The balanced equation for the reaction between vanillin and sodium borohydride i
ID: 874748 • Letter: T
Question
The balanced equation for the reaction between vanillin and sodium borohydride is :
4 C8H8O3 + BH4- + 4 H2O ? 4 C8H10O3 + H3BO3 + OH-
The balanced reaction between sodium borohydride and hydrochloric acid is :
BH4- + H+ + 3 H2O ? + H3BO3 + 4 H2
1. What is the theoretical yield of vanillyl alcohol (in mg)?
2. If the yield were 100%, how many mmol of H3BO3 would be produced before the HCl is added?
Value Units Vanillin 387.0 mg Sodium Borohydride 89.3 mg Concentration of NaOH 2.00 M Volume of NaOH 2.02 mL Concentration of HCl 2.50 M Volume of HCl 3.67 mL Volume of Rinse Water 1.00 mLExplanation / Answer
1. What is the theoretical yield of vanillyl alcohol (in mg)?
Given : mass of vanillin . We use it to get moles.
Mass of vanillin = 387.0 mg = 0.3870 g
We know mol vanillin = Mass in g / Molar mass = 0.387 g / 152.15 g per mol
= 0.00254 mol vanillin
We know from the balanced reaction
Mol ratio of vanillin to vanillyl alcohol is 4:4 or 1 : 1
So we calculate mol of vanilly alcohol by using mol of vanillin
Mol vanillyl alcohol = 0.00254 mol vanillin * 1 mol vanillyl alcohol / 1mol vanillin
= 0.00254 mol vanillyl alcohol
Theoretical yield = Maxiumum mass calculated from the mass of reactant
= mol vanillyl alcohol * molar mass
= 0.00254 mol vanillyl alcohol * 154.16 g per mol
= 0.3921 g vanillyl alcohol
Theoretical yield of vanillyl alcohol = 0.3921 g
2. If the yield were 100%, how many mmol of H3BO3 would be produced before the HCl is added?
We calculate moles of H3BO3 by using moles of vanillin
Mol H3BO3 = mol vanillin *1mol H3BO3 /4 mol vanillin (mol ratio of vanillin to H3BO3 4:1 )
= 0.00254 mol vanillin * 1 mol H3BO3 / 4 mol vanillin
=0.000636 mol H3BO3
We convert mol to m mol
= 0.000636 mol * 1000 m mol / 1mol
= 0.636 m mol of H3BO3
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