{1\text{ mol Fe}}{55.845\text{ grams Fe}}*\frac{2\text{ mol Al}}{2\text{ mol Fe}}**\frac{26.981538\text{ grams Al}}{1\text{ mol Al}}=7.25\text{ grams Al}$ $15.0\text{ grams Fe}*\frac{1\text{ mol Fe}}{55.845\text{ grams Fe}}*\frac{1\text{ mol $Al_2O_3$}}{2\text{ mol Fe}}*\frac{101.96128\text{ grams $Al_2O_3$}}{1\text{ mol $Al_2O_3$}}=13.7\text{ grams $Al_2O_3$}$
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