Air/fuel ratio on molar (volume) basis for combustion of methane with theoretical quantity of air with be

ANS:A - 9.5:1

To determine the air/fuel ratio for the combustion of methane (CH4​) with the theoretical quantity of air, we need to consider the stoichiometry of the reaction. The balanced chemical equation for the combustion of methane is: CH4​+2O2​→CO2​+2H2​O From the equation, we see that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water vapor. Now, for complete combustion, the reaction needs the stoichiometric amount of oxygen. The molar ratio of oxygen to methane is 2:1, which means 2 moles of oxygen are required for every mole of methane. Therefore, the air/fuel ratio on a molar (volume) basis is given by: Air/Fuel Ratio=2:1 So, the correct answer is 12.5:112.5:1.