Question

acetic anhydride reacts with h2o to form acetic acid, 2CH3 COOH.

a) Assuming you start with 42 mL of acetic anhydride and an excess of water, what is the theoretical yield of acetic acid?

b) If you obtain 12 mL of acetic acid, what is the % yield for the reaction?

Density of acetic anhydride 1.08 g/mL

density of acetic acid 1.05 g/mL

Answer 1

Acetic anhydride + H2O excess -----> 2CH3COOH

Density of acetic anhydride 1.08 g/mL

d = m / V = mass / Violume

1.08 = m / 42

m = 45.36

Moles of acetic anhydride = Mass / Molar mass of acetic anhydride = 45.36 / 102 = 0.4447

1 mole acetic anhydride gives 2 moles of acetic acid

Moles of acetic acid formed = 2*0.4447 = 0.8894

Mass of acetic acid = Moles*Molar mass = 0.8894*60 = 53.36 gms (Theoritical yield)

density of acetic acid 1.05 g/mL

d = 1.05 = m / 12

m = 12.6 gms of acetic acid obtained

% yield of the reaction = Actual yield*100 / Theoritical yield = 12.6*100 / 53.36 = 23.61%