The enthalpy of neutralizatino for the reaction of a strong acid with a strong base is -56 kJ/mol of water produced. How much energy will be released when 230.0 mL of 0.400 M HCl is mixedwith 150.5 mL of 0.500 M NaOH?
How much water is produced in the reaction?
How do I figure out how much water is produced in the reaction? Would I change the HCl and the NaOH to moles?
Yes. The one with the least moles will determine how much water is made.
To figure out how much water is produced in the reaction, you need to determine the limiting reagent between HCl and NaOH. The limiting reagent is the reactant that is completely consumed, thus determining the maximum amount of product that can be formed.
To find the limiting reagent, we can compare the number of moles of HCl and NaOH.
First, convert the volumes of the solutions to moles using the given molarity (concentration):
Number of moles of HCl = (0.400 M) x (0.2300 L) = 0.0920 moles
Number of moles of NaOH = (0.500 M) x (0.1505 L) = 0.0753 moles
From the calculations, we can see that NaOH has fewer moles than HCl. Therefore, NaOH is the limiting reagent, and the amount of water produced will be determined by the moles of NaOH.
Since the balanced equation for the reaction is:
HCl + NaOH → H2O + NaCl
We know that the ratio of water to NaOH is 1:1. Thus, the amount of water produced will be 0.0753 moles.
To determine the actual energy released in the reaction, you can use the enthalpy change of neutralization (-56 kJ/mol of water produced) given in the question. Multiply this value by the moles of water produced (0.0753 moles) to find the total energy released in the reaction.