Question

Use C8H18 as the formula for gasoline and 0.71g/mL as its density. If a car gets 31.2 miles per gallon, what volume of CO2 measured at 28°C and 732 torr is produced on a trip of 235 miles? Assume complete combustion of gasoline.

Answer 1

`\boxed{\text{4.5 m}^(3)}`

Step-by-step explanation:

Step 1. Calculate the volume of gasoline used.

`V = \text{235 mi} * \frac{\text{1 gal}}{\text{31.2 mi}} * \frac{\text{3.875 L}}{\text{1 gal}} = \text{28.51 L}`

Step 2. Calculate the moles of octane used.

`n = \text{28.51 L} * \frac{\text{1000 mL}}{\text{1 L}} * \frac{\text{0.71 g}}{\text{1 mL}} * \frac{\text{1 mol}}{\text{114.23 g}} = \text{177 mol}`

Step 3. Calculate the moles of CO₂ formed

C₈H₈ + 10O₂ ⟶ 8CO₂ + 4H₂O

n/mol: 177

`n = \text{177 mol C$_8$H$_8$} * \frac{\text{8 mol CO$_2$}}{\text{1 mol C$_8$H$_8$}} = \text{1420 mol CO$_2$}`

Step 4. Calculate the volume of CO₂

`p =\text{732 torr} * \frac{\text{1 atm}}{\text{760 torr}} = \text{0.9632 atm}`

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 28 °C = 301.15 K

`pV = nRT\\\\V = (nRT)/(p)\\\\V =(177 * 0.08206 * 301.15)/(0.9632) = \text{4500 L = 4.5 m}^(3)\\\\\text{The volume of CO$_2$ released is }\boxed{\textbf{4.5 m}^{\mathbf{3}}}`