The enthalpy change in the combustion of the hydrocarbon octane is delta H = -5.48 x 10^3 kJ/mol C8H18 (l). How much heat, in kilojoules, is liberated per gallon of octane burned? (Density of octane = 0.703 g/mL; 1 gal = 3.785 L.)
Convert 1 gallon to liters to mL, then use density to convert to grams and divide by molar mass to convert to moles.
Then moles x 5.48E3 kJ/mol = ??
-1.28x10^5 kJ
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To calculate the heat liberated per gallon of octane burned, we need to first determine the number of moles of octane consumed and then multiply it by the enthalpy change.
1. Start by calculating the mass of octane in grams:
- Density of octane = 0.703 g/mL
- 1 gallon = 3.785 L
- Therefore, the mass of octane in one gallon = 3.785 L * 0.703 g/mL = 2.66 g/gal
2. Convert the mass of octane in grams to moles, using the molar mass of octane:
- Molar mass of octane (C8H18) = (12.01 g/mol x 8) + (1.01 g/mol x 18) = 114.23 g/mol
- Moles of octane in one gallon = 2.66 g/gal / 114.23 g/mol = 0.0233 mol
3. Finally, calculate the heat liberated per gallon by multiplying the moles of octane by the enthalpy change:
- Heat liberated per mole = -5.48 x 10^3 kJ/mol
- Heat liberated per gallon = (0.0233 mol) * (-5.48 x 10^3 kJ/mol) = -127.64 kJ/gal
Therefore, approximately 127.64 kilojoules of heat are liberated per gallon of octane burned.