Chemistry
posted by Annie on .
Naoh (aq) + HCl (aq) > NaCl (aq) + H2O (l)
Using the dT determined in part 1, calculate the heat capacity of the calorimeter. Enthalpy of reaction= 58.3 kj/mol, Cs (NaCl = 3.91j/g C)density of 1 M NaCl = 1.037g/mL. dT= 12.88 degrees Celcius.50 mL of 2.00 M HCl and 50 ml of 2.05 M NaOH were used

What's Cs NaCl = 3.91 J/g mean?
Is dT delta T. 
yes dt is delta T and i think it is the specific heat capacity of NaCl but im not sure

You have 100 mL solution.
Use density to calculate mass from mass = volume x density.
mols NaCl formed = M x L = 0.2M x 0.050 = 0.1 mol. delta H rxn = 58.3 kJ/mol or 5.83 kJ for this reaction which is q in the following..
q = [mass NaCl soln x specific heat of NaCl soln x dT] + [Ccal x dT] = 0
Substitute and solve for Ccal. 
Originally the enthalpy is negative then how does it become postive and I have followed the steps and I came up 47.36 J/g degrees C. Also I don't understand why enthalpy is q if you could please explain that, and thankyou so much.

I had to let that equation equal something and I didn't want to type that line again so I let it equal q. The usual expression for specific heat and dT calculations is
mass H2O x specific heat H2O x (TfinalTinitial) = 0. We plug in the numbers and solve for Tinitial or Tfinal or specific heat or mass H2O. In this case, however, we have all of those values and the equation equals q or any other letter I choose. I changed the sign because we are adding heat to the solution, not taking heat out; + signs mean we add heat. Another way of looking at it is that the  sign means the neutralization is an exothermic one and that it is being used to release heat and that ADDS heat to the solution.
I went through the calculations in a hurry and came up with 47.17 J/g for Ccal I would round that to 47.2. 
Thankyou so much for all the help and the explanation and your time.

I have one more question, why is the volume of NaCl 0.05 L when we are calculating the mols and 0.1 L when we are calculating mass?

The volume of NaCl is never 0.100.
The mass of the SOLUTION (it is a solution of NaCl) comes from 50 mL + 50 mL = 100 mL, then
mass of solution = volume x density = 100 mL x 1.037 g/mL = about 104 or so grams for the salt solution.
The volume of NaCl is never 0.05 L.
The reaction is HCl + NaOH ==> NaCl + H2O
You have mols HCl = M x L = 2.00M x 0.050 L = 0.100 mol NaCl formed from this and the HCl is the limiting reagent.
For the NaOH we have 2.05M x 0.050 = 0.102 mols. Of course only 0.100 mol NaCl is formed.