What is the Molar Enthalpy of Neutralization in kJ/mol if 5.00 moles of HCl(aq) (acid) neutralized 5.00 moles of NaOH(aq) (base) and released 6649J of heat? *Make sure you report your answer using the correct sign*
DrBob222 gave me some help and I worked it out and got 1329.8 but my teacher says its still not right. HELP!
Is this a made up problem? You reported J/mol and the problem asks for kJ/mol. Since the reaction is exothermic your prof may want the negative sign included also.
dH = -6649 J/5 mol = -1329.8 J/mol or -1.3298 kJ/mol and since you are limited to 3 significant figures I would round to 1.33 kJ/mol.
I still don't like that answer; the correct value for most acids/bases is 57.1 kJ/mol but if this is a made up problem that's beside the point. Let me know how this turns out. If I'm making an error I need to know it.
To solve this problem, we can use the formula:
Molar Enthalpy of Neutralization = Heat released (in J) / Number of moles of HCl reacted
First, we need to convert the heat released from joules (J) to kilojoules (kJ):
6649 J ÷ 1000 = 6.649 kJ
Now, we can substitute the values into the formula:
Molar Enthalpy of Neutralization = 6.649 kJ / 5.00 moles
Molar Enthalpy of Neutralization = 1.33 kJ/mol
So, according to my calculations, it appears that your answer of 1.3298 kJ/mol is very close. However, it seems like you rounded incorrectly. The correct answer, using the correct sign, would be 1.33 kJ/mol. Double-check your rounding and ensure you are reporting the answer with the correct number of significant figures.