Water Vapor Adsorber Material Balance

[@LexiLueger]

For part 2: If the pellets are absorbing 25% of the water, then 75% of the original mole % will be the outlet streams molar fraction. I'm confused why a mass balance is needed for this...?

[@kukkusharma1394]

In the second part.....2 different equations and both are equal to N(h2o)...... ??

[@LearnChemE]

This screencast has been reviewed by faculty from other academic institutions.

[@uetian7537]

dear sir i need some help on energy and mass balance of a specific design ?

[@marisareiss7077]

How did he get 20 mol/hr in the first question? I plugged in n to be 1, so N(H2O)=.05. What do I do next?

[@LearnChemE]

N(H2O) = 1, then you solve for n.

[@planomathandscience]

You're using n and N the wrong way round.

[@alat7580]

@@planomathandscience Thanks!

[@СерикТабусов-ь8н]

5:24 don know how did you get 3% i got 5

[@nancynandha235]

Where did you get 99 from in the mole fraction part?

[@LearnChemE]

The 99 came from solving the equation shown: 1 mol = (20 mol/hr)(5 hr) - (n_out)(5 hr). Rearranging to isolate n_out equals 99/5.

[@kukkusharma1394]

This equation is confusing..becoz first u write this N (h2o)=20*5-n (out)*5.....ok...but then the same N(h2o)=20*5*.04-19.8*5*X .......does that not counter both equations

[@invalide]

I did the second part differently: total(in) = 20 mol/hr 0.2 mol/hr of water vapour is absorbed away, so total(out) = 19.8 mol/hr water vapour(in) = 0.8 mol/hr 0.2 mol/hr of water vapour is absorbed away, so vapour(out) = 0.6 mol/hr mole fraction of water vapour in output stream = [ water vapour(out) / total(out) ] * 100 = [0.6/19.8]*100 = 3%

[@irembaskaya5517]

Can anybody explain me the first question please?

[@LearnChemE]

The first question is trying to find the flow rate into the system, n, so a mass balance is used.