To calculate the enthalpy change for the reduction of CuO using hydrogen gas, we need to know the standard enthalpy of formation for each compound involved in the reaction. The standard enthalpy of formation Hf values are as follows:CuO s : -155.2 kJ/molH2 g : 0 kJ/mol since it's an element in its standard state Cu s : 0 kJ/mol since it's an element in its standard state H2O l : -285.8 kJ/molNow, we can use Hess's Law to calculate the enthalpy change for the reaction:Hrxn = [Hf products ] - [Hf reactants ]Hrxn = [Hf Cu + Hf H2O ] - [Hf CuO + Hf H2 ]Hrxn = [0 + -285.8 ] - [ -155.2 + 0]Hrxn = -130.6 kJ/molNow, we need to determine the number of moles of CuO and H2 involved in the reaction. The molar mass of CuO is 63.5 Cu + 16.0 O = 79.5 g/mol. Moles of CuO = mass / molar mass = 2.50 g / 79.5 g/mol = 0.0314 molAt STP standard temperature and pressure , 1 mole of gas occupies 22.4 L. Moles of H2 = volume / molar volume = 0.500 L / 22.4 L/mol = 0.0223 molSince the balanced chemical equation has a 1:1 stoichiometry between CuO and H2, we can see that H2 is the limiting reactant. Therefore, we will base our calculations on the moles of H2.Enthalpy change for the reaction based on the moles of H2:H = Hrxn moles of H2 = -130.6 kJ/mol 0.0223 mol = -2.91 kJThe enthalpy change for the reduction of 2.50 g of CuO using 0.500 L of H2 gas at STP is approximately -2.91 kJ.