I could not find a way to reply to the first one so I had to make a new post.

I understand how you got the .05 mol of H2O because of the balance equation

HCl + NaOH -> H2O + NaCl

I was confused as to why this would not work to solve for mol of H2O produced with 50 mL of 1.0 M HCl:

50g * 1 mol/36.5 g HCl
= 1.37 mol of H20

I understand what you wrote from the last post, but I do not know why the way I did it would not work, when they get two entirely different answers.

Do you know a site where I would be able to look up the heats of formation of elements?

Thanks so much for your help.

If you had 50 Grams of HCl you would have 1.37 Moles of HCL..
From the equation you would then have 1.37 Moles of H20 formed if you have the same experiment running.
1.37 Moles * 18 grams/ mole = grams of H20 formed in the reaction.

In the experiment you have .05 Liters of a solution of 1 mole/liter of HCL.. or
.050 Moles * 36.5 Grams/ mole... grams of HCL

Note the units need to cancel out.

Web site for heats of formation for the 'elements'... Heat of formation for an element is zero.

From Masterton & Slowinski Chemical principles
Table 4.1 Heats of formation Kcal/mole@ 25C
and 1 atm

HCL(g) -22.1
NaCL -98.2
NaOH -102.0
H20(l) -68.3