What will be the ratio of methanol with 0.1N KOH as catalyst for converting 1ml of lipid extract into biodiesel?
As it is for a reaction you only need a 3:1 ratio (methanol: triglyceride), however it is advisable to use an excess of methanol to avoid the reversibility of the reaction, it can be 4:1, 6:1, etc.
Generally, molar ratio of methanol to oil is 6:1, which is 100 percent excess methanol used for the reaction. Molar mass of the oil can be calculated using fatty acid profile of the oil. The molar mass of 1 mole of methanol is 32.04 g. For instance, molar mass of one mole of oil calculated based on fatty acid profile is 835 g, then ratio of methanol to oil is 6:1 used in the reaction means
methanol to be mixed is 6 multiplied by 32.04 g per mole equal to 192.24 g. So this amount of methanol to be mixed with 835 g of oil for biodiesel production.
For optimize the methanol requirements for better biodiesel yield, you can conduct a set of experiments for biodiesel production using different molar ratio of methanol to oil from 3:1 to 10:1.
The biodiesel yield is depends on reaction temperature, molar ratio, catalyst amount and free
fatty acid of oil used for the experiments.
What do you mean by 0.1N KOH as catalyst? If you are about to dissolve KOH in methanol, than it makes sense, but usually a mass ratio of 1:100 KOH:triglyceride is used. 3:1 is the stochiometric ratio, but the conversion will not surpass 80%, for total conversion you need excess methanol and two transesterification reactzions (at least)
Short answer : Oil to methanol molar ratio = 1:6
Description : Oils from various feedstocks are basically triglycerides molecules composed three ester moiety attached to a glycerine one. Which means from each molecule of oil we can produce maximum 3 molecules of biodiesel.
1 mol Oil + 3 mol of methanol= 3 mol biodiesel + 1 mol of glycerol
Hence, stoichiometrically for a trans-esterification process oil to methanol molar ratio is 1:3. but ,Trans-esterification being a bi-directional chemical reaction and a rather slow one, more reactant is to be added to shift the equilibrium towards product side in order to ensure complete conversion of oil to biodiesel following Le Chatelier principle. Generally, an oil to methanol molar ratio of 1:6 to 1:10 is employed.
Molar mass of methanol is 32.04g and molar mass of oils can be determined from the methyl ester composition. Different oil have different composition and hence different molecular weight. If we assume an average chain length of the ester as 18, will get the molecular weight of oil as 812g implies that, stoichiometrically 812g of oil need 96.12g(32.04*3) of methanol but for safe side, we add 192.24g-320.4g.(1:6-1:10)
When I produce biodiesel,from personal experience, I use 300 ml methanol (240g) for 1L of oil (900g) for all oils except coconut where the average chain length of coconut oil is 14. Extra methanol should be accounted for the same mass of coconut oil compared to other oils.
most papers in the open literature suggests 6:1 and in the continuous production it can be 9:1
In biodiesel production, molar ratio from oil to alcohol and mass ratio from catalyst to oil need to be optimized for each feedstock. In general, oil to alcohol ratio is 3, as per stoichiometry, But, since alcohol is an excess reactant, it should be more than its stoichiometric requirement. catalyst to oil mass ratio is set minimum at 1% (w/w). Again, it depends. For more clear explanation, please refer Chapter Process Modelling and Simulation of Biodiesel Synthesis Reac...
Hi. The ratio in details you will find the following article :
https://aip.scitation.org/doi/abs/10.1063/1.4898473?journalCode=apc
The best ratio is 6: 1 ( methanol: oil)
If you want to get more details please refer my article below
https://www.researchgate.net/publication/357326563_Biodiesel-Alkaline_Transesterification_Process_for_Methyl_Ester_Production
Best of luck
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