Answer
a. 3.66 g of $H_2O$ are necessary to react with 28.0 g of $NO_2$.
b. 3.44 g of $NO$ are produced from 15.8 g of $NO_2$.
c. 7.53 g of $HNO_3$ are produced from 8.25 g of $NO_2$.
Work Step by Step
a. $$ \frac{ 3 \space moles \space NO_2 }{ 1 \space mole \space H_2O } \space and \space \frac{ 1 \space mole \space H_2O }{ 3 \space moles \space NO_2 }$$
$ NO_2 $ : ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 46.01 g/mol
$$ \frac{1 \space mole \space NO_2 }{ 46.01 \space g \space NO_2 } \space and \space \frac{ 46.01 \space g \space NO_2 }{1 \space mole \space NO_2 }$$
$ H_2O $ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol
$$ \frac{1 \space mole \space H_2O }{ 18.02 \space g \space H_2O } \space and \space \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O }$$
$$ 28.0 \space g \space NO_2 \times \frac{1 \space mole \space NO_2 }{ 46.01 \space g \space NO_2 } \times \frac{ 1 \space mole \space H_2O }{ 3 \space moles \space NO_2 } \times \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O } = 3.66 \space g \space H_2O $$
b. $$ \frac{ 3 \space moles \space NO_2 }{ 1 \space mole \space NO } \space and \space \frac{ 1 \space mole \space NO }{ 3 \space moles \space NO_2 }$$
$ NO_2 $ : ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 46.01 g/mol
$$ \frac{1 \space mole \space NO_2 }{ 46.01 \space g \space NO_2 } \space and \space \frac{ 46.01 \space g \space NO_2 }{1 \space mole \space NO_2 }$$
$ NO $ : ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 30.01 g/mol
$$ \frac{1 \space mole \space NO }{ 30.01 \space g \space NO } \space and \space \frac{ 30.01 \space g \space NO }{1 \space mole \space NO }$$
$$ 15.8 \space g \space NO_2 \times \frac{1 \space mole \space NO_2 }{ 46.01 \space g \space NO_2 } \times \frac{ 1 \space mole \space NO }{ 3 \space moles \space NO_2 } \times \frac{ 30.01 \space g \space NO }{1 \space mole \space NO } = 3.44 \space g \space NO $$
c. $$ \frac{ 3 \space moles \space NO_2 }{ 2 \space moles \space HNO_3 } \space and \space \frac{ 2 \space moles \space HNO_3 }{ 3 \space moles \space NO_2 }$$
$ NO_2 $ : ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 46.01 g/mol
$$ \frac{1 \space mole \space NO_2 }{ 46.01 \space g \space NO_2 } \space and \space \frac{ 46.01 \space g \space NO_2 }{1 \space mole \space NO_2 }$$
$ HNO_3 $ : ( 1.008 $\times$ 1 )+ ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 3 )= 63.02 g/mol
$$ \frac{1 \space mole \space HNO_3 }{ 63.02 \space g \space HNO_3 } \space and \space \frac{ 63.02 \space g \space HNO_3 }{1 \space mole \space HNO_3 }$$
$$ 8.25 \space g \space NO_2 \times \frac{1 \space mole \space NO_2 }{ 46.01 \space g \space NO_2 } \times \frac{ 2 \space moles \space HNO_3 }{ 3 \space moles \space NO_2 } \times \frac{ 63.02 \space g \space HNO_3 }{1 \space mole \space HNO_3 } = 7.53 \space g \space HNO_3 $$
