How much air (moles) is required for combustion of one mole of ethanol fuel that is 100% ethanol (CH3CH2OH) ? Molecular weight of CH3CH2OH=46g Molecular weight of O2=32g Molecular weight of N2=28g Molecular weight of air =29g Air composition (by volume or by molar composition): O2:21% and N2:79% Note: Moles of air = moles of O2+ moles of N2 Assume the following combustion reaction: 6CH3CH2OH+36O2+135.36N212CO2+18H2O+135.35N2 0. For the above problem, what is the air to fuel ratio for combustion of ethanol (moles of air/moles of ethanol and grams of air/grams of ethanol)