## Chemistry: Molecular Approach (4th Edition)

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.