To calculate the heat released during the neutralization reaction, we need to determine the amount of heat generated by the reaction and then use the specific heat capacity and density to find the heat released.First, let's determine the moles of HCl and NaOH involved in the reaction:Moles of HCl = Volume of HCl x Concentration of HCl Moles of HCl = 50 mL x 0.1 M = 5 mmolMoles of NaOH = Volume of NaOH x Concentration of NaOH Moles of NaOH = 50 mL x 0.1 M = 5 mmolSince the moles of HCl and NaOH are equal, the reaction will go to completion, and all the HCl and NaOH will react to form water and NaCl.The balanced chemical equation for the reaction is:HCl aq + NaOH aq NaCl aq + H2O l The enthalpy change H for the neutralization reaction between HCl and NaOH is approximately -57.3 kJ/mol.Now, we can calculate the heat released q during the reaction:q = moles of HCl x Hq = 5 mmol x -57.3 kJ/mol q = -286.5 kJSince the heat released is negative, this indicates that the reaction is exothermic, and heat is released to the surroundings.Now, we can use the specific heat capacity c and density to find the heat released in J/gC:The total volume of the solution is 100 mL, and since the density is 1.0 g/mL, the mass of the solution is 100 g.q = mcTT = q / mc T = -286.5 kJ / 100 g x 4.18 J/gC T = -286.5 x 10^3 J / 100 g x 4.18 J/gC T = -6.85 CThe heat released during the neutralization reaction is 286.5 kJ, and the temperature change is -6.85 C.