calculation of neutralization heat

[anto007]

Hello everyone, I have this exercise: "The neutralization heat of HCl by NaOH is ΔH°r = -56.2 kJ / mol. How much heat is released when 125 mL of 1,750 M HCl is mixed with 195 mL of 0.667 M NaOH?"
Would you help me figure out how to fix it?
First I determined n
n(NaOH)=M*V= 0.667 mol/L*0.195 L=0.13 mol
n(HCl)=M*V=1.750 mol/L*0.125 L=0.219 mol

but i don't know how to continue, thanks a lot for your help

[ChenBeier]

The unit of heat released is kJ/ mol. So what is the maximum mol what react according NaOH + HCl => NaCl + H₂O ?

[anto007]

NaOH + HCl => NaCl + H₂O

1mol + 1 mol => 1 mol + 1 mol

right?

[ChenBeier]

You have to calculate how much mol you have with the datas given. Do you you have 1:1 or something else. But you have to get the ratio 1:1.

[anto007]

ah ok, I understand what you were asking me
the total moles I calculated, adding n(NaOH) and n(HCl), are 0.349 mol
but they are not in a 1: 1 ratio, with these data NaOh is the limiting reagent.
Q=-ΔH°r (heat released)
Q=-ΔH°r*n= -(-56.2 kJ/mol)*0.13 mol= 7.306 kJ
it's correct?

[ChenBeier]

Correct.

[anto007]

Great, thanks so much

[adamusa]

good

[ChenBeier]

Please no reply to 2 year old thread