To calculate the standard enthalpy change of neutralization H for the reaction between HCl and NaOH, we can use the formula:H = -q / nwhere q is the heat released during the reaction and n is the number of moles of the limiting reactant.First, we need to determine the limiting reactant. Since both solutions have the same volume 50 mL and concentration 1 M , they have the same number of moles:moles of HCl = moles of NaOH = volume concentration = 50 mL 1 mol/L = 0.050 molSince both reactants have the same number of moles, neither is limiting.Next, we need to calculate the heat released q during the reaction. We can use the formula:q = mcTwhere m is the mass of the solution, c is the specific heat capacity of the solution, and T is the temperature change. Assuming the solution is mostly water, we can use the specific heat capacity of water, which is 4.18 J/gC. The total volume of the solution is 100 mL, and since the density of water is approximately 1 g/mL, the mass of the solution is approximately 100 g.q = 100 g 4.18 J/gC 10C = 4180 JNow we can calculate the standard enthalpy change of neutralization:H = -q / n = -4180 J / 0.050 mol = -83600 J/molThe standard enthalpy change of neutralization for the reaction between HCl and NaOH is approximately -83.6 kJ/mol.