Vanillin used at start 1275g C8H8O3 MW152149 Sodium Borohydr

Vanillin used at start= 1.275g C8H8O3 MW=152.149

Sodium Borohydride used= 2.5mL of 3.42M in 1.0M NaOH NaBH4 mw= 33.8

Vanillyl Alcohol= C8H10O3 MW=154.165

3a) mol of vanillin reacted=mass/molar mass=1.275g/152.149g/mol=0.00838 mol

from the balanced equation, mol of vanillin:mol of vanillyl alcohol=4:4=1:1

So, if vanillin is the limiting reagent then, mol of vanillyl=mol of vanillin=0.00838 mol

3b) mol of sodium borohydride=0.0025 L*3.42mol/L=0.00855 mol

from the balanced equation, mol of sodium borohydride:mol of vanillyl alcohol=1:4

or, mol of sodium borohydride*4=mol of vanillyl alcohol*1

So, if sodium borohydride is the limiting reagent then, mol of vanillyl=4*mol of sodium borohydride=4*0.00855 mol=0.0342mol

c)The lesser reagent (vanillin) (0.00838mol) is the limiting reagent.