Consider the combustion of ethanol c2h5oh with air. assume the air is dry and comprised of 21% oxygen and 79% nitrogen on a molar basis. a. determine the air/fuel ratio on a molar basis. b. determine the air/fuel ratio on a mass basis.

a) 14.285

b) 8.956

Explanation:

Given :

The combustion of the ethanol is with air

Air is 21% oxygen

and, 79% nitrogen

thus, for 1 O₂ we have  (79/21)N₂

thus,

the stochiometric equation for the combustion is as:

C₂H₅OH + 3[O₂ + (79/21)N₂]   ⇒   2CO₂ + 3H₂O + 3 × (79/21)N₂

Now,

the molecular weight of the fuel (C₂H₅OH) = (2× 12) + (5 × 1) + 16 + 1 = 46 g/mol

Molecular weight of the air = (2 × 16) + ((79/21) × 28) = 137.33 g/mol

a) air/fuel ratio on a molar basis

we have

air-fuel ratio = moles of air / moles of fuel

or

air-fuel ratio = [3 × 1 + 3 × (79/21)] / 1 = 14.285

b) air/fuel ratio on a mass basis = Mass of air / mass of fuel

or

air/fuel ratio on a mass basis = (number of moles of air × molar mass of air) / (number of moles of fuel × molar mass of fuel)

on substituting the values, we have

air/fuel ratio on a mass basis = (3 × 137.33) / (1 × 46) = 8.956

this is true

explanation:

i looked it up