Answer
a. 33.0 %
b. 72.0 %
Work Step by Step
a.
$ 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 g/mol
$$ \frac{1 \space mole \space Fe }{ 55.85 \space g \space Fe } \space and \space \frac{ 55.85 \space g \space Fe }{1 \space mole \space Fe }$$
$$ 65.0 \space g \space Fe_2O_3 \times \frac{1 \space mole \space Fe_2O_3 }{ 159.70 \space g \space Fe_2O_3 } \times \frac{ 2 \space moles \space Fe }{ 1 \space mole \space Fe_2O_3 } \times \frac{ 55.85 \space g \space Fe }{1 \space mole \space Fe } = 45.5 \space g \space Fe $$
Calculation of percent yield: $$\frac{ 15.0 \space g \space Fe }{ 45.5 \space g \space Fe } \times 100\% = 33.0 \%$$
b.
$ CO $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 28.01 g/mol
$$ \frac{1 \space mole \space CO }{ 28.01 \space g \space CO } \space and \space \frac{ 28.01 \space g \space CO }{1 \space mole \space CO }$$
$ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol
$$ \frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \space and \space \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 }$$
$$ 75.0 \space g \space CO \times \frac{1 \space mole \space CO }{ 28.01 \space g \space CO } \times \frac{ 3 \space moles \space CO_2 }{ 3 \space moles \space CO } \times \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 } = 118 \space g \space CO_2 $$
Calculation of percent yield: $$\frac{ 85.0 \space g \space CO_2 }{ 118 \space g \space CO_2 } \times 100\% = 72.0 \%$$
