To calculate the enthalpy change for the combustion of methane, we can use the following balanced chemical equation:CH g + 2 O g CO g + 2 HO l Now, we can use Hess's Law to calculate the enthalpy change for this reaction:H = Hf products - Hf reactants For the products, we have 1 mole of CO and 2 moles of HO. For the reactants, we have 1 mole of CH and 2 moles of O. The enthalpy of formation of O is 0 kJ/mol since it is in its standard state.H = [ 1 mol CO -393.5 kJ/mol CO + 2 mol HO -285.8 kJ/mol HO ] - [ 1 mol CH -74.8 kJ/mol CH + 2 mol O 0 kJ/mol O ]H = -393.5 kJ + -571.6 kJ - -74.8 kJ H = -965.1 kJ + 74.8 kJH = -890.3 kJ/molNow, we need to find the enthalpy change for 2.5 moles of methane:H_total = H moles of CHH_total = -890.3 kJ/mol 2.5 molH_total = -2225.75 kJThe enthalpy change when 2.5 moles of methane gas is completely burned in excess oxygen gas is -2225.75 kJ.