To calculate the change in enthalpy H for the reduction of iron III oxide to iron metal, we can use the following equation:H = Hf products - Hf reactants The balanced chemical equation for the reduction of iron III oxide with hydrogen gas is:Fe2O3 s + 3H2 g 2Fe s + 3H2O l Given that Hf[Fe2O3] = -824.2 kJ/mol and Hf[Fe] = 0 kJ/mol, we can calculate the enthalpy change for the reaction. Note that the Hf[H2] = 0 kJ/mol since it is an element in its standard state.First, we need to find the enthalpy change for the formation of water. The standard enthalpy of formation for liquid water is -285.8 kJ/mol.Now, we can calculate the enthalpy change for the reaction:H = [2 Hf[Fe] + 3 Hf[H2O] ] - [Hf[Fe2O3] + 3 Hf[H2] ]H = [2 0 + 3 -285.8 ] - [-824.2 + 3 0 ]H = [-857.4] - [-824.2]H = -33.2 kJThe change in enthalpy for the reduction of 2 moles of iron III oxide to iron metal in the presence of hydrogen gas is -33.2 kJ.