Question

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

Answer 1

Answer:

5.3072 Liters of  volume of ethanol is required.

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 :

$E=32.9 kJ/mL\times 3,785 mL=124,526.5 kJ$

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

$E=n\times 1,368 kJ/mol$

$n=\frac{E}{1,368 kJ/mol}=\frac{124,526.5 kJ}{1,368 kJ/mol}$

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

$Volume=\frac{Mass}{density}$

$V=\frac{m}{d}=\frac{4,187.38 g}{0.789 g/mL}=5,307.2 mL= 5.3072 L$

5.3072 Liters of  volume of ethanol is required.