To calculate the enthalpy change, we first need to write the balanced chemical equation for the reaction:CH4 g + 2O2 g CO2 g + 2H2O g Next, we need to find the standard enthalpies of formation Hf for each compound involved in the reaction. The values are as follows:Hf CH4 = -74.8 kJ/molHf O2 = 0 kJ/mol since it is an element in its standard state Hf CO2 = -393.5 kJ/molHf H2O = -241.8 kJ/molNow, we can calculate the enthalpy change H for the reaction using the formula:H = [Hf products ] - [Hf reactants ]H = [1 mol CO2 -393.5 kJ/mol + 2 mol H2O -241.8 kJ/mol ] - [1 mol CH4 -74.8 kJ/mol + 2 mol O2 0 kJ/mol ]H = -393.5 - 2 241.8 + 74.8 kJ/molH = -802.3 kJ/molNow, we need to find the moles of CH4, CO2, and H2O involved in the reaction:moles of CH4 = 2.00 g / 12.01 g/mol C + 4 1.01 g/mol H = 2.00 g / 16.04 g/mol = 0.1247 molmoles of CO2 = 1.50 g / 12.01 g/mol C + 2 16.00 g/mol O = 1.50 g / 44.01 g/mol = 0.03409 molmoles of H2O = 0.75 g / 2 1.01 g/mol H + 16.00 g/mol O = 0.75 g / 18.02 g/mol = 0.04164 molSince the balanced equation shows that 1 mol of CH4 produces 1 mol of CO2 and 2 mol of H2O, we can assume that the reaction went to completion and all the CH4 was consumed. Therefore, we can use the moles of CH4 to find the enthalpy change for the given amounts of reactants and products:H for 0.1247 mol CH4 = -802.3 kJ/mol 0.1247 mol = -100.1 kJSo, the enthalpy change for the reaction of 2.00 g of methane with excess oxygen to form 1.50 g of carbon dioxide and 0.75 g of water is -100.1 kJ.