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)#?
1 Answer
This can't be done without data on the enthalpy of reaction... And
#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
Now, since energy is an extensive quantity, it scales with the size of the system. Therefore, when we find the mols of
#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
#nDeltaH_"rxn" = 0.400 cancel"mols" xx -"120.0 kcal"//cancel"mol"#
#= color(blue)(q_"rxn" = -"48.02 kcal")#
or we say that