need to find the molar heat of neutralisation with 2M Hcl and 1M NaOH (1 Viewer)

[lindanguyen88]

i got a problem with chem calculations,
can anyone show me how to calculate the molar heat of neutralisation?
basically i have
2M of HCl
1M of NaOH
these can be in the ratios like 80:20 , 60:40 , 50:50 , 40:60 , 20:80
these are what i have to test and graph.

anyways when i find the intersection between the graph,

say it was 33mL of 1M NaOH and 66mL of HCl, and the heat change was say
9 degrees. how would i calculate that??
becuase the books i've looked at they all have 1M to 1M ratio and nothing different like 2M to 1M ratio and i'm not sure how to calculate it.
need help asap!!! prac assessment tomorrow!!!
much thanks!

 

[Riviet]

HCl + NaOH -> NaCl + H₂O

n_HCl=cV
=2x0.066
=0.132

n_NaOH=cV
=1x0.033
=0.033

But from equation, 1 mole of HCl reacts with 1 mole of NaOH, so HCl is in excess. This means we use the number of moles of NaOH for our calculation molar heat of neutralisation.

total mass = 66 + 33 = 99g

∆H=mC∆T
=99x4.18x9
=3724.38J
=3724.38/(1000x0.033) KJ/mol
=113 KJ/mol

Note that the actual experiental value should be around 50-60 KJ/mol. Hope that helps. Good luck with the prac assessment too.

 

[bazool]

You have a prac assesment on heat of neutralisation? Isn't that a bit weird, our chem teacher said it's not on the syllabus anymore

 

[hart2hart]

really? wow i see. oh about the prac i did it omg it was soooo rushed. in the end i got 73% wished i got more. suprised i did get it anyways. i realised well found out that sulfuric acid was much easier to do cuz it was diprotic so it would have \been 50 50 where it was neutralised. ah well hope i do better in my trials thanks alot!!!

 

[hart2hart]

really? wow i see. oh about the prac i did it omg it was soooo rushed. in the end i got 73% wished i got more. suprised i did get it anyways. i realised well found out that sulfuric acid was much easier to do cuz it was diprotic so it would have \been 50 50 where it was neutralised. ah well hope i do better in my trials thanks alot!!!

oh btw lindanguyen88 is also me, heheh someone hacked into that account after yea, thought it must hav been weird that someone else had replied for them lols. wel thanks anyways heheh byess