6.0 moles of Al $335.07\ g$
Work Step by Step
From stoichiometry: $3.0\ mol\ Fe_2O_3\times2\ mol\ Al/1\ mol\ Fe_2O_3=6.0\ mol\ Al$ $3.0\ mol\ Fe_2O_3\times2\ mol\ Fe/1\ mol\ Fe_2O_3\times 55.845\ g/mol\ Fe=335.07\ g$
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