Answer
a. Na is the limiting reactant.
b. Na is the limiting reactant.
c. $Br_2$ is the limiting reactant.
d. Na is the limiting reactant.
Work Step by Step
a.
Find the amount of product if each reactant is completely consumed.
$$ 2 \space moles \space Na \times \frac{ 2 \space moles \ NaBr }{ 2 \space moles \space Na } = 2.0 \space moles \space NaBr $$
$$ 2 \space moles \space Br_2 \times \frac{ 2 \space moles \ NaBr }{ 1 \space mole \space Br_2 } = 4.0 \space moles \space NaBr $$
Since the reaction of $ Na $ produces less $ NaBr $ for these quantities, it is the limiting reactant.
b.
Find the amount of product if each reactant is completely consumed.
$$ 1.8 \space moles \space Na \times \frac{ 2 \space moles \ NaBr }{ 2 \space moles \space Na } = 1.8 \space moles \space NaBr $$
$$ 1.4 \space moles \space Br_2 \times \frac{ 2 \space moles \ NaBr }{ 1 \space mole \space Br_2 } = 2.8 \space moles \space NaBr $$
Since the reaction of $ Na $ produces less $ NaBr $ for these quantities, it is the limiting reactant.
c.
Find the amount of product if each reactant is completely consumed.
$$ 2.5 \space moles \space Na \times \frac{ 2 \space moles \ NaBr }{ 2 \space moles \space Na } = 2.5 \space moles \space NaBr $$
$$ 1 \space mole \space Br_2 \times \frac{ 2 \space moles \ NaBr }{ 1 \space mole \space Br_2 } = 2.0 \space moles \space NaBr $$
Since the reaction of $ Br_2 $ produces less $ NaBr $ for these quantities, it is the limiting reactant.
d.
Find the amount of product if each reactant is completely consumed.
$$ 12.6 \space moles \space Na \times \frac{ 2 \space moles \ NaBr }{ 2 \space moles \space Na } = 12.6 \space moles \space NaBr $$
$$ 6.9 \space moles \space Br_2 \times \frac{ 2 \space moles \ NaBr }{ 1 \space mole \space Br_2 } = 14.0 \space moles \space NaBr $$
Since the reaction of $ Na $ produces less $ NaBr $ for these quantities, it is the limiting reactant.
