Sulfur is burned with air (21 mole % O_2, 79% N_2) to form SO_2, and in a second

Question

Sulfur is burned with air (21 mole % O_2, 79% N_2) to form SO_2, and in a second unit the SO_2 is oxidized to form SO_3. Calculate the moles of oxygen theoretically required for the overall process per mole of sulfur fed. How many moles of air does this correspond to? Suppose air is fed to the combustion furnace at a rate 25% in excess of that theoretically required, and that the combustion reaction is complete but the oxidation only proceeds to a 90% fractional conversion of SO_2. Calculate the molar composition (mole fractions of each component) of the gas leaving the oxidation reactor.

Explanation / Answer

The reaction is S+O2--------->SO2 and SO2 +0.5O2---->SO3

Basis : 1 mole of Sulfur, as per the stoichiometry of the reactions, moles of oxygen required= 1+0.5= 1.5 moles

Air contains 21% O2 and 79% N2, moles of Air required= 1.5/0.21 =7.14

Air is supplied at 25% excess, moles of air supplied= 7.14*1.25= 8.925 moles

since combustion reaction is complete, moles of SO2 formed = 1 mole ( from combustion of S+O2--->SO2)

1 mole of SO2 reacts and the reaction is only 90% complete. Hence moles of SO3 formed= 0.9 moles, moles of SO2 remaining = 0.1 moles, Oxygen used= 0.9/2=0.45 ( SO2+0.5O2----->SO3, moles of oxygen per mole of SO2=0.5)

Oxygen remaining = Oxygen supplied- oxygen consumed = 8.925*0.21- 1.45=0.42425

N2 in the produce= 8.925*0.79=7.05

Products ( moles) : SO2= 0.1, SO3= 0.9, O2= 0.42425 and N2= 7.05

total moles = 0.1+0.9+0.42425+7.05=8.47425

Mole fraction = moles of the component / total moles

Mole fractions : SO2= 0.1/8.47425= 0.0118, SO3= 0.9/8.47425= 0.1062, O2= 0.42425/8.47425=0.051, N2= 7.05/8.47425= 0.8319

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