Answer
There must be used 21.5 g of $Fe_2O_3$ and 7.26 g of $Al$.
This reaction produces 13.7 g of $Al_2O_3$.
Work Step by Step
$ Fe $ : 55.85 g/mol
$$ \frac{1 \space mole \space Fe }{ 55.85 \space g \space Fe } \space and \space \frac{ 55.85 \space g \space Fe }{1 \space mole \space Fe }$$
$ Fe_2O_3 $ : ( 55.85 $\times$ 2 )+ ( 16.00 $\times$ 3 )= 159.70g/mol
$$ \frac{1 \space mole \space Fe_2O_3 }{ 159.70 \space g \space Fe_2O_3 } \space and \space \frac{ 159.70 \space g \space Fe_2O_3 }{1 \space mole \space Fe_2O_3 }$$
$$ 15.0 \space g \space Fe \times \frac{1 \space mole \space Fe }{ 55.85 \space g \space Fe } \times \frac{ 1 \space mole \space Fe_2O_3 }{ 2 \space moles \space Fe } \times \frac{ 159.70 \space g \space Fe_2O_3 }{1 \space mole \space Fe_2O_3 } = 21.5 \space g \space Fe_2O_3 $$
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$ Al $ : 26.98 g/mol
$$ \frac{1 \space mole \space Al }{ 26.98 \space g \space Al } \space and \space \frac{ 26.98 \space g \space Al }{1 \space mole \space Al }$$
$$ 15.0 \space g \space Fe \times \frac{1 \space mole \space Fe }{ 55.85 \space g \space Fe } = 0.269 \space mole \space Fe$$ $$0.269 \space mole \space Fe \times \frac{ 2 \space moles \space Al }{ 2 \space moles \space Fe } \times \frac{ 26.98 \space g \space Al }{1 \space mole \space Al } = 7.26 \space g \space Al $$
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$ Al_2O_3 $ : ( 26.98 $\times$ 2 )+ ( 16.00 $\times$ 3 )= 101.96 g/mol
$$ \frac{1 \space mole \space Al_2O_3 }{ 101.96 \space g \space Al_2O_3 } \space and \space \frac{ 101.96 \space g \space Al_2O_3 }{1 \space mole \space Al_2O_3 }$$
$$ 15.0 \space g \space Fe \times \frac{1 \space mole \space Fe }{ 55.85 \space g \space Fe } \times \frac{ 1 \space mole \space Al_2O_3 }{ 2 \space moles \space Fe } \times \frac{ 101.96 \space g \space Al_2O_3 }{1 \space mole \space Al_2O_3 } = 13.7 \space g \space Al_2O_3 $$
