Answer
a. 12.5 moles of $H_2O$
b. 80.5 g of $O_2$
c. 167 g of $CO_2$
d. 86.8 %
Work Step by Step
a.
$$ 2.50 \space moles \space C_4H_{10} \times \frac{ 10 \space moles \ H_2O }{ 2 \space moles \space C_4H_{10} } = 12.5 \space moles \space H_2O $$
b.
$ C_4H_{10} $ : ( 1.008 $\times$ 10 )+ ( 12.01 $\times$ 4 )= 58.12 g/mol
$$ \frac{1 \space mole \space C_4H_{10} }{ 58.12 \space g \space C_4H_{10} } \space and \space \frac{ 58.12 \space g \space C_4H_{10} }{1 \space mole \space C_4H_{10} }$$
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
$$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$
$$ 22.5 \space g \space C_4H_{10} \times \frac{1 \space mole \space C_4H_{10} }{ 58.12 \space g \space C_4H_{10} } \times \frac{ 13 \space moles \space O_2 }{ 2 \space moles \space C_4H_{10} } \times \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 } = 80.5 \space g \space O_2 $$
c.
$ C_4H_{10} $ : ( 1.008 $\times$ 10 )+ ( 12.01 $\times$ 4 )= 58.12 g/mol
$$ \frac{1 \space mole \space C_4H_{10} }{ 58.12 \space g \space C_4H_{10} } \space and \space \frac{ 58.12 \space g \space C_4H_{10} }{1 \space mole \space C_4H_{10} }$$
$ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol
$$ \frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \space and \space \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 }$$
$$ 55.0 \space g \space C_4H_{10} \times \frac{1 \space mole \space C_4H_{10} }{ 58.12 \space g \space C_4H_{10} } \times \frac{ 8 \space moles \space CO_2 }{ 2 \space moles \space C_4H_{10} } \times \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 } = 167 \space g \space CO_2 $$
d. $$percent \space yield = \frac{145 \space g \space CO_2}{167 \space g \space CO_2} = 86.8\%$$
