Final answer:
To find the percent of benzene in the feed that leaves as vapor in an equilibrium flash, we can use a graph and the given composition of the liquid. By calculating the mole fraction of benzene and the mole fraction of toluene leaving as liquid, and using the provided graph, we can determine that approximately 17% of benzene in the feed leaves as vapor.
Step-by-step explanation:
To solve this problem, we can use the concept of equilibrium flash using a graph. We are given that the liquid contains 60 mol% toluene and 40 mol% benzene, and we need to find the percent of benzene in the feed that leaves as vapor. First, we need to calculate the mole fraction of benzene from the given composition. Assuming a total of 100 moles of the solution, we have 60 moles of toluene and 40 moles of benzene. The mole fraction of benzene (x) is given by:
x = moles of benzene / total moles of solution = 40 / 100 = 0.4
Next, we need to find the mole fraction of toluene that leaves as liquid. We are given that 90% of the toluene leaving in the feed leaves as liquid. So, the mole fraction of toluene leaving as liquid (y) is given by:
y = moles of toluene leaving as liquid / total moles of feed = 0.9
Now, using the graph provided, we can plot the points (x,y) and draw a straight line through them. The point on the graph where this line intersects the x-axis represents the percent of benzene in the feed that leaves as vapor. From the graph, we can see that the corresponding percent of benzene in the feed is approximately 17%. Therefore, 17% of benzene in the feed leaves as vapor.