Chapter 7 - Chemical Reactions and Quantities - 7.8 Limiting Reactants and Percent Yield - Questions and Problems - Page 270: 7.75

a. $H_2$ is the limiting reactant for these quantities. a. $H_2$ is the limiting reactant for these quantities. a. $N_2$ is the limiting reactant for these quantities.

a. Find the amount of product if each reactant is completely consumed. Conversion factors: $$ \frac{ 1 \space mole \space N_2 }{ 2 \space moles \space NH_3 } \space and \space \frac{ 2 \space moles \space NH_3 }{ 1 \space mole \space N_2 }$$ $$ \frac{ 3 \space moles \space H_2 }{ 2 \space moles \space NH_3 } \space and \space \frac{ 2 \space moles \space NH_3 }{ 3 \space moles \space H_2 }$$ $$ 3.0 \space moles \space N_2 \times \frac{ 2 \space moles \ NH_3 }{ 1 \space mole \space N_2 } = 6.0 \space moles \space NH_3 $$ $$ 5.0 \space moles \space H_2 \times \frac{ 2 \space moles \ NH_3 }{ 3 \space moles \space H_2 } = 3.3 \space moles \space NH_3 $$ Since the reaction of $ H_2 $ produces less $ NH_3 $ for these quantities, it is the limiting reactant. b. Find the amount of product if each reactant is completely consumed. $$ 8.0 \space moles \space N_2 \times \frac{ 2 \space moles \ NH_3 }{ 1 \space mole \space N_2 } = 16 \space moles \space NH_3 $$ $$ 4.0 \space moles \space H_2 \times \frac{ 2 \space moles \ NH_3 }{ 3 \space moles \space H_2 } = 2.7 \space moles \space NH_3 $$ Since the reaction of $ H_2 $ produces less $ NH_3 $ for these quantities, it is the limiting reactant. c. Find the amount of product if each reactant is completely consumed. $$ 3.0 \space moles \space N_2 \times \frac{ 2 \space moles \ NH_3 }{ 1 \space mole \space N_2 } = 6.0 \space moles \space NH_3 $$ $$ 12.0 \space moles \space H_2 \times \frac{ 2 \space moles \ NH_3 }{ 3 \space moles \space H_2 } = 8.00 \space moles \space NH_3 $$ Since the reaction of $ N_2 $ produces less $ NH_3 $ for these quantities, it is the limiting reactant.