Question

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

Answer 1

Answer:

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 = $\frac{250}{0.96}$ = 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 = $\frac{260.416}{16}$ = 16.276 moles

Moles of ethane = $\frac{260.416}{30}$ = 8.6805 moles

Mole of ammonia = $\frac{260.416}{17}$ = 15.318 moles

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

Now,

Molar Flow Rate = $\frac{40.274}{4}$ = 40.276 / 4 = 10.068 moles/h

Therefore,

No.of moles of ammonia in the exit stream = $\frac{10.416}{17}$ = 0.592 moles

Therefore, in the exit stream ,

Mole fraction of ammonia = $\frac{0.592}{0.592 + 8.6805 + 16.276)}$

Mole fraction of ammonia = 0.0232