How many liters of 1.2M solution can be prepared with 0.50 moles of C6H12O6
1 answer:
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Answer:
2.5 % butane, 42.2 % pentane and 55.3 % hexane
Explanation:
Hello,
In this case, the mass balance for each substance is given by:
Whereas accounts for the fractions at the outlet distillate and
for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:
So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:
The total distillate flow:
And the total bottoms flow:
Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:
Then, the molar fraction of pentane and hexane:
Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.
NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.
Regards.