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

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Chemistry

Assuming you use 2.5 g vanillin and 5.0 mL of a 3.42 M NaBH4solution, what is the theoretical yield of vanillyl alcohol inmoles? in grams?

Answer & Explanation Solved by verified expert
4.3 Ratings (539 Votes)

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
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