Question

Map Ammonia is produced by reacting nitrogen and hydrogen according to the reaction below A feed stream consisting of 2.400% argon (by mole) and stoichiometric proportion of the reactants is fed into the system at the rate of 100 mol/min. The components enter a reactor, and then all ammonia is separated from the other components and leaves the process. The other components are recycled back to the feed stream, with a portion being purged from the system. The mole percentage of argon in the purge stream is 5.200%, and the conversion per pass in the reactor is 11.30%. What is the production rate of ammonia? Number mol NH,/ min What is the overall process fractional conversions of nitrogen and hydrogen? Number What is the flow rate of the recycle stream? Number mol/min

Please be clear with the answers and put the answers in boxes

Answer 1

Theory: We need to apply the mole balance in this problem.

Solution: Step 1:Feed = 100 mol/min

Argon in feed = 100*2.4/100 = 2.4 mol/min

Rest of the feed = 100-2.4 = 97.6

rest of the feed has N2 and H2 in stochiometric proportion.

That means N2 moles/H2 moles = 1/3

Lets assume N2 = X moles then H2 = 3X moles

so X +3X = 97.6 or X = 24.4 moles

so N2 moles = 24.4 moles

and H2 moles = 24.4*3 = 73.2 moles

Step 2: Argon balance:

Argon in inlet = Purge flow rate*mole fraction of Ar in purge

2.4 = P*5.2/100

or P = 46.15 mol/min

So purge flow rate is 46.15 mol/min

N2 & H2 moles in purge = 46.15*(100-5.2)/100 = 43.75 mol/min

unreaced N2 and H2 will also be in stochimetric proportion and since no N2 and H2 is going to product (NH3)

so N2/H2 mole ratio in purge = 1/3

lets assume N2 moles =Y then H2 moles = 3Y

Y+3Y = 43.75

so Y = 10.94 mol/min and 3Y = 32.82 mol/min

Step 3: Moles of N2 reacted = N2 moles to system -N2 moles oout of system

24.4-10.94 = 13.46 mol/min

moles of NH3 formed = 2*mole of N2 reacted = 2*13.46 = 26.92 mol/miin

Step 4: Overall convesrion of N2 = moles of N2 reacted/Nitrogen in system

= 13.46/24.4 = 0.5517

Step 5: Lets assume recycle flow is R mol/min

then N2 in recycle = R*mole fraction of N2 in purge = R*10.94/46.15 = 0.237R

total N2 moles to reactor = N2 from fresh feed +N2 from recycle = 24.4+0.237R

Conversion per pass = moles of N2 reacted/mole of N2 to reactor

0.113 = 13.46/(24.4+0.237R)

or 24.4+0.237R = 119.13

or R = 399.7 mol/min

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