How do I calculate enthalpy change calculations questions?

Below is a question I'm struggling with.



> 
> When $\pu{0.1 mol}$ of aqueous potassium hydroxide was added to $\pu{0.1 mol}$ of nitric acid, $\pu{5200 J}$ were transferred to the surroundings. What is the enthalpy change, in $\pu{kJ mol-1}$ for this reaction?
> 
> 
> $$\ce{KOH(aq) + HNO3(aq) -> KNO3(aq) + H2O(l)}$$
> **A** -52  
> 
> **B** -26  
> 
> **C** +26  
> 
> **D** +52
> 
> 
> 


This is how I tried to solve the question.


$$Q = \pu{5200 J} = \pu{5.2 kJ}$$
\begin{align}
\mathrm{Moles} &= 0.1 + 0.1 \\
&= \pu{0.2 mol}
\end{align}


$$\Delta H = \frac{Q}{\mathrm{mol}}$$
$$\Delta H = -\frac{5.2}{0.2} = \pu{-26 kJ mol-1}$$


The answer is actually **A**. What is the reason for this?




You divide by 0.1 mol only. Think of it this way. Let's change this question to using 0.1 mol $\ce{KOH}$ and 0.2 mol $\ce{HNO3}$. Technically, this reaction should only produce $\pu{5200 J}$ still because you only used 0.1 mol of $\ce{KOH}$ which is the limiting reagent as the reagents are in a 1:1 ratio. By your logic, however, this could compute to $\pu{\frac{5200 J}{0.3 mol}} = \pu{17.3 kJ}$ which is now different than you answer of $\ce{-26 kJ}$. Therefore, you don't consider the moles as the sum of the moles of reactants. Instead, you have to look at what's limiting in this specific calculation. In this case, they are 1:1 ratio and there is 0.1 mol of each so just divide $\pu{5200 J}$ by 0.1 mol.


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