Thank you for the comment. I looked in this book and Reklaitis calculates yield in the same way this video does, which is net rate of production (actual) divided by the maximum possible rate of production (theoretical).
Why do we use a mass flow rate for the outlet stream, instead of calculating a molar flow rate using average molecular mass of the stream and the specific gravities?
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Why is it 1.58 again at the end? There are two reactions so that confuses me with A
It is because A and B are in a stoichiometric ratio of 1:1 for production of desired product!
based on 'introduction to material and energy balances' G.V. Reklaitis, the correct answer of fractional yield above is 2/2(1.58-0.33)x100% = 80%
Thank you for the comment. I looked in this book and Reklaitis calculates yield in the same way this video does, which is net rate of production (actual) divided by the maximum possible rate of production (theoretical).
@@LearnChemE not divided by the whole feed, but the amount of feed which is consumed in all reactions, take a look example 3.15 and 3.16
Why do we use a mass flow rate for the outlet stream, instead of calculating a molar flow rate using average molecular mass of the stream and the specific gravities?
This was very helpful, I appreciate it.
Brilliant video, you covered so much and did it so well!
Thank you so much, this is a life-saver!
why do we set m2 = 100 kg/h? could we not set to 100 mol/h and skip the conversion step?
Because the outlet stream composition is mass based (wt%)
Nice work
Well done. Thank you.
Thanks a lot