Answer
Please see the work below.
Work Step by Step
We know that
$20.4g\space Al\times \frac{1mol\space Al}{26.98g\space Al}\times \frac{3mol\space I_2}{2mol\space Al}\times \frac{253.8g\space I_2}{1mol\space I_2}=288g\space I_2$
Chemistry 12th Edition
by Chang, Raymond; Goldsby, Kenneth
Published by
McGraw-Hill Education
ISBN 10:
0078021510
ISBN 13:
978-0-07802-151-0
Chapter 3 - Mass Relationships in Chemical Reactions - Questions & Problems - Page 108: 3.47
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