What is the theoretical yield of vanillyl alcohol (in mg)? Enter Your Answer: 41

Question

What is the theoretical yield of vanillyl alcohol (in mg)?

Enter Your Answer:        418.87 Correct

How to solve this part?

QUESTION -> If the yield were 100%, how many mmol of H3BO3 would be produced before the HCl is added?

Question 8

If the actual yield is 220.3 mg, what is the percentage yield?

Enter Your Answer:        52.6 Correct

I got the other 2 parts right, no idea how to do that one part, any help?

Value Units Vanillin 413.4.0 mg Sodium Borohydride 73.1 mg Concentration of NaOH 2.00 M Volume of NaOH 2.23 mL Concentration of HCl 2.50 M Volume of HCl 3.37 mL Volume of Rinse Water 1.00 mL

Explanation / Answer

413.4mg of Vanillin(MW = 152 g/mol) = 413.4/152 = 2.72 mmol
73.1mg of Sodium Borohydride(MW= 38 g/mol) = 73.1/38= 1.924 mmol

Stoichiometry of reaction:    4C8H8O3(vanilin) + NaBH4 + 4H2O -> 4C8H10O3 + H3BO3 + NaOH
1 millimole of NaBH4 requires 4 millimoles of Vanilin
1.924 mmoles of NaBH4 will require = 4*1.924= 7.7 mmoles of vanilin
But 2.72 mmoles of Vanilin is present. So Vanilin is the limiting reactant

1 mole of vanilin produces 1 mole of vanillyl alcohol
2.72 mmoles of vanilin will produce 2.72 mmoles of vanilly alcohol
Molar mass of vanillyl alcohol, MW = 154 g/mol
theoretical yield of vanillyl alcohol =2.72*154 = 418.87 mg

(b) If the yield were 100%, how many mmol of H3BO3 would be produced before the HCl is added?
1 mmole of vanilin produces 1 mmole of H3BO3
2.72 mmoles of vanilin will produce 2.72 mmoles of H3BO3

Question 8
actual yield is 220.3 mg,
percentage yield = actual yield/theoretical yield*100 = 220.3/418.87*100= 52.6%

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