ANS:C - 500

In a recycle process, some of the product (B) is recycled back into the reactor along with fresh feed (A). This is often done to increase conversion or yield. Given that the single pass once-through conversion of A to B is 20%, it means that out of the A that enters the reactor, only 20% is converted to B in a single pass. Let's denote the rate of fresh feed A as F (mols/hr) and the rate of recycle B as R (mols/hr). Since the single pass once-through conversion of A to B is 20%, the rate of production of B in a single pass is 20% of the rate of fresh feed A. So, the rate of production of B in a single pass = 0.20 * F However, since some of the B is recycled back into the reactor, the total rate of B entering the reactor in each pass is the sum of the B produced in a single pass and the recycled B: Total rate of B entering the reactor = (0.20 * F) + R Now, if the process is in a steady state, the rate of B leaving the reactor must be equal to the rate of B entering the reactor. In other words, the rate of B leaving the reactor equals the rate of fresh feed A multiplied by the desired conversion: Rate of B leaving the reactor = 0.20 * F Therefore, equating the rate of B entering the reactor with the rate of B leaving the reactor, we get: (0.20 * F) + R = 0.20 * F Solving for R: R = 0.20 * F - 0.20 * F = 0 This indicates that there's no need for recycle in this scenario because the desired conversion is achieved in a single pass.