[vmelkon]

Looking at few sites and videos, they mention mixing NaOH into methanol and the result is sodium methoxide. (here is one example : http://journeytoforever.org/biodiesel_aleksnew.html)
I don't see how that is possible.
The reaction would be NaOH + Methanol => CH3-O-Na + H2O
which means it produces water.
but sodium methoxide actually reacts with H2O to give NaOH and methanol according to other documentation. It must be kept absolutely dry.

So who is correct? Perhaps some amateurs "think" that a reaction occurs and produces sodium methoxide.

Another thing I didn't quite understand in making biodiesel is that at the end of the reaction, you end up with glycerol and the biodiesel. You end up with methanol because it is in great excess.
The question is, where is the methanol? Dissolved in the glycerol or/and biodiesel layer?
Where is the NaOH?

You can distill the methanol from biodiesel and end up with a NaOH residue, I guess. Then you can reuse it.
You can perhaps do the same to glycerol. Distill the methanol and recover the NaOH as a solid.

[marquis]

The oils that are being used to make biodiesel are triglycerides (glycerine reacted with three fatty acids).  There are three fatty acids to each molecule of glycerine.

The reaction is similar to FAME (fatty acid methyl ester) analysis.

You take the triglycerol and react it to form glycerine and the fatty acids.  Since the fatty acids have limited volatility, you form the methyl ester.  This increases the volatility and makes biodiesel more usable.  The methyl ester of the fatty acids is what is used as biodiesel.

I'm not a biodiesel expert.  But this is how I put the reactions together. 

Hope it helps and good luck.

[Benzenelover]

From what I understand, the NaOH is used to break apart the fatty acid from the glycerine, with a simple ester-base reaction. The fatty acid takes the Na, where the glycerine takes the OH. The newly freed fatty acid, now holding the Na in place of it's usual H, doesn't last too long. It quickly reacts with the methanol in the solution, to form a new ester, which is much more simple, and makes a much better diesel fuel than the triglyceride. The Na from the fatty acid then goes off to react and form NaOH, going back to it's original form, thus classifying it as a catalyst in this case.

The theory involving the NaOH stripping methanol of it's hydrogen, and creating water, is likely invalid, as any water present in the reaction will react with more NaOH and the triglyceride to form soap. This is the reason why triglycerides must be dried by boiling off the water before, before conversion to biodiesel.

[fledarmus]

No, esterification reactions do not occur under basic conditions. Once the carboxylic acid is deprotonated, it is not reactive.

The solution must be neutralized at least, and preferably partially acidified before the acid will react with the methanol.

[vmelkon]

My main question was, if you mix NaOH in methanol, does it form sodium methoxide?

The solution must be neutralized at least, and preferably partially acidified before the acid will react with the methanol.

I'm not exactly sure what steps take place at the molecular level, but the procedure for making biodiesel doesn't call for acidification.
They first blend NaOH and methanol.
Then they add it to the vegetable oil and blend. They heat it to 55 C for a few hours. Let it settle. It phase separates with glycerol at the bottom and biodiesel at the top.

[stewie griffin]

I remember looking into how to make biodiesel when I thought it would be fun just to do in the lab and say I did it. I looked into it and remember having this same question.
I never found a "correct" answer but I think I came up with something reasonable. The pKa of methanol and water are both about the same (~15.7) so there's an equilibrium between sodium methoxide and sodium hydroxide set up once you add the NaOH to the MeOH. That makes one think "Hey I can't really have pure NaOMe in this situation!". But if you add pure NaOH to pure MeOH then I supposed you could say because of Le Chatlier's principle we have a good amount of NaOMe. We have so much MeOH at the start of the reaction that the equilibrium in the following rxn is pushed far to the right: NaOH + MeOH ----> NaOMe + H2O
There's still an equilibrium but I would think that with the relative amounts of MeOH and NaOH in the system that it's more correct to say we have NaOMe.