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 125.0 m3/hr of waste gas (ρ = 0.0407 mol/L) initially having a mole fraction of 0.190 NH3, and wishes to remove 90.0% of the initial amount of NH3. The maximum concentration of ammonia in water at this temperature is 0.4000 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.

a) What rate is NH3 is being removed from the feed gas (in mol/hr)?

b) What is the minimum flow rate of water (ρ = 0.990 g/mL) required to scrub out 90.0% of the incoming NH3 (in L/hr)?

c) What is the mole fraction of NH3 in the scrubbed waste gas stream?

(I need show works plz)

Answer 1

moles of ammonia waste is given as = 0.0407 moles per 1 liter

mol fraction of NH3 = 0.19

Therefore moles of NH3 = 0.19 x0.0407 = 0.007733 per liter

90 % of it = 0.007326 x0.9 = 0.0069597 moles per liter

moles of NH3 produced per hr = 125 x1000 x 0.0065934 = 869.96 ( 135 dm3 = 1350000 liters)

Hence NH3 being removed from feed gas = 869.96 mol /hr------------------- (Answer a)

maximum concentration of ammonia in water at this temperature is = 0.4 mol NH3/ mol of water

Therefore for 869.96 mol of NH3 water required will be = ( 890.11 /0.4) = 2174.9062 mol

Molecular weight of NH4 = 18 g/mol

And ρ = 0.990 g/mL given, Hence;

= 2174.9062 x18 gm = 39148.31 gm = 39148.31 /0.99 = 39543.3 ml = 39.54 liters

Water flow rate is 39.54 liters/hr---------------------------------------------- (Answer b)

10 % NH3 left in waste = 0.1 x0.007733 = 0.0007733

mol fraction of NH3 = (0.0007733/0.0407) = 0.018-------------------------------(Answer c)