Answer
a. 0.900 mole of $O_2$
b. 18.0 g of $H_2O$
c. 257 g of $CO_2$
d. Percent yield of carbon dioxide: 72.4 %
Work Step by Step
a. $$ 0.225 \space mole \space C_3H_4 \times \frac{ 4 \space moles \ O_2 }{ 1 \space mole \space C_3H_4 } = 0.900 \space mole \space O_2 $$
b. $ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
$$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$
$ H_2O $ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol
$$ \frac{1 \space mole \space H_2O }{ 18.02 \space g \space H_2O } \space and \space \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O }$$
$$ 64.0 \space g \space O_2 \times \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \times \frac{ 2 \space moles \space H_2O }{ 4 \space moles \space O_2 } \times \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O } = 18.0 \space g \space H_2O $$
c. $ C_3H_4 $ : ( 1.008 $\times$ 4 )+ ( 12.01 $\times$ 3 )= 40.06 g/mol
$$ \frac{1 \space mole \space C_3H_4 }{ 40.06 \space g \space C_3H_4 } \space and \space \frac{ 40.06 \space g \space C_3H_4 }{1 \space mole \space C_3H_4 }$$
$ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol
$$ \frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \space and \space \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 }$$
$$ 78.0 \space g \space C_3H_4 \times \frac{1 \space mole \space C_3H_4 }{ 40.06 \space g \space C_3H_4 } \times \frac{ 3 \space moles \space CO_2 }{ 1 \space mole \space C_3H_4 } \times \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 } = 257 \space g \space CO_2 $$
d.
$$percent \space yield = \frac{186 \space g \space CO_2}{257 \space g \space CO_2} \times 100\% = 72.4\%$$
