Answer
a. 1 mol
b. 4 mol
c. 9.17 mol
Work Step by Step
a. Find the amount of product if each reactant is completely consumed.
$$ 3 \space moles \space Mn \times \frac{ 1 \space mole \ Mn_3O_4 }{ 3 \space moles \space Mn } = 1 \space mole \space Mn_3O_4 $$
$$ 3 \space moles \space O_2 \times \frac{ 1 \space mole \ Mn_3O_4 }{ 2 \space moles \space O_2 } = 1.67 \space moles \space Mn_3O_4 $$
Since the reaction of $ Mn $ produces less $ Mn_3O_4 $ for these quantities, it is the limiting reactant.
b. Find the amount of product if each reactant is completely consumed.
$$ 4 \space moles \space Mn \times \frac{ 1 \space mole \ Mn_3O_4 }{ 3 \space moles \space Mn } = 1 \space mole \space Mn_3O_4 $$
$$ 7 \space moles \space O_2 \times \frac{ 1 \space mole \ Mn_3O_4 }{ 2 \space moles \space O_2 } = 3.5 \space moles \space Mn_3O_4 $$
Since the reaction of $ Mn $ produces less $ Mn_3O_4 $ for these quantities, it is the limiting reactant.
c. Find the amount of product if each reactant is completely consumed.
$$ 27.5 \space moles \space Mn \times \frac{ 1 \space mole \ Mn_3O_4 }{ 3 \space moles \space Mn } = 9.17 \space moles \space Mn_3O_4 $$
$$ 43.8 \space moles \space O_2 \times \frac{ 1 \space mole \ Mn_3O_4 }{ 2 \space moles \space O_2 } = 21.9 \space moles \space Mn_3O_4 $$
Since the reaction of $ Mn $ produces less $ Mn_3O_4 $ for these quantities, it is the limiting reactant.
