Calculate the enthalpy change for the precipitation reaction that occurs when 50.0 mL of 0.100 M AgNO3 (aq) and 25.0 mL of 0.100 M NaCl (aq) are mixed together. The final temperature of the solution is 25.0°C, and the initial temperature of both solutions was 25.0°C. The heat capacity of the solution is 4.18 J/(g °C) and the density of each solution is 1.00 g/mL. Assume that the reaction goes to completion and that the density and specific heat capacity of the resulting solution are the same as those of pure water.

Question

Calculate the enthalpy change for the precipitation reaction that occurs when 50.0 mL of 0.100 M AgNO3 (aq) and 25.0 mL of 0.100 M NaCl (aq) are mixed together. The final temperature of the solution is 25.0°C, and the initial temperature of both solutions was 25.0°C. The heat capacity of the solution is 4.18 J/(g °C) and the density of each solution is 1.00 g

Answer 1

First, we need to determine the limiting reactant in the precipitation reaction. The balanced chemical equation for the reaction is:AgNO3  aq  + NaCl  aq   AgCl  s  + NaNO3  aq Now, we can calculate the moles of each reactant:Moles of AgNO3 =  0.100 mol/L  *  50.0 mL  *  1 L/1000 mL  = 0.00500 molMoles of NaCl =  0.100 mol/L  *  25.0 mL  *  1 L/1000 mL  = 0.00250 molSince there are fewer moles of NaCl, it is the limiting reactant. The reaction will produce 0.00250 mol of AgCl precipitate.Next, we can calculate the mass of the solution:Mass of the solution =  50.0 mL + 25.0 mL  *  1.00 g/mL  = 75.0 gSince the initial and final temperatures of the solution are the same  25.0C , there is no temperature change, and thus no heat is exchanged with the surroundings. Therefore, the enthalpy change for the reaction is 0 J.