- Hits: 235
Acetic anhydride Production
Production of acetic anhydride: (chemical reactions and calculations)
Reactions:
- Sodium acetate + Acetic Acid = water + Carbone Dioxide + Sodium acetate
- NaHCO₃(s) + CH₃COOH(l) CO₂(g) + H₂O(l) + CH₃COONa(s)
- 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).
- 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.
- 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.
- 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.