Joules to kJ/mol conversion
Accepted book enthalpies are in the units $\mathrm{kJ/mol}$. 

Simply convert $\mathrm J$ to $\mathrm{kJ}$ by dividing by $1000$. Since $1\ \mathrm{kJ} = 1000~\mathrm{J}$. 

The moles of dilute aquaous solutions is best determined by volume (L) times concentration (mol/L).$$n = V \times c$$ You would use the moles of the product. If there is, in fact, "$0.2\ \mathrm{mol}$ of the reactants present" then there would be $0.2\ \mathrm{mol}$ of the products present. since the chemical reactions are $\ce{1 mol + 1 mol -&gt; 1 mol + 1 mol}$.  (You are not interested in the number of atoms per mole, so you would not use Avogadro's number.)

$$q= C_\mathrm{sol} m\Delta T$$ where $C_\mathrm{sol}= 4.06~\mathrm{J/(g \ ^\circ C)}$ (Specific heat capacity for dilute aqueous solutions)

$$q= C_\mathrm{sol} m\Delta T / n_\mathrm{product}$$  ($n = \text{moles product}$)

Remember that Hess's Law states that 


  if a reaction can be expressed as the sum of two or more reactions, then the heat flow for the overall reaction is equal to the sum of the heat flows.


Hence, the sum of the first 2 reactions should be equal to the 3rd reaction.  


$\displaystyle \ce{NaOH + HCl -&gt; NaCl + H2O}$
$\displaystyle \ce{NH4OH + NaCl -&gt; NaOH + NH4Cl}$  (Reversed and Enthalpy sign changed)
$\displaystyle\ce{HCl + NH4OH -&gt; NH4Cl + H2O}$


Note how $\ce{NaOH}$ and $\ce{NaCl}$ cross out leaving the third reaction.