Question

The automobile fuel called E85 consists of 85% ethanol and 15% gasoline. E85 can be used in so-called flex-fuel vehicles (FFVs), which can use gasoline, ethanol, or a mix as fuels. Assume that gasoline consists of a mixture of octanes (different isomers of C8H18), that the average heat of combustion of C8H18(l) is 5400 kJ/mol, and that gasoline has an average density of 0.70 g/mL. The density of ethanol is 0.79 g/mL.

Part C Assume that the density and heat of combustion of E85 can be obtained by using 85% of the values for ethanol and 15% of the values for gasoline. How much energy could be released by the combustion of 3.0 L of E85?

Answer 1

Part A: By using the information given, calculate the energy produced by combustion of 3.0 L of gasoline.

Given that; the automobile fuel called E85 consists of 85% ethanol and 15% gasoline

Here the volume of the automobile fuel = 3.0 L

Volume of gasoline

3.0L *85%= 3.0*85/100=2.55 L

And density = mass / volume

Mass = volume * density

= 2.55 L*1000 ml/1.0L*0.70 g/ml

= 1785 g

Now moles of gasoline = amount in g / molar mass

= 1785 g/ 114 g/ mole

= 15.66 mole

Here that the average heat of combustion of C8H18(l) is 5400 kJ/mol

Then total heat = 15.66 mole *5400 kJ/mol

= 84553 KJ/ mol

= 8.5*10^4 KJ/ mol

Given that; the automobile fuel called E85 consists of 85% ethanol and 15% gasoline

Here the volume of the automobile fuel = 3.0 L

Volume of ethanol

3.0L *15%= 3.0*85/100= 0.45 L

And density = mass / volume

Mass = volume * density

= 0.45 L*1000 ml/1.0L*0.79 g/ml

= 355.5 g

Now moles of ethanol = amount in g / molar mass

= 355.5 g/ 46 g/ mole

= 7.73 mole ethanol

Here that the average heat of combustion of ethanol is 1370 kJ/mol

Then total heat by ethanol = 7.3 mole *1370 kJ/mol

=10587.7 KJ/ mol

= 1.06*10^4 KJ/ mol

E total = E ethanol + E octane

= 8.5*10^4 KJ/ mol +1.06*10^4 KJ/ mol

= 9.56 *10^4 KJ/ mol