I\'m triying to figure out the theoretical yield of sodium oleate (C18H33NaO2)?

Question

I'm triying to figure out the theoretical yield of sodium oleate (C18H33NaO2)? the synthesis starts with .5g of glycerol trioleate (olive oil) and .5g of NaOH which produces glycerol and sodium oleate. I figured I would find moles of olive oil and moles of the NaOH to find thr limiting reagent and then use the limiting reagent to find the yield (g) of sodium oleate BUT am not sure how the glycerol factors in ? am I needing more information?


glycerol trioleate(mol wt:885.43, .5g) + NaOH (mol wt: 40, .5g) ------>glycerol (mol wt:92.09) + sodium oleate (wol wt:304.44)


Calculate the theoretical yield of the sodium oleate? please show all the steps :) I am suppose to assume the the I am using pure glycerol trioleate.


Thanks!

Explanation / Answer

C57H104O6+3NaOH>3C18H33NaO2+C3H8O3 MOLES OF GYLEROL TRIOLEATE=.5/885.43 MOLES OF NaOH=.5/40 GYCEROL TRIOLEATE IS LIMITING SO MOLES OF SODIUM OLEATE FORMED=3*.5/885.43=0.001694 MAS=304.44*MOLES=0.516g

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