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 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 liquicd phase. Water is an appropriate liquid solvent for this process. The chemical plant produces 110.0 m/hr of waste gas (p 0.0407 mo/L) initially having a mole fraction of 0.140 NH, and wishes to remove 90.0% of the initial amount of NH3. The maximum concentration of ammonia in water at this temperature is 0.5500 mol NH3/ 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 NH3 being removed from the feed gas? Number mol/ hr What is the minimum flow rate of water ( = 0.990 g/mL) required to scrub out 90.0% of the incoming NH3? Number L/hr What is the mole fraction of NH3 in the scrubbed waste gas stream? Number mol NH, mol waste gas

Explanation / Answer

Molar flow rate of waste gas = density x volumetric flow

= 0.0407 mol/L x 110 m3/hr x 1000L/m3

= 4477 mol/hr

Molar flow of NH3 in waste gas = 0.140 x 4477 mol/hr

= 626.78 mol/hr

Removal rate of NH3 = 0.90*626.78 mol/hr

= 564.102 mol/hr

Part b

Minimum flow of water

= molar flow of NH3 in water x 1mol water/0.55 mol NH3

= 564.102 mol/hr x 1mol water/0.55 mol NH3

= 1025.64 mol/hr x 1 mL/0.990g x 18g/mol x 1L/1000 mL

= 18.648 L/hr

Part C

Molar flow Of NH3 in feed waste gas = 626.78 mol/hr

Molar flow of NH3 in scrubbed gas = 0.10*626.78 mol/hr

= 62.678 mol/hr

Molar flow of waste gas (excluding NH3) = 4477 - 626.78

= 3850.22 mol/hr

Mol fraction of NH3 in scrubbed gas

= moles of NH3 in scrubbed gas/total moles of scrubbed gas

= 62.678/(62.678 + 3850.22)

= 0.01602

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