The feed to the process (F) c0ntains \'A\' and \'B\' in stoichiomectric proporti

Question

The feed to the process (F) c0ntains 'A' and 'B' in stoichiomectric proportions. It also contains 0.5 mol% of impurity 'I'

The mixture (x) entering the reactor (i.e feed and recycle) contains 'A' and 'B' in stociometric proportions, but the level of impurity 'I' must not exceed 4 mol%.

In the reactor, the reaction goes to 60% completion. Pure product (C) is removed completely in the separator. The remainder of the reaction products (i.e. unconverted 'A' and 'B' along with any impurity) is recycled. In order not to exceed the maximum quantity of impurity in the reactor feed, some of the recycle is purged (P), the remainder (R) going back to the reactor.

On the basis of 100kmol of 'A' and 'B' entering the reactor at (x) carry out mass balances to find the flows of

(i) Feed (F)

(ii) Recycle (R)

(iii) Purge (P)

Explanation / Answer

100kmol of A and B is entering in the reactor i.e from the chemical equation 25 kmol of A and 75 kmol of B entered in the reactor. So we get 50 mol of C for 100% reaction so for 60% reaction we get 30k mol of C if we consider the impurities for 25k mol of A we get 25 x 0.5=12.5 kmol impuritiesfor 100% ; k7.5 mol impurities for 60%,so from 30 kmol of C remove 7.5 kmol then pure C=30-7.5 =22.5 C So feed F= 25 kmol of A+ 75k mol B Recycle =R=40% of A and B= 10 kmol of A +30 kmol of B Purge P= 7.5k mol impurities Product C= 22.5 kmol

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