<h3>
Answer:</h3>
266.997 pounds
<h3>
Explanation:</h3>
The balanced equation for the combustion of octane is;
2C₈H₁₈(l) + 25O₂(g) → 16CO₂(g) + 18H₂O(g)
Assuming that gasoline is purely octane:
Amount of gasoline available is 15.0 gallon
We are required to calculate the amount of CO₂ released to the atmosphere;
This can be done in several steps;
Step 1: Mass of octane in grams
Amount of octane = 15.0 gallons
But, 1 gallon = 3785.41 ml
Therefore;
15 gallon is equivalent to;
= 15 ×3785.41 ml
= 56,781.15 ml
We are given the density of octane (0.692 g/ml) and thus we can calculate the mass of octane.
Mass = density × volume
= 0.692 g/ml × 56,781.15 ml
= 39,292.56 g
Step 2: Number of moles of octane
Number of moles = Mass/ Molar mass
Molar mass of octane = 114.23 g/mol
Number of moles of Octane = 39,292.56 g ÷ 114.23 g/mol
= 343.978 moles
Step 3: Moles of carbon dioxide released
From the equation;
2 moles of octane completely burns in air to yield 16 moles of CO₂
Therefore;
343.978 moles of octane produces;
= (343.978 moles/2) 16
= 2751.824 moles of CO₂
Step 4: Mass of CO₂ in pounds
1 mole of CO₂ has 44.01 g
Therefore;
2751.824 moles contains;
= 2751.824 moles × 44.01 g/mol
= 121,107.77 g
But, 1 pound = 453.592 g
Therefore;
Mass of CO₂ in pounds = 121,107.77 g ÷ 453.592 g
= 266.997 pounds
Therefore, the mass of CO₂ released to the atmosphere is 266.997 pounds
