A chemical plant produces 115.0 m3/hr of waste gas (p -oo407 molL) initially hav

Question

A chemical plant produces 115.0 m3/hr of waste gas (p -oo407 molL) initially having a mole fraction of Map 0.150 NH3, and wishes to remove 90.0% of the initial amount of NH3. The maximum concentration of ammonia in water at this temperature is 0.3000 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. You will be using the minimum amount of water required for this process. Label the process flow diagram below. Label any quantities as "unknown" if they are not given above, and you have to perform a calculation to obtain the numeric value. Label any concentration as "none" if it does not exist in the stream. Place a label in every blank. The component of waste gas that is not NH3 is abbreviated below as "A" mol/hr scrubbed waste gas mol NH3/mol waste gas mol A/mol waste gas 0.9 L/hr water mol water/mol waste gas none mol/hr water 0.150 Scrubber 115.0 mol/hr effluent liquid nknown mol NH3/mol effluent liquid m3/hr waste ga 0.3000 mol/hr waste gas A mol water/mol effluent liquid mol NH3 mol waste gas mol Amole effluent liquid mol A/mol waste gas

Explanation / Answer

Incoming waste gas = 115 m3/hr = 115000 L/hr

Total number of moles in waste gas = 0.0407*115000 = 4680.5 mol/hr

Mole Fraction of NH3 = 0.15, So incoming NH3 = 4680.5*0.15 = 702.075 mol/hr

90% of NH3 has to be removed and capacity of water to dissolve NH3 = 0.3 mol/mol of water

So moles of water req = 0.9 * 702.075 / 0.3 = 2106.225 mol/hr

M.W of water = 18g/mol. So mass of water req = 37912.05 g/hr

Assuming density of water = 1g/cc Volume of water req = 37912.05 cc = 37.912 L/hr (Answer)

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