Answer
a. 12 moles of $O_2$
b. 96 g of $O_2$
c. 68.7 g of $CO_2$
Work Step by Step
a. $$ 4.0 \space moles \space C_2H_6O \times \frac{ 3 \space moles \ O_2 }{ 1 \space mole \space C_2H_6O } = 12 \space moles \space O_2 $$
b.$ 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
$$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$
$$ 88 \space g \space CO_2 \times \frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \times \frac{ 3 \space moles \space O_2 }{ 2 \space moles \space CO_2 } \times \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 } = 96 \space g \space O_2 $$
c. Calculate or find the molar mass for $ C_2H_6O $:
$ C_2H_6O $ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 46.07 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 50.0 \space g \times \frac{1 \space mole}{ 46.07 \space g} = 1.09 \space moles$$
- Calculate or find the molar mass for $ O_2 $:
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 75.0 \space g \times \frac{1 \space mole}{ 32.00 \space g} = 2.34 \space moles$$
Find the amount of product if each reactant is completely consumed.
$$ 1.09 \space moles \space C_2H_6O \times \frac{ 2 \space moles \ CO_2 }{ 1 \space mole \space C_2H_6O } = 2.18 \space moles \space CO_2 $$
$$ 2.34 \space moles \space O_2 \times \frac{ 2 \space moles \ CO_2 }{ 3 \space moles \space O_2 } = 1.56 \space moles \space CO_2 $$
Since the reaction of $ O_2 $ produces less $ CO_2 $ for these quantities, it is the limiting reactant.
- Calculate or find the molar mass for $ CO_2 $:
$ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 1.56 \space mole \times \frac{ 44.01 \space g}{1 \space mole} = 68.7 \space g$$
