To calculate the enthalpy change for the complete oxidation of 12.5 g of methane gas, we need to first determine the balanced chemical equation for the reaction. The complete oxidation of methane CH4 involves reacting it with oxygen O2 to produce carbon dioxide CO2 and water H2O :CH4 g + 2 O2 g CO2 g + 2 H2O l Next, we need to determine the number of moles of methane involved in the reaction. The molar mass of methane is:CH4 = 12.01 g/mol C + 4 * 1.01 g/mol H = 16.05 g/molNow, we can calculate the moles of methane:moles of CH4 = mass / molar mass = 12.5 g / 16.05 g/mol 0.778 molesNow we need to calculate the enthalpy change for the reaction using the enthalpy of formation values given:H_reaction = [1 * H_f CO2 + 2 * H_f H2O ] - [H_f CH4 + 2 * H_f O2 ]Since the enthalpy of formation for an element in its standard state is zero, H_f O2 = 0. The enthalpy of formation for CO2 g can be calculated using the given values for CH4 g and H2O l :H_f CO2 = H_f H2O - H_f CH4 - H_combustion CH4 The enthalpy of combustion of methane H_combustion CH4 is the enthalpy change for the complete oxidation of one mole of methane. It can be calculated using the heat capacity of water 4.18 J/gC and the mass and temperature change of the water:q = mcTq = 150 g * 4.18 J/gC * 35C - 25C = 6270 J = 6.27 kJSince 0.778 moles of methane are combusted, the enthalpy of combustion for one mole of methane is:H_combustion CH4 = 6.27 kJ / 0.778 moles 8.06 kJ/molNow we can calculate the enthalpy of formation for CO2 g :H_f CO2 = -285.8 kJ/mol - -74.8 kJ/mol - 8.06 kJ/mol -202.94 kJ/molFinally, we can calculate the enthalpy change for the reaction:H_reaction = [1 * -202.94 kJ/mol + 2 * -285.8 kJ/mol ] - [-74.8 kJ/mol + 2 * 0]H_reaction = -202.94 kJ/mol - 571.6 kJ/mol + 74.8 kJ/mol -699.74 kJ/molSince there are 0.778 moles of methane, the total enthalpy change for the complete oxidation of 12.5 g of methane is:H_total = 0.778 moles * -699.74 kJ/mol -544.19 kJ