the heat of neutralization for a strong acid in dilute water solution is about 60 kJ/mol of H+. What quantity of heat in kJ is produced when 100 mL of 3 M HCl is mixed with 100 mL of 1 M KOH?

0.100 L x 3 M HCl = 0.3 mol HCl

0.100 L x 1 M KOH = 0.1 mol KOH.
So it looks like 0.1 mole will be neutralized and it is 60 kJ/mol so
0.1 x 60 = ??

To find the quantity of heat produced when 100 mL of 3 M HCl is mixed with 100 mL of 1 M KOH, we can use the concept of heat of neutralization.

The heat of neutralization is the heat released when one mole of a substance reacts with another to form a neutral product. In this case, we are given that the heat of neutralization for a strong acid (HCl) in a dilute water solution is approximately 60 kJ/mol.

First, we need to determine the number of moles of HCl and KOH used in the reaction.

Number of moles of HCl = (volume in liters) x (molarity)
= (100 mL / 1000) L x 3 M
= 0.1 L x 3 M
= 0.3 moles

Number of moles of KOH = (volume in liters) x (molarity)
= (100 mL / 1000) L x 1 M
= 0.1 L x 1 M
= 0.1 moles

Since the stoichiometry of the reaction between HCl and KOH is 1:1, meaning they react in a one-to-one ratio, the moles of HCl and KOH are equal.

Therefore, the number of moles of HCl and KOH used in the reaction is 0.1 moles each.

Now, we can calculate the quantity of heat produced using the heat of neutralization.

Quantity of heat = (heat of neutralization) x (number of moles)
= 60 kJ/mol x 0.1 moles
= 6 kJ

Hence, when 100 mL of 3 M HCl is mixed with 100 mL of 1 M KOH, approximately 6 kJ of heat is produced.