The enthalpy of neutralization 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 185.0 mL of 0.400 M HCl is mixed with 152.5 mL of 0.500 M NaOH?
0.1850*0.400 = ?moles HCl
0.1525*0.500 = x moles NaOH
I have 0.07625 moles NaOH and 0.07400 moles HCl; therefore, 0.07400 moles H2O should be formed.
56 kJ/mol x moles H2O formed = heat released.
-4.1 KJ
That's what I have.
To calculate the energy released during the neutralization reaction, we need to use the balanced equation and the stoichiometry of the reaction.
The balanced equation for the neutralization reaction between HCl (strong acid) and NaOH (strong base) is:
HCl + NaOH -> NaCl + H2O
From the balanced equation, we see that one mole of water is produced for every one mole of HCl and NaOH reacted.
Step 1: Calculate the number of moles of HCl and NaOH.
To find the moles, we'll use the formula:
moles = concentration (M) x volume (L)
For HCl:
moles of HCl = concentration of HCl x volume of HCl
= 0.400 M (0.185 L)
= 0.0740 moles
For NaOH:
moles of NaOH = concentration of NaOH x volume of NaOH
= 0.500 M (0.1525 L)
= 0.0763 moles
Step 2: Determine the limiting reactant.
The limiting reactant is the one that is completely consumed in the reaction, determining the maximum amount of products that can be formed. It can be identified by comparing the mole ratio of the reactants from the balanced equation.
From the balanced equation, the mole ratio of HCl to NaOH is 1:1.
Since the ratio of moles of HCl to NaOH is approximately 0.0740:0.0763, HCl is the limiting reactant.
Step 3: Calculate the moles of water produced.
Since HCl is the limiting reactant, it will produce the same number of moles of water.
moles of water = moles of HCl = 0.0740 moles
Step 4: Calculate the energy released.
The enthalpy of neutralization is given as -56 kJ/mol of water produced.
Energy released = enthalpy of neutralization x moles of water
= -56 kJ/mol x 0.0740 moles
= -4.14 kJ
Therefore, approximately 4.14 kJ of energy will be released when 185.0 mL of 0.400 M HCl is mixed with 152.5 mL of 0.500 M NaOH.