Answer
a. Al
b. $O_2$
c. Al
d. Al
Work Step by Step
a.
Find the amount of product if each reactant is completely consumed.
$$ 1 \space mole \space Al \times \frac{ 2 \space moles \ Al_2O_3 }{ 4 \space moles \space Al } = 0.5 \space mole \space Al_2O_3 $$
$$ 1 \space mole \space O_2 \times \frac{ 2 \space moles \ Al_2O_3 }{ 3 \space moles \space O_2 } = 0.7 \space mole \space Al_2O_3 $$
Since the reaction of $ Al $ produces less $ Al_2O_3 $ for these quantities, it is the limiting reactant.
b.
Find the amount of product if each reactant is completely consumed.
$$ 4 \space moles \space Al \times \frac{ 2 \space moles \ Al_2O_3 }{ 4 \space moles \space Al } = 2.0 \space moles \space Al_2O_3 $$
$$ 2.6 \space moles \space O_2 \times \frac{ 2 \space moles \ Al_2O_3 }{ 3 \space moles \space O_2 } = 1.7 \space moles \space Al_2O_3 $$
Since the reaction of $ O_2 $ produces less $ Al_2O_3 $ for these quantities, it is the limiting reactant.
c.
Find the amount of product if each reactant is completely consumed.
$$ 16 \space moles \space Al \times \frac{ 2 \space moles \ Al_2O_3 }{ 4 \space moles \space Al } = 8.0 \space moles \space Al_2O_3 $$
$$ 13 \space moles \space O_2 \times \frac{ 2 \space moles \ Al_2O_3 }{ 3 \space moles \space O_2 } = 8.7 \space moles \space Al_2O_3 $$
Since the reaction of $ Al $ produces less $ Al_2O_3 $ for these quantities, it is the limiting reactant.
d.
Find the amount of product if each reactant is completely consumed.
$$ 7.4 \space moles \space Al \times \frac{ 2 \space moles \ Al_2O_3 }{ 4 \space moles \space Al } = 3.7 \space moles \space Al_2O_3 $$
$$ 6.5 \space moles \space O_2 \times \frac{ 2 \space moles \ Al_2O_3 }{ 3 \space moles \space O_2 } = 4.3 \space moles \space Al_2O_3 $$
Since the reaction of $ Al $ produces less $ Al_2O_3 $ for these quantities, it is the limiting reactant.
