To calculate the heat of neutralization, we first need to determine the amount of heat released during the reaction q using the formula:q = mcTwhere m is the mass of the solution, c is the specific heat capacity, and T is the temperature change.First, we need to find the mass of the solution. Since the density of water is approximately 1 g/mL, we can assume that the density of the solution is also close to 1 g/mL. Therefore, the mass of the solution is:m = 50 mL HCl + 50 mL NaOH * 1 g/mL = 100 gNow we can calculate the heat released:q = 100 g 4.18 J/gC 8.2 C = 3427.6 JNext, we need to determine the moles of HCl and NaOH that reacted. Since the volume and concentration of both solutions are the same, we can calculate the moles of one reactant and assume that the same amount of moles reacted for the other reactant:moles of HCl = 50 mL 2.0 mol/L = 0.1 molNow we can calculate the heat of neutralization per mole:Heat of neutralization = q / moles of HCl = 3427.6 J / 0.1 mol = 34276 J/molSo, the heat of neutralization when 50 mL of 2.0 M hydrochloric acid reacts with 50 mL of 2.0 M sodium hydroxide solution is approximately 34,276 J/mol.