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 %
