Answer
$$35.1 \space kg \space CO_2$$
Work Step by Step
1. This is a combustion reaction; therefore, propane will react with $O_2$ to produce carbon dioxide and water:
$$C_3H_8 + O_2 \longrightarrow CO_2 + H_2O$$
- Balance the amount of carbon:
$$C_3H_8 + O_2 \longrightarrow 3 CO_2 + H_2O$$
- Balance the amount of hydrogen:
$$C_3H_8 + O_2 \longrightarrow 3 CO_2 + 4H_2O$$
- Balance the amount of oxygen:
$$C_3H_8 + 5O_2 \longrightarrow 3 CO_2 + 4H_2O$$
2. Calculate the mass of propane in 18.9 L:
$$18.9 \space L \times \frac{1000 \space mL}{1 \space L} \times \frac{0.621 \space g}{1 \space mL} = 1.17\underline{37} \times 10^{4} \space g$$
3. Find the mass of carbon dioxide produced.
$ C_3H_8 $ : ( 1.008 $\times$ 8 )+ ( 12.01 $\times$ 3 )= 44.09 g/mol
$$ \frac{1 \space mole \space C_3H_8 }{ 44.09 \space g \space C_3H_8 } \space and \space \frac{ 44.09 \space g \space C_3H_8 }{1 \space mole \space C_3H_8 }$$
$ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol
$$ \frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \space and \space \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 }$$
$$ 1.17\underline{37} \times 10^4 \space g \space C_3H_8 \times \frac{1 \space mole \space C_3H_8 }{ 44.09 \space g \space C_3H_8 } \times \frac{ 3 \space moles \space CO_2 }{ 1 \space mole \space C_3H_8 } \times \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 } \times \frac{1 \space kg}{1000 \space g}= 35.1 \space kg \space CO_2 $$
