Answer
This reaction can produce 70.9 g of $Al_2O_3$.
Work Step by Step
$ Al $ : 26.98 g/mol
$$ \frac{1 \space mole \space Al }{ 26.98 \space g \space Al } \space and \space \frac{ 26.98 \space g \space Al }{1 \space mole \space Al }$$
$ Al_2O_3 $ : ( 26.98 $\times$ 2 )+ ( 16.00 $\times$ 3 )= 101.96 g/mol
$$ \frac{1 \space mole \space Al_2O_3 }{ 101.96 \space g \space Al_2O_3 } \space and \space \frac{ 101.96 \space g \space Al_2O_3 }{1 \space mole \space Al_2O_3 }$$
$$ 50.0 \space g \space Al \times \frac{1 \space mole \space Al }{ 26.98 \space g \space Al } \times \frac{ 2 \space moles \space Al_2O_3 }{ 4 \space moles \space Al } \times \frac{ 101.96 \space g \space Al_2O_3 }{1 \space mole \space Al_2O_3 } = 94.5 \space g \space Al_2O_3 $$
$$actual \space yield =\frac{ percent \space yield \times theoretical \space yield }{100\%}$$$$actual \space yield =\frac{( 75.0 \%)\times ( 94.5 \space g \space Al_2O_3 )}{100\%} = 70.9 \space g \space Al_2O_3 $$