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 do I figure out how much water is produced in the reaction? Would I change the HCl and the NaOH to moles?
see below.
To figure out how much water is produced in the reaction, you need to first determine the moles of HCl and NaOH that are being mixed together.
To do this, you can use the formula:
moles = concentration (M) x volume (L)
For HCl, the concentration is 0.400 M and the volume is 230.0 mL. Convert the volume to liters by dividing by 1000:
volume (L) = 230.0 mL / 1000 = 0.230 L
Now, calculate the moles of HCl:
moles of HCl = 0.400 M x 0.230 L = 0.092 mol
Similarly, for NaOH, the concentration is 0.500 M and the volume is 150.5 mL. Convert the volume to liters:
volume (L) = 150.5 mL / 1000 = 0.1505 L
Calculate the moles of NaOH:
moles of NaOH = 0.500 M x 0.1505 L = 0.075 mol
Now, we can use stoichiometry to determine the moles of water produced. From the balanced chemical equation, we know that the stoichiometric ratio between HCl and water is 1:1, and between NaOH and water is also 1:1.
Since the reactants are a strong acid and a strong base, we can assume that the reaction goes to completion and that all the HCl and NaOH react completely.
Therefore, the moles of water produced are equal to the smaller amount of moles between HCl and NaOH:
moles of water produced = min(0.092 mol, 0.075 mol) = 0.075 mol
Finally, to determine the energy released during the neutralization reaction, you can use the enthalpy of neutralization, which is given as -56 kJ/mol of water produced.
Calculate the energy released:
Energy released = enthalpy of neutralization x moles of water produced
= -56 kJ/mol x 0.075 mol
= -4.2 kJ
Therefore, the amount of energy released when 230.0 mL of 0.400 M HCl is mixed with 150.5 mL of 0.500 M NaOH is -4.2 kJ.