Answer
a. Fe
b. Fe
c. Fe
Work Step by Step
a. Find the amount of product if each reactant is completely consumed.
Conversion factors:
$$ \frac{ 4 \space moles \space Fe }{ 2 \space moles \space Fe_2O_3 } \space and \space \frac{ 2 \space moles \space Fe_2O_3 }{ 4 \space moles \space Fe }$$$$ \frac{ 3 \space moles \space O_2 }{ 2 \space moles \space Fe_2O_3 } \space and \space \frac{ 2 \space moles \space Fe_2O_3 }{ 3 \space moles \space O_2 }$$
$$ 2.0 \space moles \space Fe \times \frac{ 2 \space moles \ Fe_2O_3 }{ 4 \space moles \space Fe } = 1.0 \space mole \space Fe_2O_3 $$
$$ 6.0 \space moles \space O_2 \times \frac{ 2 \space moles \ Fe_2O_3 }{ 3 \space moles \space O_2 } = 4.0 \space moles \space Fe_2O_3 $$
Since the reaction of $ Fe $ produces less $ Fe_2O_3 $ for these quantities, it is the limiting reactant.
b.
$$ 5.0 \space moles \space Fe \times \frac{ 2 \space moles \ Fe_2O_3 }{ 4 \space moles \space Fe } = 2.5 \space moles \space Fe_2O_3 $$
$$ 4.0 \space moles \space O_2 \times \frac{ 2 \space moles \ Fe_2O_3 }{ 3 \space moles \space O_2 } = 2.7 \space moles \space Fe_2O_3 $$
Since the reaction of $ Fe $ produces less $ Fe_2O_3 $ for these quantities, it is the limiting reactant.
c.
$$ 16.0 \space moles \space Fe \times \frac{ 2 \space moles \ Fe_2O_3 }{ 4 \space moles \space Fe } = 8.00 \space moles \space Fe_2O_3 $$
$$ 20.0 \space moles \space O_2 \times \frac{ 2 \space moles \ Fe_2O_3 }{ 3 \space moles \space O_2 } = 13.3 \space moles \space Fe_2O_3 $$
Since the reaction of $ Fe $ produces less $ Fe_2O_3 $ for these quantities, it is the limiting reactant.
