calculation of neutralization heat

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anto007
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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
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ChenBeier
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Re: calculation of neutralization heat

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The unit of heat released is kJ/ mol. So what is the maximum mol what react according NaOH + HCl => NaCl + H2O ?
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anto007
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Re: calculation of neutralization heat

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NaOH + HCl => NaCl + H2O

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

right?
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Re: calculation of neutralization heat

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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.
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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.
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ChenBeier
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Re: calculation of neutralization heat

Post by ChenBeier »

Correct.
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anto007
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Re: calculation of neutralization heat

Post by anto007 »

Great, thanks so much :D
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Re: calculation of neutralization heat

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good
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Re: calculation of neutralization heat

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Please no reply to 2 year old thread
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