Question

A mixture containing 20 mole % butane, 35 mole % pentane and rest

Answer 1

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:

$Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB$

Whereas $y$ accounts for the fractions at the outlet distillate and $x$ for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

$0.9=\frac{y_bD}{z_bF}$

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:

$y_bD=0.9*z_bF=0.9*0.2*100mol=18mol$

The total distillate flow:

$y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol$

And the total bottoms flow:

$F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol$

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

$x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025$

Then, the molar fraction of pentane and hexane:

$x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422$

$x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553$

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.

Answer 2

Answer:

The percentage composition of the Bottoms is

- 2.46% Butane.

- 42.25% Pentane.

- 55 29% Hexane.

Explanation:

The feed is eventually separated into distillate and bottoms at the end of the day.

If the total number of moles in the feed = F, and we assume an initial basis of 100 mol

Total number of moles in the distillate = D

Total number of moles in the bottoms = B

Since distillation is a physical separation technique, with no chemical reaction expected,

The overall balance of the system,

F = 100 = B + D (eqn 1)

In the feed, there is 20 mole% of butane, 35 mole% of pentane and the rest, that is, 45 mole% of hexane.

Butane = 0.20F moles = 0 2×100 = 20 moles

Pentane = 0.35F moles = 0.35×100 = 35 moles

Hexane = 0.45F moles = 0.45×100 = 45 moles

In the distillate, there is 95 mole% of butane, 4 mole% of pentane and the rest, that is, 1 mole% is hexane

Butane = 0.95D moles

Pentane = 0.04D moles

Hexane = 0.01D moles

The composition of the Bottoms isn't known.

But, it is given that the distillate is expected to contain 90% of the butane in the feed

Component balance for butane, based on this information

Butane in the distillate = 90% of butane in feed

0.95D = 90% × 0.20F = 0.18F

Butane in the distillate = 0.95D = 0.18F

D = 0.1895F = 0.1895 × 100 = 18.95 moles

The composition of the distillate can then be rewritten as

Butane = 0.95D moles = 0.95×18.95 = 18 0025 moles

Pentane = 0.04D moles = 0.04×18.95 = 0.758 moles

Hexane = 0.01D moles = 0.01×18.95 = 0.1895 moles

From the overall balance,

100 = B + D

B = 100 - D = 100 - 18.95 = 81.05 moles

Hence, the amount of each component in the Bottoms now will be the amount in the feed minus the amount in the distillate

Butane

20 - 18.0025 = 1.9975 moles

Percent compositon = (1.9975/81.05) = 0.0246 = 2.46%

Pentane

35 - 0.758 = 34.242 moles

Percent composition = (34.242/81.05) = 0.4225 = 42.25%

Hexane

45 - 0.1895 = 44.8105 moles

Percent composition = (44.8105/81.05) = 0.5529 = 55.29%

Please note that, irrespective of the assumed basis for the total number of moles in the feed, the molar composition of the bottoms obtained, remains the same.

Hope this Helps!