4. a gas containing equal parts methane, ethane and ammonia flows at a constant rate through a laboratory water- based absorption unit, which absorbs 96% of the ammonia and retains all the water. no methane or ethane is absorbed and no water evaporates. initially there was 5 l of water in the absorber, at the end of 4 hours the liquid mass is 5.25 kg. calculate the molar flowrate of the gas stream into the absorber, and the mole fraction of ammonia in the exit gas stream

Molar flow rate is 10.068 mole/h

Mole fraction is 0.0232

Solution:

As per the question:

Amount of ammonia absorbed by water = 96% = 0.96

Initially, the weight = 5 kg

The final weight = 5.25 kg.

Therefore,

96% of the initial amount of Ammonia is represented by:

Final weight - Initial weight

5.25 – 5.00 = 0.250 kg = 250 g

250 g represents 96% of the initial ammonia.

Now,

Initially, weight of ammonia = = 260.416 g

Now,

Ammonia present in the exit gas stream is given by:

260.416 - 250 = 10.416 g

Now,

Weight of methane = Weight of ethane = 260.416 g

So,

Moles of methane = = 16.276 moles

Moles of ethane = = 8.6805 moles

Mole of ammonia = = 15.318 moles

Total no. of moles = 16.276 + 8.6805 + 15.318 = 40.274 moles

Now,

Molar Flow Rate = = 40.276 / 4 = 10.068 moles/h

Therefore,

No.of moles of ammonia in the exit stream = = 0.592 moles

Therefore, in the exit stream ,

Mole fraction of ammonia =

Mole fraction of ammonia = 0.0232

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