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

Answer:

$\boxed{\text{4.5 m}^{3}}$

Explanation:

Step 1. Calculate the volume of gasoline used.

$V = \text{235 mi} \times \dfrac{\text{1 gal}}{\text{31.2 mi}} \times \dfrac{\text{3.875 L}}{\text{1 gal}} = \text{28.51 L}$

Step 2. Calculate the moles of octane used.

$n = \text{28.51 L} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{0.71 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{114.23 g}} = \text{177 mol}$

Step 3. Calculate the moles of CO₂ formed

            C₈H₈ + 10O₂ ⟶ 8CO₂ + 4H₂O

n/mol:   177

$n = \text{177 mol C _8 H _8 } \times \dfrac{\text{8 mol CO _2 }}{\text{1 mol C _8 H _8 }} = \text{1420 mol CO _2 }$

Step 4. Calculate the volume of CO₂

$p =\text{732 torr} \times \dfrac{\text{1 atm}}{\text{760 torr}} = \text{0.9632 atm}$

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 28 °C = 301.15 K

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