Question

An experimental fuel is a mixture of 89% ethanol and 11% gasoline by volume. Calculate how much energy in the form of heat is available from 100 mL of this mixture and compare it with the amount from pure gasoline. Assume the densities of ethanol and gasoline are 0.789 and 0.737 g/mL, respectively, and that n-nonane is an appropriate model hydrocarbon for gasoline Number heat from experimental fuel kJ Number heat from gasoline kJ

Answer 1

Heat of combustion from ethanol= -1367 kJ/mol

Heat of combustion from nonane (gasoline)= -6160 kJ/mol

Molecular weight of ethanol= 12*2+6+16= 46 grams/mole

Molecular weight of nonane= 12*9+20= 128 grams/mole

In 100mL mixture, 89% ethanol = 89 mL ethanol => 89 * 0.789 => 70.22 grams =>70.22/ 46 => 1.53 moles

11% gasoline = 11 mL nonane= 11 * 0.737 => 8.107 grams => 8.107/128 => 0.06 moles

Whereas for pure gasoline, 100 mL => 100 * 0.737 => 73.7 grams => 73.7/128 => 0.58 moles

Heat available from mixture= 1.53 * 1367 + 0.06 * 6160 => 2091.51 + 369.6 => 2461.111 kJ

From pure gasoline, 0.58 * 6160 => 3546.8 kJ

So, pure gasoline has will provide more heat.