Hess's law help?

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Hess's law help?

Chemistry
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Determine the enthalpy change for NaOH (s) + HCl (aq) → NaCl (aq) + H2O (l) using Hess's Law.
it's in the conclusion, question 1
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@mathmate I didn't want to bother you with Hess's law, I'm sorry ;-;
@jagr2713 wanna take a go?
Enthalpy of products - Enthalpy of reactants
If A->B + H1 (part I) B+C-> D + H2 (part II) then A+C -> D + H1+H2 That's basically what Hess's law is about, applied to your situation. So use Hess's Law to find \(NaOH_{(s)} + HCl_{(aq)}~~ ->~~ NaCl_{(aq)}+H_2O_{(l)} + H\)
so which are what? -1,006.6 kJ/moles NaOH and -48.72 kJ/moles NaOH as the products and 3.087 kJ / 0.06338 mol NaOH and 50.33 kJ / 0.05 mol NaOH as the reactants?
nothing else would make sense in that place, that is what I'm most confused on
The two equations required are the revised equations in conclusion (2) and conclusion (3). When you add them up, and cancel equivalent reactant and product, you will end up with the one-step equation. Do not forget to add the enthalpy changes (or kJ/mol)
so NaOH(s)- -> Na+ + OH- + 48.7 kJ/mol + (I still haven't gotten the balanced chemical reaction and enthalpy change for Part II) = NaOH(s)+HCl(aq) −> NaCl(aq)+H2O(l)+H ?
that's it? Without any numbers?
Well, you will complete the balanced equation for part II, then add up the two, cancel equivalent reactant and product, and that would be the final equation. This procedure is based on Hess's law, which says that enthalpy is a state value, so no matter how you arrive at the product, the enthalpy of the product will be the same.
how about how to get the balanced equation from part II? Like, the steps to getting it?
For part I 1. It should be NaOH(s) +H2O(l)>>NaOH(aq)+H2O(l) |dw:1441517160519:dw|
It is a bit messy and contains some errors ( I crossed it out and corrected it.) as I am gaming and doing this lol. (Lost focus) Supposed to be HCl not H2O, supposed to be NaCl not NaOH, supposed to be negative 1055.32kJmol not -1055.32kJ/mol. The phase forms were accidentally misplaced
After all if you put NaOH in water, there will be dissolved NaOH and water. (NaOH would not suck all the water out). The equation is also balanced that way compared to yours.
thank you so much for a visual

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