To solve this problem, we need to first write the balanced chemical equation for the reaction:CH4 + 2O2 CO2 + 2H2ONext, we need to determine the number of moles of each reactant:Moles of CH4 = 5 grams / 16.04 g/mol = 0.3119 molesMoles of O2 = 20 grams / 32.00 g/mol = 0.625 molesNow, we need to determine the mole ratio between the reactants:Mole ratio of CH4 to O2 = 0.3119 moles CH4 / 0.625 moles O2 = 0.4987Since the balanced equation requires 1 mole of CH4 for every 2 moles of O2, the mole ratio should be 0.5. Since the actual mole ratio 0.4987 is very close to the required ratio, we can conclude that both reactants are almost completely consumed, and neither is significantly in excess.However, since the actual mole ratio is slightly less than the required ratio, we can conclude that CH4 is the limiting reactant, and O2 is the excess reactant.To find out how much of the excess reactant remains, we can calculate the amount of O2 that was actually consumed:Amount of O2 consumed = 0.3119 moles CH4 * 2 moles O2 / 1 mole CH4 = 0.6238 moles O2Now, we can subtract the amount of O2 consumed from the initial amount of O2:Remaining O2 = 0.625 moles O2 - 0.6238 moles O2 = 0.0012 moles O2Finally, we can convert the remaining moles of O2 back to grams:Remaining O2 grams = 0.0012 moles O2 * 32.00 g/mol = 0.0384 gramsSo, the limiting reactant is methane CH4 , and there are approximately 0.0384 grams of oxygen O2 remaining after the reaction is complete.