## Chemistry: Molecular Approach (4th Edition)

a. Al b. $O_2$ c. Al d. Al
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.