In: Chemistry
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 A: By using the information given, calculate the energy produced by combustion of 1.5 L of gasoline.
Part B: By using the information given as well as data in Appendix C, calculate the energy produced by combustion of 1.5 L of ethanol. Consider that water is in the gaseous state.
Part A: By using the information given, calculate the energy produced by combustion of 1.5 L of gasoline.
Given that; the automobile fuel called E85 consists of 85% ethanol and 15% gasoline
Here the volume of the automobile fuel = 1.5 L
Volume of gasoline
1.5L *85%= 1.5*85/100= 1.275 L
And density = mass / volume
Mass = volume * density
= 1.275 L*1000 ml/1.0L*0.70 g/ml
= 892.5 g
Now moles of gasoline = amount in g / molar mass
= 892.5 g/ 114 g/ mole
= 7.829 mole
Here that the average heat of combustion of C8H18(l) is 5400 kJ/mol
Then total heat = 7.829 mole *5400 kJ/mol
= 42276.32 KJ/ mol
= 4.23*10^4 KJ/ mol
Part B: By using the information given as well as data in Appendix C, calculate the energy produced by combustion of 1.5 L of ethanol. Consider that water is in the gaseous state.
Given that; the automobile fuel called E85 consists of 85% ethanol and 15% gasoline
Here the volume of the automobile fuel = 1.5 L
Volume of ethanol
1.5L *15%= 1.5*85/100= 0..225 L
And density = mass / volume
Mass = volume * density
= 0.225 L*1000 ml/1.0L*0.79 g/ml
= 177.75 g
Now moles of ethanol = amount in g / molar mass
= 177.75 g/ 46 g/ mole
= 3.86 mole ethanol
Here that the average heat of combustion of ethanol is 1370 kJ/mol
Then total heat by ethanol = 3.86 mole *1370 kJ/mol
= 5293.86 KJ/ mol
= 5.29*10^3 KJ/ mol