calculation of neutralization heat

Chemistry and homework help forum.

Organic Chemistry, Analytical Chemistry, Biochemistry, Physical Chemistry, Computational Chemistry, Theoretical Chemistry, High School Chemistry, Colledge Chemistry and University Chemistry Forum.

Share your chemistry ideas, discuss chemical problems, ask for help with scientific chemistry questions, inspire others by your chemistry vision!

Please feel free to start a scientific chemistry discussion here!

Discuss chemistry homework problems with experts!

Ask for help with chemical questions and help others with your chemistry knowledge!

Moderators: expert, ChenBeier, Xen

Post Reply
User avatar
anto007
Sr. Member
Sr. Member
Posts: 32
Joined: Tue Aug 10, 2021 10:13 am
Location: Italy

calculation of neutralization heat

Post by 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
The mind is like a parachute, it only works if it opens. -A. Einstein
User avatar
ChenBeier
Distinguished Member
Distinguished Member
Posts: 916
Joined: Wed Sep 27, 2017 7:25 am
Location: Berlin, Germany

Re: calculation of neutralization heat

Post by ChenBeier »

The unit of heat released is kJ/ mol. So what is the maximum mol what react according NaOH + HCl => NaCl + H2O ?
User avatar
anto007
Sr. Member
Sr. Member
Posts: 32
Joined: Tue Aug 10, 2021 10:13 am
Location: Italy

Re: calculation of neutralization heat

Post by anto007 »

NaOH + HCl => NaCl + H2O

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

right?
The mind is like a parachute, it only works if it opens. -A. Einstein
User avatar
ChenBeier
Distinguished Member
Distinguished Member
Posts: 916
Joined: Wed Sep 27, 2017 7:25 am
Location: Berlin, Germany

Re: calculation of neutralization heat

Post by 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.
User avatar
anto007
Sr. Member
Sr. Member
Posts: 32
Joined: Tue Aug 10, 2021 10:13 am
Location: Italy

Re: calculation of neutralization heat

Post by anto007 »

ah ok, I understand what you were asking me :wink:
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?
Last edited by anto007 on Sat Aug 28, 2021 5:15 am, edited 1 time in total.
The mind is like a parachute, it only works if it opens. -A. Einstein
User avatar
ChenBeier
Distinguished Member
Distinguished Member
Posts: 916
Joined: Wed Sep 27, 2017 7:25 am
Location: Berlin, Germany

Re: calculation of neutralization heat

Post by ChenBeier »

Correct.
User avatar
anto007
Sr. Member
Sr. Member
Posts: 32
Joined: Tue Aug 10, 2021 10:13 am
Location: Italy

Re: calculation of neutralization heat

Post by anto007 »

Great, thanks so much :D
The mind is like a parachute, it only works if it opens. -A. Einstein
Post Reply