Answer
9.3 g of $HBr$ are needed to dissolve 3.2 g of pure iron.
This reaction produces 0.12 g of $H_2$
Work Step by Step
1.
$ Fe $ : 55.85 g/mol
$$ \frac{1 \space mole \space Fe }{ 55.85 \space g \space Fe } \space and \space \frac{ 55.85 \space g \space Fe }{1 \space mole \space Fe }$$
$ HBr $ : ( 79.90 $\times$ 1 )+ ( 1.008 $\times$ 1 )= 80.91 g/mol
$$ \frac{1 \space mole \space HBr }{ 80.91 \space g \space HBr } \space and \space \frac{ 80.91 \space g \space HBr }{1 \space mole \space HBr }$$
$$ 3.2 \space g \space Fe \times \frac{1 \space mole \space Fe }{ 55.85 \space g \space Fe } \times \frac{ 2 \space moles \space HBr }{ 1 \space mole \space Fe } \times \frac{ 80.91 \space g \space HBr }{1 \space mole \space HBr } = 9.3 \space g \space HBr $$
2.
$$ \frac{1 \space mole \space H_2 }{ 2.016 \space g \space H_2 } \space and \space \frac{ 2.016 \space g \space H_2 }{1 \space mole \space H_2 }$$
$$ 3.2 \space g \space Fe \times \frac{1 \space mole \space Fe }{ 55.85 \space g \space Fe } \times \frac{ 1 \space mole \space H_2 }{ 1 \space mole \space Fe } \times \frac{ 2.016 \space g \space H_2 }{1 \space mole \space H_2 } = 0.12 \space g \space H_2 $$
