Answer
(a)$$C_3H_8 + 5O_2\longrightarrow 3CO_2+4H_2O$$
(b) $482 \space g \space CO_2$
Work Step by Step
(a)
1. Balance the amount of carbons in each side:
$$C_3H_8 + O_2\longrightarrow 3CO_2+H_2O$$
2. Balance the amount of hydrogens in each side:
$$C_3H_8 + O_2\longrightarrow 3CO_2+4H_2O$$
3. Balance the amount of oxygens in each side:
$$C_3H_8 + 5O_2\longrightarrow 3CO_2+4H_2O$$
- The equation is completely balanced.
(b)
1. According to the balanced equation, each mole of propane $(C_3H_8)$ produces 3 moles of carbon dioxide:
$$\frac{1 \space mole \space C_3H_8}{3 \space moles \space CO_2} \space and \space \frac{3 \space moles \space CO_2}{1 \space mole \space C_3H_8} $$
2. Calculate the molar mass for carbon dioxide:
$ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol
3. Use this information as conversion factors to find the amount of grams of $CO_2$ produced
$$3.65 \space moles \space C_3H_8 \times \frac{3 \space moles \space CO_2}{1 \space mole \space C_3H_8} \times \frac{44.01 \space g \space CO_2}{1 \space moles \space CO_2} = 482 \space g \space CO_2$$
