If anyone could move this to the chemical engineering subject (Can\'t seem tovbe

Question

If anyone could move this to the chemical engineering subject (Can't seem tovbe anle to select it). thank you

In the process sketched in Figure P6.3.18, NazCO3 is produced by the reaction Na2SCaC03 NazCO3CaS The reaction is 90% complete on one pass through the reactor, and the amount of CaCO3 entering the reactor is 50% in excess of that needed. Calculate on the basis of 1000 lb/hr of fresh feed, all streams, molar and mass compositions. Assign a letter for each stream. Write the results in a table. Fresh 1000 lb/hr, Feed Pure Na2S 40% Na2S 60% H2O ?? Reactor ,30% CaCO 8000 Na2CO3 20% H20 1923 lb/hr 20% CaS soln 30% CaCO3 soln. CaCO3 soln. 60 lb/hr CaCO Total:

Explanation / Answer

SOLUTION:

Fresh feed of Na2S and H2O contain 1000*0.4= 400 lb/hr of Na2S

moles of Na2S= 400/78= 5.128 lb moles/hr

moles of Na2CO3 formed= 5.128 lb moles/hr. molar masss of Na2CO3= 106

mass of Na2CO3 formed = 5.128*106= 544 lb/hr but tihs is only 80% in the final solution. Hence mass of Na2CO3 solution formed/hr product = 544/0.8= 680 lb/hr

moles of CaS formed = 5.128 moles/hr. mass of CaS formed = 5.128*72= 369 lb/hr, this correspond to 20% CaS. mass of CaS solution= 369/0.2 = 1845 lb/hr

let R= recycled moles of Na2S/hr

moles of Na2S entering = 5.128 +R

moles of CaCO3 required= 5.128+R

moles of CaCO3 supplied = 1.5*(5.128+R)

moles of Na2S reacted = 0.9*(5.128+R)  

moles of CaCO3 reacted = 0.9*(5.128+R)

moles of CaCO3 remaining = (1.5-0.9)*(5.128+R) =0.6*(5.128+R)

moles of CaCO3 removed = 60/100 =0.6 lb moles/hr

moles of CaCO3 remaining = 0.6*(5.128+R)-0.6 = 0.6*(4.128+R)

this is recycled aliong with 30% of 1923 lb/hr of fresh CaCO3

hence 5.769 ( fresh CaCO3)+ 0.6*(4.128+R) = 1.5*(5.128+R) ( CaCO3 supplied to the reactor)

5.769 +2.476 +0.6R= 7.692+1.5R

8.245-7.692 = 0.9R

R=0.6144 moles/hr of Na2S that is 0.6144*78 lb/hr of =47.9 lb/hr Na2S.

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