1 pts D Question 6 C3H6 + CO + H2-C4H8O In an existing process, 180 kgmol/h of C

Question

1 pts D Question 6 C3H6 + CO + H2-C4H8O In an existing process, 180 kgmol/h of C,H, (P) is mixed with 420 kgmol/h of a mixture containing 50% CO (C) and 50% H2 (H) and with a recycle stream containing propylene and then fed to a reactor, 30% of the propylene fed to the reactor is consumed. The reactor effluent is sent to a separator. The desired product butanal (B) is removed in one stream, unreacted CO and H, are removed in a second stream, and unreacted P is recovered and recycled. Reactor Separator P, C, -?,? ?. ?, ? 7 Suppose you have chosen the input-output diagram as your system. What are the material balance equations? C2- C(cons)- C6 P1- P(cons)

Explanation / Answer

Both H2 and CO are entering through 2 and leaving from 6 after being consumed in the reaction

Hence H(2)-H (consumed)= H(6) and C(2)-C( consumed)= C(6)

What ever C4H8O entering the separator will have to leave, B4= B7.

B generated is removed from separator through B7. B( gen)= B7.

From the reaction stoichiometry, 1 mole of P gives rise to 1 mole of B7, Hence P1= B7.

Fresh feed is added to ensure what ever is consumed during the course of reaction is replaced.

Hence P(1)= P ( consumed) and hence P1=0 is also not correct

Since P4 is 0.7*(Fresh + recycle), this is not correct. C2 entering is consumed and then withdrawn. Hence C2 =C6 is not correct.

Writing balance of C3H6 across the reactor = {P( from5)+ 180(Fresh feed)}*0.3= moles of C3H6 consumed = moles of C4H8O formed =180

P (5)= 420 kmoles/hr

Hence 1,2,3,6,7 and 8 are correct statements.

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