To calculate the standard enthalpy change for the neutralization reaction, we first need to determine the heat released during the reaction q and then divide it by the number of moles of the limiting reactant.The balanced chemical equation for the neutralization reaction between hydrochloric acid HCl and sodium hydroxide NaOH is:HCl aq + NaOH aq NaCl aq + H2O l Since both solutions have the same concentration 0.1 M and volume 50 mL , they will react in a 1:1 ratio, and neither is a limiting reactant.First, let's find the number of moles of HCl and NaOH:moles = Molarity Volume in L moles of HCl = 0.1 M 0.050 L = 0.005 molmoles of NaOH = 0.1 M 0.050 L = 0.005 molNow, let's calculate the heat released during the reaction q :q = mcTwhere m is the mass of the solution, c is the specific heat capacity, and T is the temperature change.Since the density of the solution is 1 g/mL, the mass of the 100 mL solution is 100 g. The temperature change T can be obtained from the experiment, but it is not provided in the problem. Therefore, we will assume that the temperature change is given as T.q = 100 g 4.18 J/gC T Now, we can calculate the standard enthalpy change H by dividing the heat released q by the number of moles of the limiting reactant in this case, either HCl or NaOH, since they have the same number of moles :H = q / molesH = 100 g 4.18 J/gC T / 0.005 molH = 418 J/C T / 0.005 molH = 83600 J/mol TSince the temperature change T is not provided in the problem, we cannot calculate the exact value of the standard enthalpy change H . However, we can express it in terms of T:H = 83600 J/mol T