Answer
a. 19.2 g of $O_2$ are necessary to react with 13.6 g of $NH_3$
b. 3.79 g of $N_2$ are produced from the reaction of 6.50 g of $O_2$
c. 54.0 g of $H_2O$ are formed from the reaction of 34.0 g of $NH_3$
Work Step by Step
a. $$ \frac{ 4 \space moles \space NH_3 }{ 3 \space moles \space O_2 } \space and \space \frac{ 3 \space moles \space O_2 }{ 4 \space moles \space NH_3 }$$
$ NH_3 $ : ( 1.008 $\times$ 3 )+ ( 14.01 $\times$ 1 )= 17.03 g/mol
$$ \frac{1 \space mole \space NH_3 }{ 17.03 \space g \space NH_3 } \space and \space \frac{ 17.03 \space g \space NH_3 }{1 \space mole \space NH_3 }$$
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
$$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$
$$ 13.6 \space g \space NH_3 \times \frac{1 \space mole \space NH_3 }{ 17.03 \space g \space NH_3 } \times \frac{ 3 \space moles \space O_2 }{ 4 \space moles \space NH_3 } \times \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 } = 19.2 \space g \space O_2 $$
b. $$ \frac{ 3 \space moles \space O_2 }{ 2 \space moles \space N_2 } \space and \space \frac{ 2 \space moles \space N_2 }{ 3 \space moles \space O_2 }$$
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
$$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$
$ N_2 $ : ( 14.01 $\times$ 2 )= 28.02 g/mol
$$ \frac{1 \space mole \space N_2 }{ 28.02 \space g \space N_2 } \space and \space \frac{ 28.02 \space g \space N_2 }{1 \space mole \space N_2 }$$
$$ 6.50 \space g \space O_2 \times \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \times \frac{ 2 \space moles \space N_2 }{ 3 \space moles \space O_2 } \times \frac{ 28.02 \space g \space N_2 }{1 \space mole \space N_2 } = 3.79 \space g \space N_2 $$
c. $$ \frac{ 4 \space moles \space NH_3 }{ 6 \space moles \space H_2O } \space and \space \frac{ 6 \space moles \space H_2O }{ 4 \space moles \space NH_3 }$$
$ NH_3 $ : ( 1.008 $\times$ 3 )+ ( 14.01 $\times$ 1 )= 17.03 g/mol
$$ \frac{1 \space mole \space NH_3 }{ 17.03 \space g \space NH_3 } \space and \space \frac{ 17.03 \space g \space NH_3 }{1 \space mole \space NH_3 }$$
$ H_2O $ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol
$$ \frac{1 \space mole \space H_2O }{ 18.02 \space g \space H_2O } \space and \space \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O }$$
$$ 34.0 \space g \space NH_3 \times \frac{1 \space mole \space NH_3 }{ 17.03 \space g \space NH_3 } \times \frac{ 6 \space moles \space H_2O }{ 4 \space moles \space NH_3 } \times \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O } = 54.0 \space g \space H_2O $$
