Answer
82.9 g of $H_2SO_4$ and 1.70 g of $H_2$
Work Step by Step
1. $ Al $ : 26.98 g/mol
$$ \frac{1 \space mole \space Al }{ 26.98 \space g \space Al } \space and \space \frac{ 26.98 \space g \space Al }{1 \space mole \space Al }$$
$ H_2SO_4 $ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 4 )+ ( 32.07 $\times$ 1 )= 98.09 g/mol
$$ \frac{1 \space mole \space H_2SO_4 }{ 98.09 \space g \space H_2SO_4 } \space and \space \frac{ 98.09 \space g \space H_2SO_4 }{1 \space mole \space H_2SO_4 }$$
$$ 15.2 \space g \space Al \times \frac{1 \space mole \space Al }{ 26.98 \space g \space Al } \times \frac{ 3 \space moles \space H_2SO_4 }{ 2 \space moles \space Al } \times \frac{ 98.09 \space g \space H_2SO_4 }{1 \space mole \space H_2SO_4 } = 82.9 \space g \space H_2SO_4 $$
2. $ H_2 $ : ( 1.008 $\times$ 2 )= 2.016 g/mol
$$ \frac{1 \space mole \space H_2 }{ 2.016 \space g \space H_2 } \space and \space \frac{ 2.016 \space g \space H_2 }{1 \space mole \space H_2 }$$
$$ 15.2 \space g \space Al \times \frac{1 \space mole \space Al }{ 26.98 \space g \space Al } \times \frac{ 3 \space moles \space H_2 }{ 2 \space moles \space Al } \times \frac{ 2.016 \space g \space H_2 }{1 \space mole \space H_2 } = 1.70 \space g \space H_2 $$
