## Chemistry: Molecular Approach (4th Edition)

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.
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.