To calculate the enthalpy of precipitation for the reaction between NaOH and CaCl2, we need to first write the balanced chemical equation for the reaction:CaCl2 aq + 2 NaOH aq Ca OH 2 s + 2 NaCl aq Next, we need to use Hess's Law, which states that the enthalpy change of a reaction is the sum of the enthalpy changes of formation of the products minus the sum of the enthalpy changes of formation of the reactants.H = Hf products - Hf reactants We are given the enthalpy of formation for NaCl and Ca OH 2. However, we need to find the enthalpy of formation for NaOH and CaCl2. The standard enthalpy of formation for NaOH is -425.6 kJ/mol, and for CaCl2 is -795.4 kJ/mol.Now we can plug in the values into Hess's Law equation:H = [1 -986.09 kJ/mol + 2 -407.14 kJ/mol ] - [1 -795.4 kJ/mol + 2 -425.6 kJ/mol ]H = -986.09 - 814.28 - -795.4 - 851.2 H = -1800.37 + 1646.6H = -153.77 kJ/molThe enthalpy of precipitation for the reaction between 0.1 M NaOH and 0.1 M CaCl2 at 25C is -153.77 kJ/mol.