Chapter 7 - Chemical Reactions and Quantities - 7.7 Mass Calculations for Reactions - Questions and Problems - Page 264: 7.64

a. 20.5 g of $NH_3$ b. 0.604 g of $H_2$ c. 67.6 g of $NH_3$

a. $$ \frac{ 3 \space moles \space H_2 }{ 2 \space moles \space NH_3 } \space and \space \frac{ 2 \space moles \space NH_3 }{ 3 \space moles \space H_2 }$$ $ H_2 $ : ( 1.008 $\times$ 2 )= 2.016 g/mol $$ \frac{1 \space mole \space H_2 }{ 2.016 \space g \space H_2 } \space and \space \frac{ 2.016 \space g \space H_2 }{1 \space mole \space H_2 }$$ $ NH_3 $ : ( 1.008 $\times$ 3 )+ ( 14.01 $\times$ 1 )= 17.03 g/mol $$ \frac{1 \space mole \space NH_3 }{ 17.03 \space g \space NH_3 } \space and \space \frac{ 17.03 \space g \space NH_3 }{1 \space mole \space NH_3 }$$ $$ 3.64 \space g \space H_2 \times \frac{1 \space mole \space H_2 }{ 2.016 \space g \space H_2 } \times \frac{ 2 \space moles \space NH_3 }{ 3 \space moles \space H_2 } \times \frac{ 17.03 \space g \space NH_3 }{1 \space mole \space NH_3 } = 20.5 \space g \space NH_3 $$ b. $$ \frac{ 1 \space mole \space N_2 }{ 3 \space moles \space H_2 } \space and \space \frac{ 3 \space moles \space H_2 }{ 1 \space mole \space N_2 }$$ $ N_2 $ : ( 14.01 $\times$ 2 )= 28.02 g/mol $$ \frac{1 \space mole \space N_2 }{ 28.02 \space g \space N_2 } \space and \space \frac{ 28.02 \space g \space N_2 }{1 \space mole \space N_2 }$$ $ H_2 $ : 2.016 g/mol $$ \frac{1 \space mole \space H_2 }{ 2.016 \space g \space H_2 } \space and \space \frac{ 2.016 \space g \space H_2 }{1 \space mole \space H_2 }$$ $$ 2.80 \space g \space N_2 \times \frac{1 \space mole \space N_2 }{ 28.02 \space g \space N_2 } \times \frac{ 3 \space moles \space H_2 }{ 1 \space mole \space N_2 } \times \frac{ 2.016 \space g \space H_2 }{1 \space mole \space H_2 } = 0.604 \space g \space H_2 $$ c.$$ \frac{ 3 \space moles \space H_2 }{ 2 \space moles \space NH_3 } \space and \space \frac{ 2 \space moles \space NH_3 }{ 3 \space moles \space H_2 }$$ $ H_2 $ : 2.016 g/mol $$ \frac{1 \space mole \space H_2 }{ 2.016 \space g \space H_2 } \space and \space \frac{ 2.016 \space g \space H_2 }{1 \space mole \space H_2 }$$ $ NH_3 $ : 17.03 g/mol $$ \frac{1 \space mole \space NH_3 }{ 17.03 \space g \space NH_3 } \space and \space \frac{ 17.03 \space g \space NH_3 }{1 \space mole \space NH_3 }$$ $$ 12.0 \space g \space H_2 \times \frac{1 \space mole \space H_2 }{ 2.016 \space g \space H_2 } \times \frac{ 2 \space moles \space NH_3 }{ 3 \space moles \space H_2 } \times \frac{ 17.03 \space g \space NH_3 }{1 \space mole \space NH_3 } = 67.6 \space g \space NH_3 $$