Question

Sulfuric acid is the chemical produced in the United States with the highest volume of production. In one of the earliest processes used to make it, an ore containing iron pyrites (FeS2) is roasted (burned) with air. The following reactions take place in the roasting furnace: The gas leaving the reactor goes to a catalytic converter in which most of the SO2 produced is further oxidized to SO3: Finally, the gas leaving the converter passes through an absorption tower, in which the SO3 is absorbed in water to produce sulfuric acid (H2SO4). (a) An ore containing 82 wt% FeS2 and 18% inerts is fed to a roasting furnace. Dry air is fed to the furnace in 40% excess of the amount theoretically required to oxidize all of the sulfur in the ore to SO3, an FeS2 oxidation of 85% is obtained, with 40% of the FeS2 converted forming sulfur dioxide and the rest forming sulfur trioxide. Two streams leave the roaster: a gas stream containing SO2, SO3, 02, and N2, and a solid stream containing unconverted pyrites, ferric oxide, and inert material in the ore. Calculate the required feed rate of air in standard cubic meters per 100 kg of ore roasted and the molar composition and volume (SCMJ100 kg of ore) of the gas leaving the roasting oven.

Answer 1

Ans

Basis : 100 kg/min of FeS2.

FeS2 in the feed= 100*0.82= 82 kg/min

Moles of FeS2= Mass/Molecular weight= 82kg/120= 0.683 kgmoles/min

Oxygen required to completely convert all the Fes2 to SO3 = (15/2)*0.683=5.125 moles/min

Air to be supplied= 5.125/0.21 =24.40moles/min

Air actually supplied = 1.4*24.4=34.16 moles

Nitrogen in air = 34.16*0.79=26.98 moes/min

Moles of oxygen= 34.16*0.21=7.18 moles/min

40% of FeS2 got converted to SO2

40% of FeS2= 0.4*0.683=0.2732 moles/min SO2 formed= 2*0.2732=0.5464 moles/min

Oxygen utilized for Conversion to SO2= (11/2)*0.2732=1.5026 moles/min

60% of FeS2 got converted to SO3. Hence FeS2 converted to SO3= 0.683*0.6=0.4098 moles/min Oxygen requires= 0.4098*15/2= 3.074 moles/min

Oxygen remaining= Oxygen supplied- (oxygen consumed)= 7.18-(1.5026+3.074)=2.6034 moles/min

SO2 formed= 0.5464 moles/min SO3 formed= 2*0.4098 moles/min=0.8196 moles/min

N2=26.98 moles/min

Fe2O3 formed= 0.2732//2 ( Due to the formation of SO2)+ 0.4098/2 ( due to the formation of SO3) = 0.3415 moles/min

Fe2O3 produced in Kg/min = 0.3415* molecular weight= 0.3415*160 =54.64 moles/min

Products : N2= 26.98 moles/min   O2= 2.6034 moles/min, SO2= 0.5464 moles/min, SO3= 0.8196

Flow rate of gases from the reaction = 26.98+2.6034+0.5464+0.8196=30.9494 moles/min

Composition : N2= 100*26.98/30.9494=87.17% O2= 100*2.6034/30.9494=8.411 SO2= 100*0.5464/30.9494=1.765 SO3= 100*0.8196/30.9494=2.65%