The following process is used to recover acetic acid (C_2 O_2 H_4) from calcium
ID: 530834 • Letter: T
Question
The following process is used to recover acetic acid (C_2 O_2 H_4) from calcium acetate (Ca (C_2 O_2 H_3)_2), by using sulphuric acid (H_2 SO_4). The reaction is given below: Ca(C_2 O_2 H_3)_2 + H_2 SO_4 rightarrow 2 C_2 O_2 H_4 + CaSO_4 The molar ratio between sulphuric acid and calcium acetate in the fresh feed is 6:5. The recycle stream contains calcium acetate only. The single pass conversion is 90%. The mass flow - rate of calcium acetate in the fresh feed is 100 kg/h. Calculate flow - rates (in kg/h) of the recycle, product and by-product streams.Explanation / Answer
Fresh feed contains : Calcium acetate= 100 kg/hr
Molar mass of calcium acetate= 158 g/mole
Moles= mass/molar mass
Moles of calcium acetate= 100/158 kg moles/hr=0.63 kgmoles/hr
Moles of sulphuric acid = 0.63*6/5 =0.756 kgmoles/hr
The reaction is Ca(C2O2H3)2+H2SO4------>CaSO4+2CH3COOH
Molar ratio of reactants calcium acetate : H2SO4 ( Theoretical) = 1:1
Actual ratio = 0.63:0.756
Excess H2SO4 is removed along with CaSO4. Moles of H2SO4 excess over and above the stoichiometric requirement = 0.756-0.63= 0.126 kgmoles/hr
This is removed along with CaSO4.
Moles of CaSO4 formed =0.63 kgmoles/hr and moles of acetic acid formed= 2*0.63= 1.26 kgmoles/hr
Let R= recycle flow rate of calcium acetate
Moles of calcium acetate entering = R+0.63
Moles of H2SO4 reacted= 0.63
Since the stoichiometric ratio of calcium acetate to sulphuric acid is 1:1
Moles of calcium acetate reacted= (R+0.63)*0.9= 0.63 =moles of H2SO4 reacted
R+0.63= 0.7
R=0.7-0.63=0.07 moles/hr
So Products : CaSO4=0.63 kgmoles/hr, molar mass = 136, mass of CaSO4 formed= 136*0.63= 85.68 kg/hr, acetic acid =1.26 kgmoles/hr, molar mass= 60, mass of acetic acid formed= 60*1.26 =75.6 kg/hr
H2SO4= 0.126 kgmoles/hr*98= 12.348 kg/hr
Flow rate of recycle calcium acetate= 0.07 moles/hr, mass flow rate of calcium acetate= 158*0.07= 11 kg/hr
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