First, we need to determine the limiting reactant. To do this, we will calculate the moles of each reactant and compare their mole ratios.Molar mass of CH4 = 12.01 C + 4 * 1.01 H = 16.05 g/molMoles of CH4 = 12.5 g / 16.05 g/mol = 0.778 molesMolar mass of O2 = 2 * 16.00 O = 32.00 g/molMoles of O2 = 50.0 g / 32.00 g/mol = 1.5625 molesNow, we will compare the mole ratios of CH4 and O2 to the balanced equation:CH4 : O2 = 1 : 20.778 moles CH4 : 1.5625 moles O2 = 1 : 2.008Since the ratio of CH4 to O2 is slightly less than the balanced equation 1 : 2 , CH4 is the limiting reactant.Now, we will calculate the mass of CO2 produced using the stoichiometry of the balanced equation:1 mole CH4 produces 1 mole CO20.778 moles CH4 will produce 0.778 moles CO2Molar mass of CO2 = 12.01 C + 2 * 16.00 O = 44.01 g/molMass of CO2 = 0.778 moles * 44.01 g/mol = 34.23 gTherefore, 34.23 grams of CO2 will be produced when 12.5 grams of methane reacts with 50.0 grams of oxygen.