Answer
a. 3.72 g of $C$
b. 84.0 g of $CO$
c. 4.20 g of $Fe$
Work Step by Step
a. $$ \frac{ 1 \space mole \space Fe_2O_3 }{ 3 \space moles \space C } \space and \space \frac{ 3 \space moles \space C }{ 1 \space mole \space Fe_2O_3 }$$
$ Fe_2O_3 $ : ( 55.85 $\times$ 2 )+ ( 16.00 $\times$ 3 )= 159.70 g/mol
$$ \frac{1 \space mole \space Fe_2O_3 }{ 159.70 \space g \space Fe_2O_3 } \space and \space \frac{ 159.70 \space g \space Fe_2O_3 }{1 \space mole \space Fe_2O_3 }$$
$ C $ : ( 12.01 $\times$ 1 )= 12.01 g/mol
$$ \frac{1 \space mole \space C }{ 12.01 \space g \space C } \space and \space \frac{ 12.01 \space g \space C }{1 \space mole \space C }$$
$$ 16.5 \space g \space Fe_2O_3 \times \frac{1 \space mole \space Fe_2O_3 }{ 159.70 \space g \space Fe_2O_3 } \times \frac{ 3 \space moles \space C }{ 1 \space mole \space Fe_2O_3 } \times \frac{ 12.01 \space g \space C }{1 \space mole \space C } = 3.72 \space g \space C $$
b. $$ \frac{ 3 \space moles \space C }{ 3 \space moles \space CO } \space and \space \frac{ 3 \space moles \space CO }{ 3 \space moles \space C }$$
$ C $ : ( 12.01 $\times$ 1 )= 12.01 g/mol
$$ \frac{1 \space mole \space C }{ 12.01 \space g \space C } \space and \space \frac{ 12.01 \space g \space C }{1 \space mole \space C }$$
$ CO $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 28.01 g/mol
$$ \frac{1 \space mole \space CO }{ 28.01 \space g \space CO } \space and \space \frac{ 28.01 \space g \space CO }{1 \space mole \space CO }$$
$$ 36.0 \space g \space C \times \frac{1 \space mole \space C }{ 12.01 \space g \space C } \times \frac{ 3 \space moles \space CO }{ 3 \space moles \space C } \times \frac{ 28.01 \space g \space CO }{1 \space mole \space CO } = 84.0 \space g \space CO $$
c. $$ \frac{ 1 \space mole \space Fe_2O_3 }{ 2 \space moles \space Fe } \space and \space \frac{ 2 \space moles \space Fe }{ 1 \space mole \space Fe_2O_3 }$$
$ Fe_2O_3 $ : ( 55.85 $\times$ 2 )+ ( 16.00 $\times$ 3 )= 159.70 g/mol
$$ \frac{1 \space mole \space Fe_2O_3 }{ 159.70 \space g \space Fe_2O_3 } \space and \space \frac{ 159.70 \space g \space Fe_2O_3 }{1 \space mole \space Fe_2O_3 }$$
$ Fe $ : ( 55.85 $\times$ 1 )= 55.85 g/mol
$$ \frac{1 \space mole \space Fe }{ 55.85 \space g \space Fe } \space and \space \frac{ 55.85 \space g \space Fe }{1 \space mole \space Fe }$$
$$ 6.00 \space g \space Fe_2O_3 \times \frac{1 \space mole \space Fe_2O_3 }{ 159.70 \space g \space Fe_2O_3 } \times \frac{ 2 \space moles \space Fe }{ 1 \space mole \space Fe_2O_3 } \times \frac{ 55.85 \space g \space Fe }{1 \space mole \space Fe } = 4.20 \space g \space Fe $$
