First, we need to determine the number of moles of hydrogen gas H2 in 2.5 g.moles of H2 = mass / molar massmoles of H2 = 2.5 g / 2.02 g/molmoles of H2 = 1.2376 molThe balanced chemical equation for the reaction of hydrogen gas with oxygen gas to produce water is:2 H2 g + O2 g 2 H2O l From the balanced equation, we can see that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water. Since we have excess oxygen gas, we don't need to worry about the moles of oxygen gas.Now, we can calculate the moles of water produced:moles of H2O = moles of H2 * 2 moles of H2O / 2 moles of H2 moles of H2O = 1.2376 mol * 2/2 moles of H2O = 1.2376 molNow we can calculate the enthalpy change for the reaction using the enthalpy of formation of water:H = moles of H2O * enthalpy of formation of H2OH = 1.2376 mol * -285.83 kJ/mol H = -353.92 kJThe enthalpy change for the reaction of 2.5 g of hydrogen gas with excess oxygen gas to produce water is -353.92 kJ.