Final answer:
The limiting reactant is HCl with 0.025 mol. Using the literature value of -55.8 kJ/mol for the enthalpy change of the neutralization of HCl and NaOH, the total enthalpy change for the reaction is -1.395 kJ.
Step-by-step explanation:
To calculate the enthalpy change (kJ) for the neutralization reaction between 25.0 mL of 1.0 M HCl and 25.0 mL of 1.2 M NaOH, first, we need to identify the limiting reactant. It's done by comparing the moles of each reactant since the molar ratio in the balanced equation is 1:1. We convert volumes to moles by using the formula moles = concentration (M) × volume (L).
Moles of HCl = 1.0 M × 0.025 L
= 0.025 mol HCl
Moles of NaOH = 1.2 M × 0.025 L
= 0.03 mol NaOH
Here, HCl is the limiting reactant because it has fewer moles, and the reaction will stop once all HCl is consumed. Next, we use the provided literature value of the reaction, which is -55.8 kJ/mol, to calculate the enthalpy change. Since we have 0.025 mol of the limiting reactant (HCl), the total enthalpy change is:
Enthalpy change = -55.8 kJ/mol × 0.025 mol
= -1.395 kJ.
Therefore, the enthalpy change for this reaction is -1.395 kJ.