How do I know if a reaction is favorable? (High School student)

I am doing a homework question where I have to react NaCl + *Something* to get NaOH, and I can't find a reliable source online as to whether a reaction is possible.
For example, I thought of NH4OH + NaCl -> NaOH + NH4Cl, but I don't know if a reaction would occur, and I decided it is appropriate to ask this question since I had the same question on different equations that I managed to do.




Your reasoning is valid -- i.e., a precipitate would force the reaction in the direction that produces $\ce{NaOH}$... but $\ce{MgCl2}$ is *highly* soluble, about 50 g/100 ml, according to this useful [Wikipedia solubility table](https://en.wikipedia.org/wiki/Solubility_table).


You can use that table to find less soluble chlorides, though... Search on *chloride* and look for an *insoluble* one.


Persist... *most* chlorides are soluble. Too bad *electricity* was not allowed... or is it?


BTW, please edit your question to *include* your comment. If you don't show your thought processes, it might appear you are looking for someone to do your homework *for* you ;-) Also, notice my slight edit: a great many reactions are *possible*, but you need to find one that is **favorable**, i.e., favors going in the direction you want (technically, equilibrium favors the product).


 0


> 
> I am doing a homework question where I have to react NaCl + Something
> to get NaOH, and I can't find a reliable source online as to whether a
> reaction is possible.
> 
> 
> 


I am glad that you could not find any, because there is no such "something" that will convert $\ce{NaCl}$ into $\ce{NaOH}$ directly, except electricity. See chloro-alkali process.


However, long before electricity were widely available, a process called Solvay process could convert NaCl to sodium carbonate. Reacting water soluble sodium carbonate with lime aq. $\ce{Ca(OH)2}$ gave us $\ce{NaOH}$ solution.


 0

A reaction is favorable, if and when the change in free enthalpy G is negative. So try to find the free enthalpies of formation of products and reagents. Do calculate the free enthalpy change, taking into account the concentrations of the different stuffs. The reaction is possible if $\Delta $G is negative. If it is positive, the reaction goes in opposite direction. This reasoning does indicate whether the reaction is possible or not. It does not give any information about the rate of the reaction, which can be very quick or rather slow.


 0

Just to little complement AChem - before electrolytic production of $\ce{NaOH}$, one of industrial methods was the combination of


* [Solvay process](https://en.wikipedia.org/wiki/Solvay_process) of soda production:


\begin{align}
\ce{NaCl(aq) + NH3(aq) + CO2(aq) + H2O(l) &-> NH4Cl(aq) + NaHCO3(s)}\\
\ce{2 NH4Cl(aq) + Ca(OH)2(s,aq) &-> CaCl2(aq) + 2 NH3(g) + 2 H2O(l)}\\
\ce{2 NaHCO3(s)&->[heat]Na2CO3(s) + H2O(g) + CO2(g)}\\
\end{align}


* Caustification of soda:


$$\ce{Na2CO3(aq) + Ca(OH)2(s,aq) -> 2 NaOH(aq) + CaCO3(g)}$$




---


There is an elegant lab method of the direct conversion, but too expensive, suitable for hardly more but demonstration:


$$\ce{2 NaCl(aq) + Ag2O(s) + H2O(l) -> 2 AgCl(s) + 2 NaOH(aq)}$$


 0

Going from $\ce{NaCl}$ to $\ce{NaOH}$ you exchange the chloride anion for hydroxide. To do this, you might think of using something that releases hydroxide ions and takes the chloride ions out of solution by forming an insoluble compound. The only soluble hydroxide whose chloride is insoluble is the extremely toxic thalium (I) hydroxide. The reaction would be as follows:


$\ce{NaCl(aq) + TlOH(aq) -> NaOH(aq) + TlCl(s)}$


Though the intended solution is most likely electrolysis, as implied by the other answers.


 0