Question

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)`?

Answer 1

`"48.0 kcal"` was released... at constant pressure.

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

`DeltaH = -"120.0 kcal"` [Source 1] [Source 2]
`DeltaH = -"12.00 kcal"` [Source]

As this is a strong acid-strong base reaction, I would never believe a mere `-"12.00 kcal"`... The `-"120.0 kcal"` is more physically reasonable. Furthermore, this should be `"kcal/mol"`, i.e. per mol of `"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 `"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"))`

`= "0.400 mols Na"_2"O"`

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

`nDeltaH_"rxn" = 0.400 cancel"mols" xx -"120.0 kcal"//cancel"mol"`

`= color(blue)(q_"rxn" = -"48.02 kcal")`

or we say that `"48.02 kcal"` was released. And in what scenario? At constant pressure.