Enthalpy change of reaction Suppose you have the following equation:

$$\ce{2NH4SCN(s) + Ba(OH)2(s) -&gt;Ba(SCN)2(s) + 2H2O(l) + 2NH3(g)}$$

You initiate this reaction using 15.22g of the $\ce{NH4SCN}$ (and excess of the barium hydroxide) in a boiling tube resting in 50g of water. The temperature rises by 11 degrees. Therefore energy = $4.2 \times 11 \times 50 = 2.31kJ$ and moles = $\frac{15.22}{76.1} = 0.2mol$

I would then assume that it would just be $\frac{2.31}{0.2} = 11.55kJmol^{-1}$

But the actual answer is 23.1 (double my answer) which I am guessing is because there are 2 moles of $\ce{2NH4SCN}$. I do not understand why this is the case, because you are using 0.2 moles and the energy raised is 2.31 so why the multiplication by 2?
The reason lies with the definition of the Heat of Reaction. It states that it is the amount of heat when 1 mole of the product is formed. As you noted, since 2 moles of $\ce{NH4SCN}$ is needed for each mole of product, then you need to multiply by 2.

Added:
If you change the number of moles of $\ce{NH4SCN}$ to 0.1 moles (to remove the stoichiometric 2), then your calculations would give you the answer.