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.