# How do you calculate the energy released when 24.8 g Na_2O reacts in this reaction: Na_2O(s) + 2HI(g) -> 2Nal(s) + H_2O(l)?

Sep 16, 2017

$\text{48.0 kcal}$ was released... at constant pressure.

This can't be done without data on the enthalpy of reaction... And $\text{Nal}$ doesn't exist. $\text{NaI}$ does. A brief google search gives conflicting results:

$\Delta H = - \text{120.0 kcal}$ [Source 1] [Source 2]
$\Delta H = - \text{12.00 kcal}$ [Source]

As this is a strong acid-strong base reaction, I would never believe a mere $- \text{12.00 kcal}$... The $- \text{120.0 kcal}$ is more physically reasonable. Furthermore, this should be $\text{kcal/mol}$, i.e. per mol of $\text{Na"_2"O}$ (since it has a stoichiometric coefficient of $1$ and is on the reactants' side).

Now, since energy is an extensive quantity, it scales with the size of the system. Therefore, when we find the mols of $\text{Na"_2"O}$, we know the scaling factor.

24.8 cancel("g Na"_2"O") xx "1 mol"/(61.979 cancel("g Na"_2"O"))

$= \text{0.400 mols Na"_2"O}$

Reading the reaction itself gives us $\text{1 mol Na"_2"O}$, $\text{2 mols HI}$, etc. As a result, the energy released by $\text{0.400 mols}$ is just 40% of the $\text{120.0 kcal}$ that was for every $\text{1 mol}$:

$n \Delta {H}_{\text{rxn" = 0.400 cancel"mols" xx -"120.0 kcal"//cancel"mol}}$

$= \textcolor{b l u e}{{q}_{\text{rxn" = -"48.02 kcal}}}$

or we say that $\text{48.02 kcal}$ was released. And in what scenario? At constant pressure.