You started off very good but messed up a few times before the end.
1. You had 400 mL of the solution (200 mL HCl + 200 mL NaOH = 400 mL) and that is 400 grams solution.
2. I note you carry out the molar masses to the nth degree, then use 4.20 for specific heat. This isn't terrible but I see it as unnecessary work.
3. You calculated a dH for HCl and a dH for NaOH. You have only one dH for the reaction and the problem means for that to be PER MOLE H2O formed. The mols H2O formed will be the lesser of the mols HCl or NaOH, so use the mols NaOH (because that is smaller)
Problem: Assume 200.0 mL of 0.400M HCl is mixed with 200.0 mL of 0.400M NaOH in a coffee-cup calorimeter. The temperature of the solutions before mixing was 25.10°C; after mixing and allowing the reaction to occur, the temperature is 27.78°C. What is the enthalpy chanfe when one moe of acid is neutralized? Assume that the desities of all solutions are 1.00g/mL and their specific heat capacities are 4.20J/gK
Work:
~200mL(1.00g/ 1mL) = 200g HCl
~200mL(1.00g/ 1mL) = 200g NaOH
~(200gHCl)(1 mole/36.4609gHCl)=
5.4853molHCl
~(200gNaOH)(1mole/39.9869gNaOH)= 5.0016molNaOH
~q=mc(delta T) x=(200g)(4.20J/G°C)(27.78-25.10) = 2251.2 J HCl amd NaOH.
~qr + qsolution = 0 x + 2251.2 = 0 x= -2251.2 J HCl and NaOH
~delta H = (-2251.2 J/ 5.4853mol HCl)= -410.4060 j/mol --> -0.4104060kJ/mol HCl
~delta H= (-2251.2J/5.0016mol NaOH)= -450.0960J/mol --> -0.4500960kJ/mol NaOH
I don't think i did this correct can someone please help me
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