Question #a1bda

The enthalpy change of reaction when a strong acid is neutralized by a strong base is actually

`DeltaH_"rxn" = -"57 kJ mol"^(-1)`

which means that `"57 kJ"` of heat are being given off, hence the minus sign, when `1` mole of strong acid is being neutralized by `1` mole of strong base.

In your case, nitric acid, a strong acid, and potassium hydroxide, a strong base, will neutralize each other in a `1:1` mole ratio as shown by the balanced chemical equation

`"HNO"_ (3(aq)) + "KOH"_ ((aq)) -> "H"_ 2"O"_ ((l)) + "KNO"_ (3(aq))`

Notice that if you have `0.5` moles of nitric acid and only `0.2` moles of potassium hydroxide, the strong base will act as a limiting reagent, i.e. it will be completely consumed before all the moles of nitric acid get the chance to take part in the react.

Therefore, this reaction will consume `0.2` moles of potassium hydroxide and `0.2` moles of nitric acid -- think `1:1` mole ratio here!

You can thus say that this reaction will give off

`0.2 color(red)(cancel(color(black)("moles HNO"_3))) * "57 kJ given off"/(1color(red)(cancel(color(black)("mole HNO"_3)))) = color(darkgreen)(ul(color(black)("11.4 kJ given off")))`

I'll leave the answer rounded to three sig figs, but keep in mind that your values do not justify three significant figures for the answer.