Ask Question
27 March, 06:05

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.

+1
Answers (1)
  1. 27 March, 06:53
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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. ...” in 📙 Engineering if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers