To calculate the enthalpy change for the reaction, we first need to determine the amount of heat released or absorbed during the reaction. We can do this using the formula:q = mcTwhere q is the heat released or absorbed, m is the mass of the solution, c is the specific heat capacity of the solution, and T is the change in temperature.First, let's find the mass of the solution. We know the volume 50.0 mL and the density 1.05 g/mL of the silver nitrate solution:mass = volume densitymass = 50.0 mL 1.05 g/mLmass = 52.5 gNext, we need to find the specific heat capacity of the solution. Since the solution is mostly water, we can assume the specific heat capacity is approximately equal to that of water, which is 4.18 J/ gC .Now we can calculate the heat released or absorbed using the formula:q = mcTq = 52.5 g 4.18 J/ gC -5.47C q = -1201.6 JSince the temperature change is negative, the reaction is exothermic, and heat is released. Therefore, q = -1201.6 J.Now we need to find the moles of AgNO3 in the reaction. We can do this using the molarity and volume of the solution:moles = molarity volumemoles = 0.150 mol/L 0.050 Lmoles = 0.0075 molFinally, we can calculate the enthalpy change H for the reaction by dividing the heat released or absorbed by the moles of AgNO3:H = q / molesH = -1201.6 J / 0.0075 molH = -160213.3 J/molThe enthalpy change for the precipitation reaction between 50.0 mL of 0.150 M silver nitrate solution and excess sodium chloride solution is approximately -160.2 kJ/mol.