C8H8O3 + NaBH4 -------> C8H10O3
no of moles of vanilin  = W/G.M.Wt
                                    Â
= 2.5/152 = 0.0164 moles
no of moles of NaBH4Â Â = molarity * volume in L
                                     Â
= 3.42*0.005Â Â = 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
                                         Â
= 0.0164*154 = 2.525g of vanilly alcohol
