1) NaOH(s)--> Na+(aq) + OH- (aq)
2) Na+(aq) + OH(aq) + H+(aq) + Cl-(aq) --> H2O + Na+(aq) + Cl-(aq)

Target
equation: NaOH(s) + H+(aq) + Cl(aq)--> H20 + Na+(aq) + Cl-(aq)

I need help solving this.. Thanks

1) NaOH(s)--> Na+(aq) + OH- (aq)
2) Na+(aq) + OH(aq) + H+(aq) + Cl-(aq) --> H2O + Na+(aq) + Cl-(aq)

Target
equation: NaOH(s) + H+(aq) + Cl-(aq)--> H20 + Na+(aq) + Cl-(aq)



Actually, it appears you have solved it.

NaOH(s)\,-->\, Na^+(aq) + OH^- (aq)

HCl\, --> \,H^+\,+\,Cl^-

OH^-\, +\, H^+ \,=\, H_2O

therefore, adding them together and eliminating redundancies on both sides of the "-->",

NaOH\,+\,HCl\,-->\,H_2O\,+\,Na^+(aq)\,+\,Cl^-(aq)

Depending on how you wish to express it, you can write it this way:

NaOH\,+\,HCl\,-->\,H_2O\,+\,NaCl(aq)

In this case, you don't emphasize that the NaCl is dissociated. If you wish to emphasize that NaOH and HCl are dissociated you can write this:

Na^+\,+\,OH^-\,+\,H^+\,+\,Cl^-\,-->\,H_2O\,+\,Na^+(aq)\,+\,Cl^-(aq)

But the Na+ and Cl- on both sides of the equation is sort of silly. They're simply "spectator ions".

NaOH\, +\, HCl\, =\, Na^+\,+\,OH^-\,+\,H^+\,+\,Cl^-\,-->\,H_2O\,+\,Na^+(aq)\,+\,Cl^-(aq)


I'm not sure how this relates to Hess's law which is about energy changes.