To calculate the heat of neutralization, we first need to determine the amount of heat released during the reaction. We can do this using the formula:q = mcTwhere q is the heat released, 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. Since the density of the solution is approximately equal to the density of water 1 g/mL , we can calculate the mass as follows:m = volume of HCl + volume of NaOH = 50 mL + 50 mL = 100 mLm = 100 mL 1 g/mL = 100 gNext, we need the specific heat capacity of the solution. The specific heat capacity of water is 4.18 J/ gC , and since the solution is mostly water, we can assume the specific heat capacity of the solution is also 4.18 J/ gC .Now we can calculate the heat released:T = final temperature - initial temperature = 39C - 25C = 14Cq = mcT = 100 g 4.18 J/ gC 14C = 5852 JNow we need to find the moles of HCl and NaOH reacted. Since the volume and concentration of both solutions are the same, they will react in a 1:1 ratio, and the limiting reactant will be completely consumed.moles of HCl = moles of NaOH = volume concentration = 50 mL 2.0 M = 0.050 L 2.0 mol/L = 0.1 molFinally, we can calculate the heat of neutralization per mole of HCl and NaOH reacted:Heat of neutralization = q / moles of HCl = 5852 J / 0.1 mol = 58520 J/molThe heat of neutralization of hydrochloric acid HCl and sodium hydroxide NaOH when 50 mL of 2.0 M HCl is reacted with 50 mL of 2.0 M NaOH, given that the initial and final temperatures of the solution are 25C and 39C, respectively, is 58,520 J/mol.