Question

Sodium salicylate (HOC6H4COONa) is used in medicine as pain killer. It can be prepared from sodium phenolate (C6H5ONa) and carbon dioxide. Most commonly, solutions of sodium phenolate are produced by treating phenol with sodium hydroxide. Suppose phenol is run into a 30% excess hot aqueous caustic soda (87% NaOH by mass) with 130 m3 dry carbon dioxide at 300 kPa and 100oC per hr. After the desired amount of carbon dioxide is absorbed and reacted, sodium salicylate solution is withdrawn. The solution has a composition of 68.21% HOC6H4COONa, 8.73% C6H5ONa, 5.24% C6H5OH, 8.92% NaOH, and 8.90% H2O. The gaseous stream contains CO2 and H2O. On a basis of 100 kg hot aqueous caustic soda, calculate:

a) kg of salicylate solution/hr

b) degree of completion of the reaction

c) composition of gaseous stream

Answer 1

Basis : 100 kg aqueous NaOH

NaOH is 30% excess

87 mass% NaOH

The diagram is shown below

A)

Amount of NaOH = 100(0.87) = 87 Kg

30% excess so

NaOH required = (87/1.3 )= 66.923 Kg

NaOH reacted = 66.923 Kg

NaOH unreacted = 87-66.923 = 20.077 Kg

Salicylate Solution contains 8.92% NaOH

So total solution flowrate = 20.077/(0.0892) = 225.0784 Kg/h

Mass percent of salicylate in solution = 68.21%

Amount of salicylate solution produced per hour = 225.0784 (0.6821) = 153.525 Kg/h

B)

The reactions occuring are

Weight fraction of unreacted C6H5ONa = 8.73%

Total solution = 225.0784 Kg/h

Unreacted C6H5ONa = 225.0784(0.0873) = 19.649 Kg/h

Mass of NaOH reacted = 66.923 Kg

M. W of NaOH = 40 kg/kmol

Moles of NaOH reacted = 66.923/40 =

1.6730 kmol/h

According to Stiochiometry of first reaction

Moles of C6H5ONa produced for 1.670 Kmol =

1.670(1) = 1.670 Kmol/h

M. W of C6H5ONa = (6) (12) +(5×1) +(16) +(23) = 116 kg/Kmol

Mass of C6H5ONa produced = 116(1.670) =193.72 Kg/h

Mass of unreacted C6H5ONa = 19.649 Kg/h

Degree of completion of reaction =

1-((unreacted C6H5ONa) /(total C6H5ONa) ) =

1-(19.649/193.72) = 0.8985

Degree of completion = 0.8985 = 89.85%

C)

Dry CO2 is at 300 KPa = 300×(1000) Pa

T = 100°C = 373 K

V =130 m3/h

According to ideal gas

n = PV/RT

n = 300(1000) (130) /(8.314× 373)

n = 12576.093 mol/h = 12.5760 Kmol/h

Mass of C6H5ONa reacted to form salicylate =

(193.72-19.649) = 174.071 Kg/h

M. W of C6H5ONa = 116 Kg/kmol

Moles of C6H5ONa reacted =( 174.071/116) = 1.50 Kmol

According to Stiochiometry of second reaction

CO2 reacted to 1.50 Kmol C6H5ONa = 1.5(1) = 1.50 Kmol

CO2 reacted = 1.50 Kmol/h

Moles CO2 unreacted = 12.5760-1.50

= 11.076 Kmol/h

M. W of CO2 = 44 Kg/Kmol

Mass of CO2 unreacted = 11.076(44)

= 487.344 Kg/h

According to Stiochiometry of first reaction H2O formed from 1.6730 Kmol NaOH = 1.6730(1) = 1.6730 Kmol/h

M. W of water = 18 Kg/kmol

Mass of water formed = 18(1.6730) =

30.114 Kg/h

Mass fraction of water in salicylate solution =

0.0890

Total solution = 225.0784 Kg/h

Water present in solution = 225.0784(0.890) = 20.03 Kg/h

Water in 100 Kg NaOH solution(87% NaOH)

= 100(1-0.87) = 13 Kg

Water produced in reaction = 30.114 kg/h

Water in gas stream = (30.114+13) -20. 03 = 23.084 Kg/h

Gases contain CO2 and H2O

Component	mass(Kg)	wt%
CO2	487.344	95.477
H2O	23.084	4.522
Total	510.428	100

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C6H5OH + NaOH → C6H5ONa+ H20

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