Answer
a. 20.0 moles of $H_2O$
b. 15.3 g of $CO_2$
c. 75.7 g of $H_2O$
Work Step by Step
a.
$$ 5.00 \space moles \space C_3H_8 \times \frac{ 4 \space moles \ H_2O }{ 1 \space mole \space C_3H_8 } = 20.0 \space moles \space H_2O $$
b.
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
$$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$
$ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol
$$ \frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \space and \space \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 }$$
$$ 18.5 \space g \space O_2 \times \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \times \frac{ 3 \space moles \space CO_2 }{ 5 \space moles \space O_2 } \times \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 } = 15.3 \space g \space CO_2 $$
c.
$ C_3H_8 $ : ( 1.008 $\times$ 8 )+ ( 12.01 $\times$ 3 )= 44.09 g/mol
$$ \frac{1 \space mole \space C_3H_8 }{ 44.09 \space g \space C_3H_8 } \space and \space \frac{ 44.09 \space g \space C_3H_8 }{1 \space mole \space C_3H_8 }$$
$ H_2O $ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol
$$ \frac{1 \space mole \space H_2O }{ 18.02 \space g \space H_2O } \space and \space \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O }$$
$$ 46.3 \space g \space C_3H_8 \times \frac{1 \space mole \space C_3H_8 }{ 44.09 \space g \space C_3H_8 } \times \frac{ 4 \space moles \space H_2O }{ 1 \space mole \space C_3H_8 } \times \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O } = 75.7 \space g \space H_2O $$
