A chemical plant produces an ammonia-rich waste gas that cannot be released as-i

Question

A chemical plant produces an ammonia-rich waste gas that cannot be released as-is into One method to reduce the ammonia concentration in the waste gas is to bubble the waste gas through a liquid solvent. Components of the gas that are highly soluble, like ammonia, will dissolve into the liquid phase. Water is an appropriate liquid solvent for this process. The chemical plant produces 110.0 m^3/hr of waste gas (rho = 0.0407 mo/L) initially having a mote fraction of 0.140 NH_3. and wishes to remove 90.0% of the initial amount of NH_3. The maximum concentration of ammonia in water at this temperature is 0.3500 mol NH_3/mol water. Neglect the absorption of other waste gas components into the water, and the evaporation of water into the waste gas stream. At what rate is NH_3 being removed from the feed gas? What is the minimum flow rata of water (rho = 0.990 g/mL) required to scrub out 90.0% of the incoming NH_3? What is the mole fraction of NH_3 in the scrubbed waste gas stream?

Explanation / Answer

Ans. 2

Part A. Given,

Waste gas production rate = 110.0 m3 / hr

                                                = 110.0 x 103 L/ hr                   ; [1 m3= 103 L]

Density of waste gas = 0.0407 mol/L

Now,

Waste gas production rate (mol/ hr) = waste gas production rate (L/hr) x density (mol/L)

                                                            = (110.0 x 103 L/ hr) x (0.0407 mol/L)

                                                            = 4.477 x 103 mol/ hr

NH3 production rate (mol/ hr) = mole fraction NH3 x waste gas production rate (mol/hr)

                                                            = 0.140 x 4.477 x 103 mol/ hr

                                                            = 6.2678 x 102 mol/ hr

Desired % removal rate of NH3 = 90 % of NH3 produced

                                                = 90% of (6.2678 x 102 mol/ hr)

                                                = 5.64102 x 102 mol/ hr

Part B: Given,

            NH3 saturation concentration in water = 0.3500 mol NH3 / mol H2O

Minimum water flow rate required =

(Desired % removal rate of NH3/ NH3 saturation concentration in water)

= (5.64102 x 102 mol NH3/ hr) / (0.3500 mol NH3 / mol H2O)

= 1.61 x 103 mol H2O/ hr

Now,

Total mass of water flowing per hour = number of moles x molar mass

                                                = 1.61 x 103 mol x (18.01 g mol-1)

                                                = 2.89 x 104 g

Volume of water flowing per hour = Mass of water flowing per hour / density of water

                                                = 2.89 x 104 g / (0.990 g/ mL)

                                                = 2.93 x 104 mL

                                                = 2.93 x 10 L                         ; [1 mL = 10-3 L]

Therefore, desired water flowrate = 2.93 x 10 L / hr = 29.3 L/ hr  

Ans. 3. Moles of waste gas produced per hr = 4.477 x 103 mol

Moles of NH3removed per hr = 5.64102 x 102 mol

Remaining moles of NH3 = moles NH3 produced per hr - moles NH3 removed per hr

                                          = 6.2678 x 102 mol - 5.64102 x 102 mol

                                          = 0.62678 x 102 mol

Remaining moles of waste gas after removal of NH3 =

(total moles of waste gas – moles of NH3 removed)

= 4.477 x 103 mol - 5.64102 x 102 mol

= 3.912898 x 103 mol

Moles fraction of NH3 = Remaining moles of NH3 / remaining moles of waste gas

                                    = 0.62678 x 102 mol / 3.912898 x 103 mol

                                    = 0.016

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