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 179b: 17


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Work Step by Step

Number of moles of the reactants: Aluminum: $20.0\ g\div 26.983 g/mol=0.741\ mol$ Iron (III) oxide: $10.0\ g\div 159.69\ g/mol=0.063\ mol$ Ratio: $0.741/0.063=11.8$ Iron (III) oxide is the limiting reactant. From stoichiometry: $0.063\ mol\times 2_{Fe}/1_{Fe_2O_3} \times 55.845\ g/mol\ Fe=7.04\ g\ Fe$ Aluminum reacted: $0.063\ mol\times 2_{Al}/1_{Fe_2O_3} \times 26.982\ g/mol\ Al=3.40\ g\ Al$ Remaining: $20.0-3.40=16.60\ g\ Al$ $\begin{smallmatrix} &2\ Al&Fe_2O_3&\rightarrow&2\ Fe&Al_2O_3\\ Initial &0.741&0.063&&0&0\\ Reacted&-0.126&-0.063&&+0.126&+0.063\\ Final&0.615&0&&0.126&0.063\\ \end{smallmatrix}$
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