Question

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

What 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, d = 0.703 g/mL) with an energy density of 32.9 kJ/mL? (1 gal = 3.785 L)

Answer 1

5.3072 Liters of volume of ethanol is required.

Step-by-step explanation:

Volume of isooctane = 1 gal = 3.785 L

3.785 L =3.785 × 1000 mL= 3,785 mL

Energy density of isooctane = 32.9 kJ/mL

Energy produced on combustion of 3,785 mL of isooctane :

`E=32.9 kJ/mL* 3,785 mL=124,526.5 kJ`

Moles of ethanol which will produce same amount of energy 'E' as 1 gallon of iosooctane = n

Ethanol’s enthalpy of combustion =
`\Delta H_(comb)=-1,368 kJ/mol`

Energy released when 1 mole of ethanol is combusted = 1,368 kJ/mol

`E=n* 1,368 kJ/mol`

`n=(E)/(1,368 kJ/mol)=(124,526.5 kJ)/(1,368 kJ/mol)`

n = 91.03 mole

Mass of 91.03 moles of ethanol , m= 91.03 mol × 46 g/mol = 4,187.38 g

Volume of ethanol = V

Density of ethanol = d = 0.789 g/mL

`Volume=(Mass)/(density)`

`V=(m)/(d)=(4,187.38 g)/(0.789 g/mL)=5,307.2 mL= 5.3072 L`

5.3072 Liters of volume of ethanol is required.