Pardon this somewhat elementary question but...
If I am making a 20:1 molar ratio of methanol to FFA...
1 mole MeOH = 32.04g
1 mole FFA = ~282.5g
Would that be (32.04 * 20)g MeOH to every 282.5g FFA?
Which equates to 640g MeOH for every 282.5g FFA.
In other words, is a molar ratio of 20:1 equal to 20 moles of MeOH to 1 mole of FFA?
Thanks in advance for the clarification.
yes that's right, BUT,
Where did you see mention of a 20:1 molar ratio of methanol to FFA? What is the purpose?
As far as transesterification goes, the academic literature suggests a 6:1 molar ratio of methanol to triglyceride.
20:1 is the typical ratio that has been used over the years for Acid Esterification not transesterification. You should be abl eto find it in similar academic papers to where the 6:1 came from.
Jon van Gerpen was certainly one who used this type of amount at the pilot plant in Iowa when he was at ISU.
ok, Thanks for the explanation.
Is the 20:1 ratio triglycerides,
and the 6:1 ratio FFA's? (or "Normal" ratios instead of "Molar")?
No, the 6:1 ratio is Methanol to Triglycerides in the base catalysed transesterification reaction.
and the 20:1 molar ratio is Methanol to FFA for the acid catalysed esterification reaction. Esterificatiom typically uses a much bigger excess than transesterification.
Research and discussion into these ratios are quite old and have been optimised over the years. It was in the last century when people where talking about 6:1 and 20:1 ratios.
but why worry about it when better solutions exist?
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everything run B100 when its warm enough
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