How to calculate the heat capacity of a calorimeter based on a neutralisation reaction? I've looked at the other similarly worded questions but I can't seem to make heads or tails out of them.

So here's my case. I need to find the heat capacity of the calorimeter (in Joule/degree Celsius), and I've been provided with the following information:

$$\ce{NaOH (aq) + HCl (aq) -&gt; NaCl (aq) + H2O}$$


$\Delta T = 11.5\ ^\circ \ \text{C}$
$\Delta_r H = -58.3 \text{ kJ/mol}$
$s_{p,\ce{NaCl}} = 3.91 \text{ J/(g}\cdot \ ^\circ \text{C)}$
$d_{1\ \text{M NaCl}}=1.0037\text{ g/mL}$
$V= 100 \text{ mL}$


So far I've calculated the mass of $\ce{NaCl}$:

$$m = d\cdot V = (1.037 \text{ g/mL})(100 \text{ mL}) = 103.7 \text{ g}$$

Then I calculated the heat capacity of $\ce{NaCl}$...at least I think that's what I did. I'm not even sure if I was supposed to do this:


$q = m\cdot s_{p,\ce{NaCl}}\cdot \Delta T$
$q = (103.7 \text{ g})(3.91 \text{J/(g} \cdot\  ^\circ \text{C}))(11.5 \ ^\circ \text{ C})$
$q = 4662.8 \text{ J}$


Now I'm confused as to the next step, and I don't know how to use $\Delta_r H$ in this problem.

Any help would be immensely appreciated.



Edit: Below I've added more things I've tried in my attempt to find the answer. 

I saw this equation in a textbook so I gave it a shot:


$q = - \Delta H\cdot n_{\text{limiting reactant}}$
$q = (- 58,300 \text{ J/mol})(1.62 \text{ mol HCl})$
$q = 94,446 \text{ J}$


But I don't know how/if this fits anywhere in the problem.

Another equation I tried: 


$q = C_P\Delta T$
$-58,300 \text{ J}= C (11.5 \ ^\circ \text{C})$
$C = 58,300 \text{ J} / 11.5 \ ^\circ \text{C}= 5069.5 \text{ J/}\ ^\circ \text{C}$


Which is at least the correct unit of the heat capacity I'm supposed to find.

Finally, I tried subtracting the heat capacity I found for $\ce{NaCl} \ (4662.8 \text{ J})$ from the supposed heat capacity of the system $(5069.5 \text{ J})$, I get $406.76 \text{ J}$

I'm kind of at a loss...
First, you have to find out the concentrations so you can calculate the amount of heat released. Since the density is given for a 1 M solution, I will assume this concentration.

Next, the mass of our solution:
$$ m = \rho \times V = 103.7~\mathrm{g}$$

Now we have to calculate the amount of energy that gets released through the reaction, but viewed from the environment and not the reaction (that's the reason for the negative sign before $q$):
$$ -q = \Delta_\text{r}H \times n = -58.3 \times 0.1 ~\mathrm{kJ\, mol^{-1}\, mol} = -5.83~\mathrm{kJ}$$

The total heat capacity of your system is given as
$$ c_\text{tot} = c_\text{s, sol}\times m + c_\text{cal}$$
where $c_\text{s, sol}$ is the specific heat capacity of your solution and $c_\text{cal}$ is the heat capacity of the calorimeter.

Now we know all the quantities needed for the following equation:
$$ q = c_\text{tot}\Delta T$$

We expand:
$$ q =( c_\text{s, sol}\times m + c_\text{cal}) \times \Delta T$$

Solve for $c_\text{cal}$:
$$ c_\text{cal} = \frac{q-c_\text{s, sol}m\Delta T}{\Delta T} = \frac{5.83~\mathrm{kJ} - 3.91~\mathrm{J\, g^{-1}\, K^{-1}} \times 103.7~\mathrm{g} \times 11.5~\mathrm{K}}{11.5~\mathrm{K}} = 101~\mathrm{J\, K^{-1}}$$

And now you know the heat capacity of the calorimeter.