To calculate the enthalpy change for the oxidation of 5 moles of hydrogen gas, we will use the following balanced chemical equation:2 H g + O g 2 HO l The standard enthalpy of formation Hf of water HO is -285.8 kJ/mol, and the standard enthalpy of formation of hydrogen gas H and oxygen gas O are both 0 kJ/mol, as they are in their elemental forms.Using Hess's Law, we can calculate the enthalpy change H for the reaction:H = [Hf products ] - [Hf reactants ]H = [2 -285.8 kJ/mol ] - [2 0 kJ/mol + 0 kJ/mol]H = -571.6 kJNow, we need to calculate the enthalpy change for the oxidation of 5 moles of hydrogen gas. Since the balanced equation shows that 2 moles of hydrogen gas react to produce 2 moles of water, we can set up a proportion:5 moles H / 2 moles H = x / -571.6 kJx = 5 moles H -571.6 kJ / 2 moles Hx = -1429 kJTherefore, the enthalpy change for the oxidation of 5 moles of hydrogen gas is -1429 kJ.