Question

Calculate the percent yield of benzyl alcohol Sue obtained when doing the following reaction. Sue started with 2.1 g of benzoic anhydride and 0.15 g of sodium borohydride (NaBH4). She obtained 0.5 g of benzyl alcohol. You must show all of your work to receive any credit.

Answer 1

Answer:

The percent yield is 116.6 %

Explanation:

Step 1: Data given

Mass of benzoic anhydride = 2.1 grams

Mass of sodium borohydride = 0.15 grams

Molar mass of benzoic anhydride = 226.23 g/mol

Molar mass of NaBH4 = 37.83 g/mol

Mass of benzyl alcohol produced = 0.5 grams

Step 2: The balanced equation

C14H10O3 + NaBH4 → C7H6O2 + C7H8O + NaB

Step 3: Calculate moles of C14H10O3

Moles C14H10O3 = Mass / molar mass

Moles C14H10O3 = 2.10 grams / 226.23 g/mol

Moles C14H10O3 = 0.00928 moles

Step 4: Calculate moles NaBH4

Moles NaBH4 = 0.150 grams / 37.83 g/mol

Moles NaBH4 = 0.00397 moles

Step 5: Calculate limiting reactant

For 1 mol of C14H10O3 we need 1 mol of NaBH4 to produce 1 mol of C7H8O

NaBH4 is the limiting reactant. It will completely be consumed ( 0.00397 moles).

C14H10O3 is in excess. There will react 0.00397 moles.

There will remain 0.00928 - 0.00397 = 0.00531 moles

Step 6: Calculate moles of C7H8O

For 1 mol of C14H10O3 we need 1 mol of NaBH4 to produce 1 mol of C7H8O

For 0.00397 of C14H10O3 we need 0.00397 mol of NaBH4 to produce 0.00397 mol of C7H8O

Step 7: Calculate mass of C7H8O

Mass of C7H8O = Moles * molar mass

Mass of C7H8O = 0.00397 moles * 108.14 g/mol

Mass of C7H8O = 0.429 grams = theoretical yield

Step 8: Calculate % yield

% yield = (actual yield/ theoretical yield) * 100%

% yield = 0.5/0.429) *100 %

% yield = 116.6 %

The percent yield is 116.6 %