Map o emical plant produces an ammonia-rich waste gas that cannot be released as

Question

Map o emical plant produces an ammonia-rich waste gas that cannot be released as-is into the environmen ne 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 125.0 m3hr of waste gas (pe 0.0407 mol/L) initially having a mole fraction of 0.140 NH3, and wishes to remove 90.0% of the initial amount of NH3-The maximum concentration of ammonia in water at this temperature is 0.4500 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 (pe 0.990 g/mL) required to scrub out 90.0% of the incoming NH3? Number L/hr at he mle fractbed wsts str Number mol NH mol waste gas

Explanation / Answer

Volumetric flow rate of waste gas = 125 m3/hr x 1000 L/m3

= 125000 L/hr

Molar flow rate of waste gas = 125000 L/hr x 0.0407 mol/L

= 5087.5 mol/hr

Molar flow rate of NH3 = 0.140 x 5087.5 = 712.25 mol/hr

Part a

90% NH3 removed

Removal flow rate of NH3 = 0.90 x 712.25 = 641.025 mol/hr

Part b

Minimum molar flow rate of water = removal rate / max concentration

= (641.025 mol NH3 /hr) / (0.45 mol NH3 / mol water)

= 1424.5 mol water / hr

Mass flow rate of water = 1424.5 mol x 18 g/mol

= 25641 g/hr

Volumetric flow rate = (25641 g/hr) / (0.990 g/mL)

= 25900 mL / hr x 1L/1000 mL

= 25.90 L/hr

Part c

Moles of waste gas = (1-0.140) x 5087.5 = 4375.25 mol/hr

Moles of NH3 after scrubbing = 712.25 - 641.025

= 71.225 mol/hr

Mol fraction of NH3 = 71.225 / 4375.25 = 0.0163

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