# Problem 2 Mercaptans, hydrogen sulfide, and other sulfur compounds are removed from natural gas by various so-called “sweetening processes” that make available otherwise useless “sour” gas. As you know H2S is toxic in very small quantities and is quite corrosive to process equipment. A proposed process to remove H2S is by reaction with SO2 H2S (g) + SO2 (g) → S (s) + H2O (g) In a test of the process, a gas stream containing 20% H2S and 80% CH4, was combined with a stream of pure SO2. The process produced 156 lb-mol of S(s), and in the product gas the ratio of SO2 to H2S was equal to 3, and the ratio of H2O to H2S was 10 a) Write the balanced reaction. b) Calculate the amount (Ib-mol) of H2S and SO2 in the feed. is in excess. d Determine the amount of product fomed (lb-mol) and its composition Product P 20% H2S–+ 80% CH4 ngoa nizs nH2o nCHa Reactor SO2 156 Ib-mol

## Problem 2 Mercaptans, hydrogen sulfide, and other sulfur compounds are removed from natural gas by various so-called “sweetening processes” that make available otherwise useless “sour” gas. As you know H2S is toxic in very small quantities and is quite corrosive to process equipment. A proposed process to remove H2S is by reaction with SO2 H2S (g) + SO2 (g) → S (s) + H2O (g) In a test of the process, a gas stream containing 20% H2S and 80% CH4, was combined with a stream of pure SO2. The process produced 156 lb-mol of S(s), and in the product gas the ratio of SO2 to H2S was equal to 3, and the ratio of H2O to H2S was 10 a) Write the balanced reaction. b) Calculate the amount (Ib-mol) of H2S and SO2 in the feed. is in excess. d Determine the amount of product fomed (lb-mol) and its composition Product P 20% H2S–+ 80% CH4 ngoa nizs nH2o nCHa Reactor SO2 156 Ib-mol

• a.

Amount of H2S unreacted = (0.2x – 104) lbmol

Amount of SO2 unreacted = (y-52) lbmol

Amount of S at outlet = 156 lbmol

Amount of CH4 at outlet = 0.8x lbmol

Amount of H2O at outlet = 104 lbmol

• b.

Hence amount of H2S in feed = 0.2*572

= 114.4 lbmol

Amount of SO2 in feed = 83.2 lbmol

• c.

To produce 156 lbmol of S , it requires (2/3)*156 lbmol of H2S that means it requires 104 lbmol of H2S

Hence H2S is the limiting reagent, therefore

Percentage excess of other reactant in excess = (83.2 – 52)*100 / 83.2

• d.

Amount of H2S unreacted = (0.2*572 – 104) = 10.4 lbmol

Amount of SO2 unreacted = (83.2-52) = 31.2 lbmol

Amount of S at outlet = 156 lbmol

Amount of CH4 at outlet = 0.8*572 = 457.6 lbmol

Amount of H2O at outlet = 104 lbmol