To calculate the enthalpy of precipitation, we first need to determine the balanced chemical equation for the reaction and the enthalpy change for the reaction H . The balanced chemical equation for the reaction between barium chloride BaCl2 and sodium sulfate Na2SO4 is:BaCl2 aq + Na2SO4 aq BaSO4 s + 2NaCl aq Next, we need to determine the limiting reactant. To do this, we can compare the moles of each reactant:Moles of BaCl2 = volume concentration = 50.0 mL 0.200 M = 10.0 mmolMoles of Na2SO4 = volume concentration = 30.0 mL 0.150 M = 4.50 mmolSince the stoichiometry of the reaction is 1:1, we can see that Na2SO4 is the limiting reactant. Therefore, the reaction will produce 4.50 mmol of BaSO4.Now we need to find the enthalpy change H for the reaction. We can use the standard enthalpy of formation Hf values for each compound in the reaction:Hf BaCl2, aq = -858 kJ/molHf Na2SO4, aq = -1381 kJ/molHf BaSO4, s = -1435 kJ/molHf NaCl, aq = -407 kJ/molUsing Hess's Law, we can calculate the enthalpy change for the reaction:H = [Hf BaSO4, s + 2Hf NaCl, aq ] - [Hf BaCl2, aq + Hf Na2SO4, aq ]H = [ -1435 kJ/mol + 2 -407 kJ/mol ] - [ -858 kJ/mol + -1381 kJ/mol ]H = -1435 - 814 - -858 - 1381 H = -2249 + 2239H = -10 kJ/molNow we can calculate the enthalpy of precipitation for the reaction:Enthalpy of precipitation = H moles of limiting reactantEnthalpy of precipitation = -10 kJ/mol 4.50 mmol Enthalpy of precipitation = -45 kJTherefore, the enthalpy of precipitation for this reaction is -45 kJ.