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 $Al_2O_3$}$
