To calculate the standard enthalpy change for the neutralization of HCl with NaOH, we first need to determine the amount of heat released during the reaction and then divide it by the moles of water formed.1. Calculate the moles of HCl and NaOH:moles of HCl = volume L concentration M moles of HCl = 0.050 L 0.100 M = 0.0050 molmoles of NaOH = volume L concentration M moles of NaOH = 0.075 L 0.075 M = 0.005625 mol2. Determine the limiting reactant:Since HCl and NaOH react in a 1:1 ratio, the limiting reactant is HCl with 0.0050 mol.3. Calculate the heat released during the reaction:The heat released q can be calculated using the formula q = mcT, where m is the mass of the solution, c is the specific heat capacity, and T is the change in temperature.First, we need to find the mass of the solution:mass = volume densitymass = 50.0 mL + 75.0 mL 1.00 g/mL = 125.0 gNext, we need to find the change in temperature T . Since we are not given the initial and final temperatures, we will assume that the reaction is exothermic and the temperature change is negligible. This is a common assumption in neutralization reactions.Now we can calculate the heat released q :q = mcTq = 125.0 g 4.18 J/gK TSince we assumed that the temperature change is negligible, we can rewrite the equation as:q = 125.0 g 4.18 J/gK 0q = 0 J4. Calculate the standard enthalpy change H :H = q / moles of water formedH = 0 J / 0.0050 molSince the heat released is 0 J, the standard enthalpy change is also 0 J/mol. This is an idealized result, as in reality, there would be a small temperature change and a non-zero enthalpy change. However, without the initial and final temperatures, we cannot calculate the exact value.