Step 1

The reaction involved is,

$C_{8} H_{18} (l) + \frac{25}{2} O_{2} (g) \to {8CO}_{2} {(g) + 9 H}_{2} O (g)$

Step 2

$\begin{array}{rcl} Volume of gasoline taken & = & 73 L \\ Density of gasoline & = & 0.77 kg/L \\ Amount of gasoline taken & = & 0.77 \times 73 \\ = & 56.21 kg \\ = & 56210 g \\ No. of moles of gasoline taken & = & \frac{Mass of gasoline}{Molar mass of gasoline} \\ = & \frac{56210}{114.2} \\ = & 492.207 \end{array}$

Step 3

Assuming that all the gasoline in the 73 L liquid tank is completely consumed, the number of moles of CO₂ produced is about eight times the moles of gasoline taken as, eight moles of carbon dioxide is obtained upon the consumption of 1 mole of gasoline.

$\begin{array}{rcl} {No. of moles of CO}_{2} produced (n) & = & 8 \times 492.207 \\ = & 8937.66 \\ Pressure (P) & = & 1.00 atm \\ Temperature (T) & = & 298.15 K \\ Gas constant (R) & = & 0.08205 {L.atm.mol}^{-1} {.K}^{-1} \\ Volume (V) & = & \frac{nRT}{P} \\ = & \frac{8937.66 \times 0.08205 \times 298.15}{1} \\ = & 218643.83 L \\ ≃ & 2.19 \times 10^{5} L \end{array}$

Step 4

The volume of CO₂ released upon the combustion of 73 L of gasoline is calculated to be 218643.83 L.

What is the volume (in L) of carbon dioxide at 1.00 atm and 298.15 K are released from a car's engine upon consumption of a 73.0 L LIQUID tank gasoline? (Gasoline density: 0.770 kg/L, Molar mass of C₈H₁₈: 114.2 g/mol) C₈H₁₈ (l) + ²⁵/₂ O₂ (g) → 8 CO₂ (g) + 9 H₂O (g)

Public Answer