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 (Tfinal-Tinitial) = 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.

To calculate the heat capacity of the calorimeter, we can use the equation:

q = (m × Cs × dT) + (n × ΔH)

Where:
q is the heat gained or lost by the system (calorimeter),
m is the mass of the solution in the calorimeter (in grams),
Cs is the specific heat capacity of the solution in the calorimeter (in J/g°C),
dT is the change in temperature (in °C),
n is the number of moles involved in the reaction,
ΔH is the enthalpy of reaction (in kJ/mol).

Let's calculate each term step by step:

1. Calculate the mass of the solution in the calorimeter:
- The volume of HCl used is 50 mL, and the density of 1 M NaCl is 1.037 g/mL.
- Therefore, the mass of the HCl solution is: 50 mL × 1.037 g/mL = 51.85 g.
(Note: The mass of the NaOH solution is also 51.85 g since they have the same volume.)

2. Calculate the number of moles involved in the reaction:
- 50 mL of 2.00 M HCl is used, which means there are (50 mL × 2.00 mol/L) = 0.100 mol of HCl.
- 50 mL of 2.05 M NaOH is used, which means there are (50 mL × 2.05 mol/L) = 0.1025 mol of NaOH.

3. Calculate the heat gained or lost by the system (calorimeter):
- In this case, the heat gained or lost by the system is the negative of the heat gained or lost by the reactants (according to Hess's Law).
- Since 1 mole of HCl reacts with 1 mole of NaOH, and the enthalpy of reaction is -58.3 kJ/mol, the heat gained or lost by the system is:
- (-58.3 kJ/mol) × 0.100 mol = -5.83 kJ.

4. Convert the heat gained or lost to joules:
- Since 1 kJ = 1000 J, the heat gained or lost is: -5.83 kJ × 1000 J/kJ = -5830 J.

5. Calculate the specific heat capacity (Cs) of NaCl:
- Given that Cs(NaCl) = 3.91 J/g°C, we will use this value directly.

6. Calculate the change in temperature (dT):
- Given that dT = 12.88°C, we will use this value directly.

7. Calculate the heat capacity of the calorimeter:
- We can rearrange the equation: q = (m × Cs × dT) + (n × ΔH) to find the heat capacity (C) of the calorimeter, which is our goal.
- Rearranging the equation gives: C = -q / (m × dT) = -(-5830 J) / (51.85 g × 12.88°C).

Calculating this expression will give you the heat capacity of the calorimeter.

Note: Make sure to carefully convert units and pay attention to significant figures when carrying out the calculations.