[CHM 115] Fwd: Re: Lab

Andrew J. Pounds pounds_aj at mercer.edu
Sun Oct 26 19:01:33 EDT 2014




-------- Original Message --------
Subject: 	Re: Lab
Date: 	Sun, 26 Oct 2014 19:00:24 -0400
From: 	Andrew J. Pounds <pounds_aj at mercer.edu>
Reply-To: 	pounds_aj at mercer.edu



The following reaction takes place in

HCl  + NaOH  -> NaCl + H2O

The density of the NaCl solution is listed on the report form.


On 10/26/14 18:56,  wrote:
>
> So for the mass of the solution for reaction three do I need to know 
> the density of HCl? It isn't listed in the lab report form so can I 
> just assume it's 1 g/mL?
>
> On Oct 26, 2014 5:44 PM, "Andrew J. Pounds" <pounds_aj at mercer.edu 
> <mailto:pounds_aj at mercer.edu>> wrote:
>
>     On 10/26/14 15:19,  wrote:
>>     Hi Dr. Pounds!
>>     I'm trying to do the lab, and we either really messed up in lab
>>     or I'm calculating the deltaH values for reactions three and four
>>     wrong. I think my mistake might be in calculating the mass of the
>>     solution for these reactions. Could you give me any pointers for
>>     calculating the mass of the solution for reactions three and four?
>>     Thanks!
>>
>
>     So, in the equation $q = m c_s \Delta T$ you need to use the mass
>     of the solution.  You know the volume and the density of the
>     solution is listed on the lab report form.  You can therefore use
>     $m = \rho V$ to compute the mass of the solutions.   The specific
>     heats for these solutions can also be found on the report form.
>
>     Now, when it comes to $\Delta H$, you have to remember that it is
>     the constant volume heat capacity on a per mole basis.   So take
>     your $q$ that you determined and divide it by the moles of substance.
>
>     In the case of the dissolution of NaOH you have roughly 2 grams of
>     NaOH.  Convert this to moles  (2 g / 40 g/mol) = 0.05.  So your
>     enthalpy of reaction for the dissolution of NaOH should be $q / 0.05$.
>
>     In the case of the reaction of NaOH (aq) with HCl (aq) you have 50
>     ml of 1 M NaOH (that would also be 0.05 moles), so divide the heat
>     by that amount to get the enthalpy like you did before.
>
>     Hopefully that give you enought to get through the rest of the
>     lab, but let me know if you have any questions.
>
>
>
>
>
>     -- 
>     Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu  <mailto:pounds_aj at mercer.edu>)
>     Professor of Chemistry and Computer Science
>     Mercer University,  Macon, GA 31207(478) 301-5627  <tel:%28478%29%20301-5627>
>     http://faculty.mercer.edu/pounds_aj
>


-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627
http://faculty.mercer.edu/pounds_aj



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