where the quantity in the solid \"out\" indicates that deposit 7 mg/min of GaAs.
ID: 926636 • Letter: W
Question
where the quantity in the solid "out" indicates that deposit 7 mg/min of GaAs. Also, we have assumed that the flow rate of the gas stream doesn't change significantly. Ammonia is used for fertilizer production and has been critical successful agriculture. A steady-state chemical cess is used to convert nitrogen (N_2) and hydrogen (H_2) to monia (NH_3) by the reaction What is the concentration of methane in the outlet gas? 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 lb_m/ft^3) enters the process at a volumetric flow rate of 31,000 ft^3/hr. A single stream leaves the process. If all of the nitrogen is consumed in the reaction, what is the molar flow rate of hydrogen in the exiting stream? N_2 + 3H_2 rightarrow 2NH_3Explanation / Answer
Given the composition of the entering stream-1 is 95 mol% N2 and 5 mol% H2.
The flow rate of stream-1 is 400 lbmol/hr
1 lbmol = 453.59 mol
Hence molar flow rate of stream -1 = 400 lbmol x (453.59 g / lbmol) /hr = 181436 mol / hour
Hence molar flow rate of N2 = 181436 mol / hour x (95/100) = 172364 mol N2/hour
Hence molar flow rate of H2 = 181436 mol / hour x (5/100) = 9072 mol H2/hour
Given the density of the entering stream containing pure H2 is 0.08 lbm / ft3
Volumetric flow rate of the entering stream-2(pure H2) is 31000 ft3 / hour
Hence mass flow rate of entering stream-2 (pure H2) = volumetric flow rate x density
= 31000 ft3 x (0.08 lbm / ft3 ) / hour
= 2480 lbm / hour H2 = 2480 lbm x (453.59 g/lbm) / hour = 1124903 g H2 /hour
Molecular mass of H2 = 2.0 g/mol
Hence molar flow rate of entering stream-2 (pure H2) = 1124903 g H2 x (1 mol H2 / 2 g H2) / hour
= 562451.6 mol H2 / hour
Hence total molar flow rate of H2 into the process = Flow rate of H2 in stream-1 + Flow rate of H2 in stream-2
= 9072 mol H2/hour + 562451.6 mol H2 / hour
= 571523.6 mol H2 / hour
The balanced chemical reaction is
N2 + 3 H2 ---- > 2 NH3
1mol, 3 mol ------ 2 mol
From the above balanced reaction it is clear that
1 mol of N2 reacts with 3 mol of H2
Hence 172364 mol N2/hour of N2 that will react with the moles of H2
= 172364 mol N2 x ( 3 mol H2 / 1 mol N2) / hour = 517092 mol H2/hr
Hence moles of H2 remain unreacted
= Total mol/hour - 517092 mol H2/hr
= 571523.6 mol H2 / hour - 517092 mol H2/hour
= 54431.6 mol H2/hour
The above unreacted H2 will flow thrugh the existing stream
Hence molar flow rate of H2 in the existing stream = 54431.6 mol H2/hour (answer)
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