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Help with Heat of Neutralization lab!

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So we have a lab on the heat of neutralization of NaOH and HCl given:

 

mass of NaOH = 48.01 g

Volume of NaOH = 1000.00 mL

Amount of HCl = 83.3 mL of 12.0 M of HCl in 1000.00 mL solution

 

And we found the initial temperature, final temperature

Temp initial = 21.5 degrees C

Temp final = 25.0 degrees C

 

And we have to find the heat of neutralization using this equation:

 

ΔHθneutralization = (-specific heat capacity of liquid water x mass (acid solution + base solution) x change in temp) / mole of limiting base or acid

 

And I should be getting an enthalpy of around -58 -ish but I'm wayyy off. Help please!!  :help:

 

 

P.S. I'm not that great at writing chem labs... :/

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Well first I figured out what the limiting reactant was; base or acid.

 

Then I graphed the data we had collected from the lab for time v. temperature and using the initial and final temperatures, found the change in temperature.

(You find the final temperature by drawing a line from the highest data point back the y-axis)

 

Then I substituted all of my data into the equation:

 

∆H(neutralization) = [-specific heat capacity (H2O) x mass (acid solution + base solution) x âˆ†T (H2O)] / n(limiting acid or base)

 

and then convert from J to kJ

 

The only part that brought me down in this lab were uncertainties so include those and remember percent error.

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