Acrylonitrile can be formed from propylene, ammonia and oxygen in the reaction 3

Question

Acrylonitrile can be formed from propylene, ammonia and oxygen in the reaction 3 To produce acrylonitrile stoichiometric ratios of propylene, ammonia, and oxygen are fed into the reactor The single pass conversion of this reaction is 31.3% of the propylene. The product of the reaction is fed into a separator where the unreacted propylene, ammonia, and oxygen are recycled back to the feed stream, and the acrylonitrile and water exit as products. and the aoryonirnie andwader e o reducamonia, and oxygen are recycled back to the feed stream. g6 mol/mir q7 mol C3He/min q8 mol NH3/min qg mol O2/min g0 mol/minV4 mol/min q5 mol/min Reactor Separator ql mol C3H6/min 2 mol NH3/min 3 mol O2/min q10 mol/min 12.2 kg C3H3N/min n mol H20/min

Explanation / Answer

1. No. of moles of C3H3N required = mass/molar mass = 12.2 x 1000g/ 53.06g/mol = 229.93 mol

If we take one mole of propylene we will get 1 mole of C3H3N, so no. of moles of propylene required = 229.93

Percentage of conversion = 31.3 %

if 100% propylene is fed in to the stream only 31.3 % only reacts other 68.7 % is recycled and again going back and joins the fresh feed.

So, each time we need to add only 31.3 % of propylene to the fresh feed, i.e.

q1 = 229.93 mol/min

q2 = 229.93 mol/min (No of moles of propylene and ammonia are same from the above equation)

q3 = 229.93 mol x (3/2) = 344.90 mol/min

So, q0 = q1 + q2 + q3 = 804.76 mol/min

2. 68.7 % of propylene is recycled

so, q7 = 229.93 x 68.7 / 31.3 = 504.67 mol/min

q8 = 504.67 mol/min (No of moles of propylene and ammonia are same from the above equation)

q9 = 504.67 mol/min x 3/2 = 757 mol/min

So, q6= q7 + q8 + q9 = 1766.34 mol/min

3. q0/q6 = 804.76 / 1766.34 = 0.456

(Please rate my answer if you find it helpful, good luck...)

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