Answer
a. 9.59 g of $O_2$ are necessary to react with 15.0 g of glycine.
b. 6.00 g of urea are produced from 15.0 g of glycine.
c. 13.2 g of $CO_2$ are produced from 15.0 g of glycine.
Work Step by Step
- Identify the conversion factors and calculate the masses:
a. $$ \frac{ 2 \space moles \space C_2H_5NO_2 }{ 3 \space moles \space O_2 } \space and \space \frac{ 3 \space moles \space O_2 }{ 2 \space moles \space C_2H_5NO_2 }$$
$ C_2H_5NO_2 $ : ( 1.008 $\times$ 5 )+ ( 12.01 $\times$ 2 )+ ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 75.07 g/mol
$$ \frac{1 \space mole \space C_2H_5NO_2 }{ 75.07 \space g \space C_2H_5NO_2 } \space and \space \frac{ 75.07 \space g \space C_2H_5NO_2 }{1 \space mole \space C_2H_5NO_2 }$$
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
$$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$
$$ 15.0 \space g \space C_2H_5NO_2 \times \frac{1 \space mole \space C_2H_5NO_2 }{ 75.07 \space g \space C_2H_5NO_2 } \times \frac{ 3 \space moles \space O_2 }{ 2 \space moles \space C_2H_5NO_2 } \times \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 } = 9.59 \space g \space O_2 $$
b. $$ \frac{ 2 \space moles \space C_2H_5NO_2 }{ 1 \space mole \space CH_4N_2O } \space and \space \frac{ 1 \space mole \space CH_4N_2O }{ 2 \space moles \space C_2H_5NO_2 }$$
$ C_2H_5NO_2 $ : ( 1.008 $\times$ 5 )+ ( 12.01 $\times$ 2 )+ ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 75.07 g/mol
$$ \frac{1 \space mole \space C_2H_5NO_2 }{ 75.07 \space g \space C_2H_5NO_2 } \space and \space \frac{ 75.07 \space g \space C_2H_5NO_2 }{1 \space mole \space C_2H_5NO_2 }$$
$ CH_4N_2O $ : ( 1.008 $\times$ 4 )+ ( 12.01 $\times$ 1 )+ ( 14.01 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 60.06 g/mol
$$ \frac{1 \space mole \space CH_4N_2O }{ 60.06 \space g \space CH_4N_2O } \space and \space \frac{ 60.06 \space g \space CH_4N_2O }{1 \space mole \space CH_4N_2O }$$
$$ 15.0 \space g \space C_2H_5NO_2 \times \frac{1 \space mole \space C_2H_5NO_2 }{ 75.07 \space g \space C_2H_5NO_2 } \times \frac{ 1 \space mole \space CH_4N_2O }{ 2 \space moles \space C_2H_5NO_2 } \times \frac{ 60.06 \space g \space CH_4N_2O }{1 \space mole \space CH_4N_2O } = 6.00 \space g \space CH_4N_2O $$
c.
$$ \frac{ 2 \space moles \space C_2H_5NO_2 }{ 3 \space moles \space CO_2 } \space and \space \frac{ 3 \space moles \space CO_2 }{ 2 \space moles \space C_2H_5NO_2 }$$
$ C_2H_5NO_2 $ : ( 1.008 $\times$ 5 )+ ( 12.01 $\times$ 2 )+ ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 75.07 g/mol
$$ \frac{1 \space mole \space C_2H_5NO_2 }{ 75.07 \space g \space C_2H_5NO_2 } \space and \space \frac{ 75.07 \space g \space C_2H_5NO_2 }{1 \space mole \space C_2H_5NO_2 }$$
$ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol
$$ \frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \space and \space \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 }$$
$$ 15.0 \space g \space C_2H_5NO_2 \times \frac{1 \space mole \space C_2H_5NO_2 }{ 75.07 \space g \space C_2H_5NO_2 } \times \frac{ 3 \space moles \space CO_2 }{ 2 \space moles \space C_2H_5NO_2 } \times \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 } = 13.2 \space g \space CO_2 $$
