The correct answer is 107.7 lb mol/ hr. I\'ve been stuck on this for a few hours

Question

The correct answer is 107.7 lb mol/ hr. I've been stuck on this for a few hours and any help would be greatly appreciated. I will spread the love of thumbs up from all of my classmates for the full step by step solution! Thanks. 18. Ammonia is used for fertilizer production and has been critical to successful agriculture. A steady-state chemical pro. cess is used to convert nitrogen (N2) and hydrogen (H2) to am. monia (NH3) by the reaction N2 3H2 2NH3 Stream 1, containing 95 mole% nitrogen and 5 mole% hydro- gen, enters the process at a rate of 400 lbmol/hr, and stream 2, containing pure hydrogen (density 0.08 lbm/fts), enters the process at a volumetric flow rate of 31,000 ft3/hr. single A leaves the process. If all of the nitrogen is consumed in the reaction, what is the molar flow rate of hydrogen in the ex- iting stream?

Explanation / Answer

According to the reaction,

1 mole N2 needs 3 moles H2 for complete reaction

On a per hour basis:

Amount of H2 entering through stream 1 = 0.05*400 = 20 lb mol

Amount of N2 entering through stream 1 = 0.95*400 = 380 lb mol

Mass of H2 entering from stream 2 = 0.08*31000 = 2480 lb

MW of H2 = 0.0044 lbm

So, moles of H2 coming from stream 2 = 2480/0.0044 = 563636.3 lb mol

So, total moles of H2 in system = 563636.3 + 20 = 563656.3 lb mol

Moles of H2 consumed in the reaction = 3*380 = 1140 lb mol

So, moles of H2 left in the exit stream, per hour = 562516.3 lb mol/hr

The answer you gave can come only when the density and volumetric flow rate of stream 2 is in units of grams, and not pounds.

Revert back if you have any queries.

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