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lord [1]
3 years ago
14

Ethanol (CH3CH2OH) has been suggested as an alternative fuel source. Ethanol’s enthalpy of combustion is Hcomb = –1368 kJ/mol. W

hat volume of ethanol (d = 0.789 g/mL) is required to produce the same amount of energy as 1 gallon of gasoline, which can be modeled as isooctane (C8H18) with an energy density of 32.9 kJ/mL? (1 gal = 3.785 L; d = 0.703 g/mL)
a. 1.41 gal
b. 0.85 gal
c. 1.00 gal
d. 1.15 gal
d. 0.90 ga
Chemistry
1 answer:
madam [21]3 years ago
8 0

Answer:

a) V = 1.41 gal

Explanation:

Ethanol:

∴ Hcomb = - 1368 KJ/mol.........( - ) exothermic reaction

∴ δ = 0.789 g/mL

∴ Mw Ethanol = 46.07 g/mol

⇒ Eδ = (1368 KJ/mol)(mol/46.07 g)(0.789 g/mL) = 23.4285 KJ/mL

isooctane:

∴ Eδ = 32.9 KJ/mL

∴ δ = 0.703 g/mL

  • 1 gal = 3785 mL

⇒ E = (32.9 KJ/mL)(3785 mL) = 124526.5 KJ

Volume of Ethanol:

⇒ (Eδ)(V) = 124526.5 KJ

⇒ V = (124526.5 KJ)/(23.4285 KJ/mL) = 5315.167 mL

⇒ V = (5315.167 mL)( gal/3785 mL) = 1.404 gal ≅ 1.41 gal

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