Answer
a. 0.667 mole of $C_5H_{12}$ must react to produce 4.00 moles of water.
b. 27.5 g of $CO_2$ are produced from 32.0 g of oxygen gas.
c. 92.9 g of $CO_2$ are produced.
Work Step by Step
a.$$ 4.00 \space moles \space H_2O \times \frac{ 1 \space mole \ C_5H_{12} }{ 6 \space moles \space H_2O } = 0.667 \space mole \space C_5H_{12} $$
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
$$ \frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \space and \space \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 }$$
$$ 32.0 \space g \space O_2 \times \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \times \frac{ 5 \space moles \space CO_2 }{ 8 \space moles \space O_2 } \times \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 } = 27.5 \space g \space CO_2 $$
c.
- Calculate or find the molar mass for $ C_5H_{12} $:
$ C_5H_{12} $ : ( 1.008 $\times$ 12 )+ ( 12.01 $\times$ 5 )= 72.15 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 44.5 \space g \times \frac{1 \space mole}{ 72.15 \space g} = 0.617 \space mole$$
- Calculate or find the molar mass for $ O_2 $:
$ O_2 $ : 32.00 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 108 \space g \times \frac{1 \space mole}{ 32.00 \space g} = 3.38 \space moles$$
Find the amount of product if each reactant is completely consumed.
$$ 0.617 \space mole \space C_5H_{12} \times \frac{ 5 \space moles \ CO_2 }{ 1 \space mole \space C_5H_{12} } = 3.08 \space moles \space CO_2 $$
$$ 3.38 \space moles \space O_2 \times \frac{ 5 \space moles \ CO_2 }{ 8 \space moles \space O_2 } = 2.112 \space moles \space CO_2 $$
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 $ : 44.01 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 2.112 \space mole \times \frac{ 44.01 \space g}{1 \space mole} = 92.9 \space g$$
