Answer
Please see the work below.
Work Step by Step
We know that
$10.3g\space Fe_2O_3\times (\frac{1mol Fe_2O_3}{159.69g})\times(\frac{1651KJ}{2mol\space Fe_2O_3})=53.2KJ $
General Chemistry 10th Edition
by Ebbing, Darrell; Gammon, Steven D.
Published by
Cengage Learning
ISBN 10:
1-28505-137-8
ISBN 13:
978-1-28505-137-6
Chapter 6 - Thermochemistry - Questions and Problems - Page 266: 6.137
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