Question

A chemical plant produces an ammonia-rich waste gas that cannot be released as-is into the environment. 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 105.0 m^3/hr of waste gas (rho = 0.0407 mol/L) initially having a mole fraction of 0.150 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 0.4000 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. What rate is NH_3 is being removed from the feed gas? What is the minimum flow rate 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?

Answer 1

Given : Volumetric flow rate of ammonia = 105 m³/hr = 7/240 m³/s

Mole density of waste gas, d_w = 0.0407 mol/L = 40.7 mol/m³

Initial mole fraction of ammonia in incoming waste gas = 0.120

Thus, mole density of ammonia in incoming waste gas = 0.120*40.7 = 4.884 mol/m³

Thus, rate at which ammonia comes in from the incoming stream = mole density of NH₃*Volumetric flow rate = 4.884*(7/240) = 0.14245 moles NH₃ per second

Since 90% ammonia gets removed, so rate of removal of ammonia = 0.9*rate at which ammonia comes in = 0.9*0.14245 = 0.128 moles NH₃ per second = 460.8 mol/h

(2) Ammonia holding capacity of water = 0.35 mol NH₃ per mole of water

So, to hold 0.128 moles ammonia, moles of water required = 1/0.35*0.128 = 0.365 moles water

Molecular mass of water = 18 g

So, mass of water required = 18*0.365 = 6.57 g

Thus, volumetric flow rate of water required = mass required/density of water = 6.57/0.99 = 6.63 ml per second = 6.63*10-6*3600 = 0.023868 m₃/hr = 23.868 L/h

(3) New mole fraction of ammonia = moles of ammonia in the outgoing stream/total moles in outgoing stream

Moles of ammonia in outgoing stream = 10% of moles of ammonia in incoming stream (because 90% gets removed ) = 0.10*initial mole density = 0.1*4.884 = 0.4884 moles

Total moles in outgoing stream = total moles in initial stream-moles of ammonia removed = 40.7-(90% of 4.884) = 36.3

Thus, new mole fraction of ammonina = 0.4884/36.3 = 0.01345