To calculate the standard enthalpy change H for the precipitation reaction, we need to first determine the heat q released during the reaction, and then divide it by the moles of the limiting reactant.The balanced chemical equation for the precipitation reaction is:AgNO3 aq + NaCl aq AgCl s + NaNO3 aq First, we need to determine the limiting reactant. Since both solutions have the same volume 50.0 mL and concentration 0.100 M , they have the same number of moles:moles of AgNO3 = moles of NaCl = 50.0 mL 0.100 mol/L = 0.00500 molSince the stoichiometry of the reaction is 1:1, both reactants are consumed at the same rate, and there is no limiting reactant in this case.Next, we need to calculate the heat q released during the reaction. We can use the formula:q = mcTwhere m is the mass of the solution, c is the specific heat capacity of the solution, and T is the temperature change.Assuming the specific heat capacity of the solution is the same as water c = 4.18 J/gC and the density of the solution is approximately the same as water 1 g/mL , we can calculate the mass of the solution:mass of the solution = volume of the solution density of the solution = 50.0 mL + 50.0 mL 1 g/mL = 100.0 gNow we can calculate the heat q released:q = 100.0 g 4.18 J/gC -2.5C = -1045 JSince the heat is negative, it means the reaction is exothermic.Finally, we can calculate the standard enthalpy change H by dividing the heat q by the moles of the limiting reactant in this case, either AgNO3 or NaCl :H = q / moles of limiting reactant = -1045 J / 0.00500 mol = -209000 J/molTherefore, the standard enthalpy change for the precipitation reaction is -209 kJ/mol.