Assuming you use 2.5 g vanillin and 5.0 mL of a 3.42 M NaBH4 solution, what...

Assuming you use 2.5 g vanillin and 5.0 mL of a 3.42 M NaBH4 solution, what is the theoretical yield of vanillyl alcohol in moles? in grams?

Expert Solution

C8H8O3 + NaBH4 -------> C8H10O3

no of moles of vanilin = W/G.M.Wt

= 2.5/152 = 0.0164 moles

no of moles of NaBH4 = molarity * volume in L

= 3.42*0.005 = 0.0171moles

limiting reagent is vanillin

1 mole of vaniline react with NaBH4 to gives 1 mole of vanilly alcohol

0.0164 moles of vaniline react with NaBH4 to gives = 1*0.0164 moles of vanilly alcohol

mass of vanilly alcohol = no of moles * gram molar mass

= 0.0164*154 = 2.525g of vanilly alcohol

queen_honey_blossom answered 1 year ago

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