Two questions but I may have more later.
1. Which initial T HCl is right? I used BOTH and NEITHER gives your answer of 4679 J.
2. Why do you think NaOH is the limiting reagent? 150 mL x 1.052M HCl = 157.8 millimoles HCl. 10.2g NaOH is 10.2/0.040 = 252 millimols. On a 1:1 basis I don't see how NaOH can be the limiting reagent.
Look at HCl being the LR and recalculate that part. See if it's closer to what you expect.
Help needed with a calorimeter lab.
I have to measure the enthalpy of a neutralization reaction between NaOH and HCl as well as find the concentration of NaOH the following ways.
- using the mass and volume of NaOH
- Using calorimetric data assuming no heat lost to calorimeter
- Using calorimetric data including heat lost to calorimeter (calorimeter constant)
-Use the theoretical value of -55.90 kJ / mol H2O formed and the number of joules of heat liberated in this experiment to calculate how many moles of H2O (n water) are formed. Then, knowing that you used 100 mL of NaOH solution, and since n water = n NaOH initial, you can calculate the molar concentration of the original solution.
concentration of HCl- 1.052
mass of NaOH- 10.1g mixed with 250ml of water
volume of HCl- 150ml
volume of NaOH used in the calorimeter -100ml
initial temperature of HCl- 24.9 degrees Celsius
initial temperature of HCl -24 degrees Celsius
calorimeter constant- 62.979J/ degree celcius
final temperature: 28.924 degrees Celcius
I could find the concentration using the mass and volume, I'm just not sure how to backtrack from finding q (heat) to concentration or if that's what I'm even supposed to do. I've been working on this for ages and I feel like giving up.
Any help please?
This is what I've attempted so far,
h=mc T
=(150+100)(4.184)(28.924-24)
=4 679J
h reaction = -4679J
NaOH seems to be the limiting factor, so
n NaOH = 100 x (10.1/250)
=4.04
h reaction/ mol= -4679J/4.04 moles
=1158
This is so far off the expected value I'm pretty sure I made some major errors in the math above.
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