Answer
See answer below.
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: $20.0-3.40=16.60\ g\ Al$
$\begin{smallmatrix}
&2\ Al&Fe_2O_3&\rightarrow&2\ Fe&Al_2O_3\\
Initial &0.741&0.063&&0&0\\
Reacted&-0.126&-0.063&&+0.126&+0.063\\
Final&0.615&0&&0.126&0.063\\
\end{smallmatrix}$
