Answer
a. $$C_2H_6O + 3O_2 \longrightarrow 2CO_2 + 3H_2O$$
b. $24 \space moles \space O_2$
c. 4.8 g of $O_2$
d. $ 239 \space g \space CO_2$ and $147 \space g \space H_2O $
Work Step by Step
a. Identify each reactant and product, and write the unbalanced equation:
Reactants: Ethanol $(C_2H_6O)$ and Oxygen $(O_2)$
Products: Carbon dioxide $(CO_2)$ and Water $(H_2O)$
$$C_2H_6O + O_2 \longrightarrow CO_2 + H_2O$$
- Balance the number of carbons:
$$C_2H_6O + O_2 \longrightarrow 2CO_2 + H_2O$$
- Balance the number of hydrogens:
$$C_2H_6O + O_2 \longrightarrow 2CO_2 + 3H_2O$$
- Balance the number of oxygens:
$$C_2H_6O + 3O_2 \longrightarrow 2CO_2 + 3H_2O$$
b. $$ 8.0 \space moles \space C_2H_6O \times \frac{ 3 \space moles \ O_2 }{ 1 \space mole \space C_2H_6O } = 24 \space moles \space O_2 $$
c. $ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol
$$ \frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \space and \space \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 }$$
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
$$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$
$$ 4.4 \space g \space CO_2 \times \frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \times \frac{ 3 \space moles \space O_2 }{ 2 \space moles \space CO_2 } \times \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 } = 4.8 \space g \space O_2 $$
d. $ C_2H_6O $ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 46.07 g/mol
$$ \frac{1 \space mole \space C_2H_6O }{ 46.07 \space g \space C_2H_6O } \space and \space \frac{ 46.07 \space g \space C_2H_6O }{1 \space mole \space C_2H_6O }$$
$ CO_2 $ : 44.01 g/mol
$$ \frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \space and \space \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 }$$
$$ 125 \space g \space C_2H_6O \times \frac{1 \space mole \space C_2H_6O }{ 46.07 \space g \space C_2H_6O } \times \frac{ 2 \space moles \space CO_2 }{ 1 \space mole \space C_2H_6O } \times \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 } = 239 \space g \space CO_2 $$
$ H_2O $ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol
$$ \frac{1 \space mole \space H_2O }{ 18.02 \space g \space H_2O } \space and \space \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O }$$
$$ 125 \space g \space C_2H_6O \times \frac{1 \space mole \space C_2H_6O }{ 46.07 \space g \space C_2H_6O } \times \frac{ 3 \space moles \space H_2O }{ 1 \space mole \space C_2H_6O } \times \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O } = 147 \space g \space H_2O $$
