In one of the Feeco International blast plant, a simplified view of the processi

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In one of the Feeco International blast plant, a simplified view of the processing of iron ore.........
Please I need the answer for this question, a detail answer with all step and showing the formulas used . thanks a lot

Problem 1 (100 points) In one of the Feeco International blast furnace plant, a simplifed view of the processing of iron ore (hematite, Fe,O, ) to Fe is thus presented in which the principal reaction is consider to be Fe(), + 3C ? 2Fe + 3CO along with the following undesired side reaction Sixty percent of a metric ton of carbon (coke) is reacted with 2000 kg of pure iron oxide, Fe O. to produces 1125.0 kg of pure iron, 183 kg of FeO, and 85.0 kg of Fe.O, Calculate following items: (a) Determine the molar composition of the output stream of species from the furnace (b) The percentage of excess carbon furnished based on the principal reaction (c) The percentage conversion of Fe O, to Fe (d) What is the selectivity in this process (of Fe with respect to Feo)? If you solve problem la corectly (exira 10 poins) Fe-so kg kmole, C-12 kgkmole, O-16 kg/kmole best

Explanation / Answer

Feed contains : 2000 kg of Fe2O3, expressed in terms of moles, moles of Fe2O3= mass/molar mass= 2000/160= 12.5 kg moles. mass of C= 60% of ton, 1 ton =1000kg, mass of C= 1000*60/100=600 kg, moles of C= mass/atomic weight= 600/12= 50 kg moles. As per the principal reaction Fe2O3+3C ------>2Fe+3CO, 1 mole of Fe2O3 requires 3 mole of C. Hence moles of C required= 3*12.5= 37.5 kg moles

moles of excess C supplied= moles of C supplied- moles of C utilized= 50-37.5= 12.5 kg moles

% excess of C =100* excess moles/stoichiometric moles= 100*12.5/37.5= 33.33%

2. Product: Fe= 1125 kg.,moles of Fe = mass/atomic weight= 1125/56= 20.089 kg moles

moles of Fe2O3 required for this production =moles of Fe formed/2= 20.089/2= 10.045 kg moles

moled of FeO formed= mass of FeO formed/ atomic weight= 183/72= 2.542 kg moles

from side reaction Fe2O3+C -------.2FeO+CO, moles of Fe2O3 used= 2.542/2= 1.271 kg moles

moled of Fe2O3 given = 85 5kg/160 kg = 0.53 kg moles

total of Fe2O3 in the products= 0.53+1.271+10.045= 11.846 kg moles

moles of Fe2O3 supplied= 12.5 kg moles. moles of Fe2O3 entriained with CO= mole s of Fe2O3 supplied- moles in the products= 12.5-11.846= 0.654 kg moles

Products ( solids): in terms of moles : Fe2O3= 0.53, FeO= 2.542 and Fe= 20.089, total moles= 20.089+0.53+2.542= 23.161

molar composition = moles of component/ total moles

Molar composition : Fe2O3= 0.53/23.161=0.023, Fe= 20.089/23.161=0.867, FeO= 2.542/23.161=0.11

Entrained gases : Co ( from Fe2O3+3C ------->2Fe+3CO) = 20.088*3/2= 30.1335, CO from Fe2O3+C ---->2FeO+CO is 1.271 kg moles=31.4045 kg moles, Entrained iron = 0.654 kg moles

Moles of Fe2O3 used= 12.5 kg moles, moles of Fe formed from principal reaction = 2*12.5= 25 kg moles

moles of Fe actually formed= 20.089

% conversion of Fe2O3 to Fe= 100*20.089/25=80.356

selectivity of FeO/ Fe= moles of FeO formed/moles of Fe formed= 20.089/2.542= 7.9

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