Ask Question
12 April, 22:37

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.

+2
Answers (1)
  1. 13 April, 02:19
    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?
Find 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 you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers