Question

True or false:

The required air-to-fuel ratio (in g/g) is atleast 20% higher for complete ethanol combustion than for one complete octane combustion.

Answer 1

Ethanol combustion reaction

C2H5OH + 3O2 -----> 2CO2 + 3H2O

Molecular weight of ethanol = 2*12 + 6*1 + 16= 46 g/mol

Molecular weight of Air = 29 g/mol

O2 required for complete combustion = 3 moles / mole of ethanol

Air conatins 21 % O2

Air required = 3 / 0.21 = 14.2857 mol

mass of air required = 14.2857 mol x 29 g/mol = 414.2857 g

mass of ethanol = 1 mol x 46 g/mol = 46 g

Air to fuel ratio = 414.2857 / 46 = 9

For octane:

Combustion reaction:

C8H18 + (25/2)O2 ----> 8CO2 + 9H2O

Molecular weight of octane = 8*12 + 18*1 = 114 g/mol

mass of air required = (12.5 / 0.21)*29 = 1726.19 g

mass of octane = 1 mol x 114 g/mol = 114 g

Air to fuel ratio = 1726.19 / 114 = 15.142

Now,

The given statement is false as Air to fuel ratio required for octane is higher than that required for ethanol.