Consider a feed stream to a Cl

Consider a feed stream to a Claus process that consists of 10.0mole% H2S and 90.0% CO2. Onethird of the stream is sent to afurnace where the H2S is burned completely with a stoichiometricamount of air fed at 1 atm and 25°C. The combustion reaction is H2S3 2 O2! SO2 H2O The product gases from this reaction are then mixedwith the remaining two-thirds of the feed stream and sent to areactor in which the following reaction goes to completion: 2H2SSO2! 3S 2H2O The gases leave the reactor at 10:0 m3/min, 320°C, and205 kPa absolute. Assuming ideal-gas behavior, determine the feedrate of air in kmol/min. Provide a single balanced chemicalequation reflecting the overall process stoichiometry. How muchsulfur is produced in kg/min?

Answer:

(a): Reaction-1: H2S + 3/2 O2 ——– > SO2 + H2O

Reaction-2: 2H2S + SO2 ——— > 3S + 2H2O

——————————————————————————–

SO2, being common to both reactant and product, will cancel out.Hence the overall single balanced chemical reaction is:

Overall reaction: 3H2S + 3/2O2 ——- > 3S + 3H2O

or 6H2S + 3O2 ——- > 6S + 6H2O (answer)

(b): For the gas leaving the reactor:

V = 10.0 m3/min

T = 320 oC = 320 + 273 = 593 K

P = 205 kPa = 2.05×105 Pa = 2.05×105N/m2

Let the moles of gas leaving the reactor be ‘n’ mol/min

Applying ideal gas equation

PV = nRT

=> n = PV/RT = (2.05×105 N/m2 x 10.0 m3) / (8.314Jmol-1K-1 x 593 K)

=> n = 4.16×102 mol/min or 416 mol/min

Hence moles of gas leaving the reactor is 416 mol/min.

Suppose the initial moles of reactant mixture entering thereactor be ‘Y’ mol/min

=> moles of H2S in the feed stream = Yx10/100 = 0.1Ymol/min

moles of CO2 in the feed stream = 0.9Y ml/min

Given 1/3 rd mol of H2S is converted to SO2.

=> moles of H2S converted to SO2 = 0.1Y / 3 = 0.03333Ymol/min

In the balanced chemical reaction, 1 mol H2S reacts with 1.5 molO2

=> 0.03333Y mol H2S will react with the moles of O2 = (1.5mol O2 / 1 mol H2S) x 0.03333Y mol H2S

= 0.05Y mol O2

=> Stoichiometric moles of air(0.21 O2 + 0.79 N2) required =0.05Y mol O2 + 0.05Y x (0.79/0.21) mol N2

= 0.05Y mol O2 + 0.188Y mol N2

= 0.238Y mol air / min

In the overall balanced reaction, 3 mol H2S forms 3 mol H2O.

Hence 0.1Y mol H2S will form 0.1Y mol H2O.

Hence the gas leaving the reactor contains 0.1Y mol H2O, 0.9Ymol CO2 and 0.188Y mol N2.

=> total moles of gas leaving the reactor = 0.1Y + 0.9Y +0.188Y moll/min = 1.188Y mol/min

Also 1.188Y mol/min = 416 mol/min

=> Y = 416/1.188 = 350.2 mol/min

Hence feed rate of air = 0.238Y mol air/min = 0.238 x 350.2 =83.3 mol = 0.0833 kmol air/min (answer)

In the balanced reaction 2 mol H2S forms 3 mol S

=> 0.1Y mol H2S will also form 0.1Y mol S

Hence moles of S formed = 0.1Y = 0.1 x 350.2 = 35.02 mol/min

= 35.02 mol x (32.065 g/mol)/min = 1123 g S/min = 1.12 Kg S /min (answer)


 
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