Mark,

I did a quick Google and found (amongst others)

http://www.arpnjournals.org/jeas/res..._0816_4863.pdf
This gives the Molar ratios and the reaction coefficients for different reaction temperatures.

Now, this should give us the ability to predict the amount of water for the given ratios of KOH and MeOH.

I believe that Molar ratios are based on the relative mass of the feedstock eg KOH should have K (40), O (16), H (1) = 57, Methanol CH3OH would have C (12) H *4 (4) O (16) = 32

In this case a molar ratio of 0.012 ( KOH:MeOH) would give mass ratios of (32*0.012):57 or 0.0067. Thus for a molar ratio of 0.012 KOH:MeOH you would need to add 6.7g of KOH per Kg of Methanol.

For a reaction temperature of 25° and a molar ratio of 0.012, this would give us:

KOH in solution 1.09%

CH3OH in solution 96.91%

CH3OK in solution 1.64%

H2O in solution 0.81%

Are my figures accurate?

Please comment on this as I have only lower school Chemistry.

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