To calculate the enthalpy change when calcium chloride dihydrate CaCl22H2O is heated to release its water of hydration, we need to consider the enthalpy change of the reaction:CaCl22H2O s CaCl2 s + 2H2O g First, we need to find the enthalpy of formation of CaCl22H2O. We know the enthalpy of formation of CaCl2 is -795.8 kJ/mol. The enthalpy of formation of water H2O is -285.8 kJ/mol. Therefore, the enthalpy of formation of CaCl22H2O can be calculated as follows:Hf CaCl22H2O = Hf CaCl2 + 2 * Hf H2O Hf CaCl22H2O = -795.8 kJ/mol + 2 * -285.8 kJ/mol Hf CaCl22H2O = -1367.4 kJ/molNow, we need to find the enthalpy change of the reaction Hrxn . To do this, we can use the following equation:Hrxn = Hf products - Hf reactants In this case, the products are CaCl2 and 2H2O, and the reactant is CaCl22H2O. Plugging in the values, we get:Hrxn = Hf CaCl2 + 2 * Hf H2O - Hf CaCl22H2O Hrxn = -795.8 kJ/mol + 2 * -285.8 kJ/mol - -1367.4 kJ/mol Hrxn = 44 kJ/molSo, the enthalpy change when calcium chloride dihydrate is heated to release its water of hydration is approximately 44 kJ/mol.