Answer
93.7 %
Work Step by Step
- Find the theoretical yield:
$ 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
$$ \frac{1 \space mole \space Fe }{ 55.85 \space g \space Fe } \space and \space \frac{ 55.85 \space g \space Fe }{1 \space mole \space Fe }$$
$$ 50.0 \space g \space Fe_2O_3 \times \frac{1 \space mole \space Fe_2O_3 }{ 159.70 \space g \space Fe_2O_3 } \times \frac{ 2 \space moles \space Fe }{ 1 \space mole \space Fe_2O_3 } \times \frac{ 55.85 \space g \space Fe }{1 \space mole \space Fe } = 35.0 \space g \space Fe $$
Calculation of percent yield: $$\frac{ 32.8 \space g \space Fe }{ 35.0 \space g \space Fe } \times 100\% = 93.7 \%$$
