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Acetic anhydride Production

Production of acetic anhydride: (chemical reactions and calculations)

 

Reactions:

  1. Sodium acetate + Acetic Acid = water + Carbone Dioxide + Sodium acetate
  • NaHCO₃(s) + CH₃COOH(l)                   CO₂(g) + H₂O(l) + CH₃COONa(s)

  

  1. Sodium acetate + Acetyl chloride = Acetic anhydride + Sodium chloride
  • CH₃COONa(s) + CH₃COCl(l)    (CH₃CO)₂(l) + NaCl(s)

 

Calculations:

Acetic anhydride needed for 1 trial in aspirin process production = 5 Liters.

Theory calculation to know the amount of the following reagents needed for acetic anhydride production: (Acetic Acid “vinegar”, Sodium acetate, Sodium Bicarbonate and Acetyl Chloride).

  1. Acetyl chloride:

                               C4H6O3

Acetic anhydride       ρ = 1.08 g/ml

                              M = 102.089 g/mol

 

  • Molar number of acetic anhydride:

ρ = m/V ⇒ m = ρ.V ⇒ 1.08 x 5000 = 5400 g.

n = m/M ⇒ 5400/102.089 = 52.896 moles.

 

  • Calculate the volume of Acetyl Chloride:

CH₃COONa(s) + CH₃COCl(l)                  (CH₃CO)₂(l) + NaCl(s)  according to the reaction the ratio between Acetyl Chloride and Acetic Anhydride is (1:1), that’s mean they have the same molar number.

                           CH₃COCl

Acetyl Chloride      ρ = 1.1 g/ml

                           M = 78.49 g/mol

ρ = m/V & m = n.M ⇒ V = n.M/ ρ ⇒ 52.896 x 78.49 / 1.1 = 3.77 L.

 

  1. Calculate the mass of Sodium Acetate:

In the same reaction mentioned above, Sodium acetate should be use it in excess to ensure the reaction is completed.

                             CH₃COONa

Sodium Acetate      ρ = 1.53 g/ml

                            M = 82.0343 g/mol

Here the ratio of the excess should be la little higher; we choose according to the reference the following ratio (1.1:1)

n = 52.895 x 1.1 = 58.1845 moles

n = m/M ⇒ m = n.M ⇒ 58.1845 x 82.0343 = 4.77 kg.

 

  • Calculate the mass of Sodium Bicarbonate:

NaHCO₃(s) + CH₃COOH(l)                    CO₂(g) + H₂O(l) + CH₃COONa(s) according to the reaction the ratio between sodium bicarbonate and sodium acetate is (1:1), that’s mean they have the same molar number.

                                      NaHCO₃

Sodium Bicarbonate     ρ = 2.2 g/ml

                                      M = 84.007 g/mol

m = n.M ⇒ 58.145 x 84.007 = 4.88 kg.

 

  1. Calculate the volume of acetic acid 5% (vinegar):

In the same reaction mentioned above, Sodium acetate should be use it in excess to ensure the reaction is completed

                         CH₃COOH

                      ρ = 1.0005 g/ml

Acetic Acid      C = 0.86 mol/l

                     M = 60.05 g/mol

 

Here the ratio of the excess should be la little higher; we choose according to the reference the following ratio (1.1:1)

The number of moles needed n x 1.1 = 58.1845 x 1.1 = 64.00295 moles.

C = n/V⇒ V= n/C = 64.00295/0.86 = 74.4 L.