A student mixes a white solid and a clear liquid together. The mixture turning orange is evidence of a chemical reaction.

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

notka56 [123]

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.

8 0

3 years ago

Read 2 more answers

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine. CCl4(g) + (1/2)O

madam [21]

The question is incomplete, here is the complete question:

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine.

$CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c=4.4\times 10^9$ at 1,000 K

Calculate Kc for the reaction $2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)$

Answer: The value of $K_c'$ for the final reaction is $1.936\times 10^{19}$

Explanation:

The given chemical equations follows:

$CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c$

We need to calculate the equilibrium constant for the equation, which is:

$2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)$

As, the final reaction is the twice of the initial equation. So, the equilibrium constant for the final reaction will be the square of the initial equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c'=(K_c)^2$

We are given:

$K_c=4.4\times 10^9$

Putting values in above equation, we get:

$K_c'=(4.4\times 10^9)^2=1.936\times 10^{19}$

Hence, the value of $K_c'$ for the final reaction is $1.936\times 10^{19}$

3 0

3 years ago

If the chemist mistakenly makes 250 mL of solution instead of the 200 mL, what molar concentration of sodium nitrate will the ch

Evgesh-ka [11]

Answer:

0.120M is the concentration of the solution

Explanation:

Assuming the mass of sodium nitrate dissolved was 2.552g

Molar concentration is an unit of concentration widely used in chemsitry defined as the moles of solute (In this case NaNO3) in 1L of solution.

To find this question we must find the moles of NaNO3 in 2.552g. With this mass and the volume (250mL = 0.250L) we can find molar concentration as follows:

Moles NaNO3 -Molar mass: 84.99g/mol-

2.552g * (1mol / 84.99g) = 0.0300 moles NaNO3

Molar concentration:

0.0300 moles NaNO3 / 0.250L =

<h3>0.120M is the concentration of the solution</h3>

7 0

3 years ago

What do you call the procedure that helps you determine the volume of an irregularly shaped object, whole using a graduated cyli

ki77a [65]

The method is called the displacement method.

You place some water in the graduated cylinder and measure its volume.
Then you add your object and measure the new volume.
The difference between the two volumes is the volume of your object.

6 0

3 years ago

What is the mass of 80.0mL of ethanol at 20 degress celcius

Oxana [17]

Answer: What is the mass, in grams, of 135 mL of ethanol? d=0.789 g/mL - the ethanol density. V=135 mL - the volume of ethanol. m=0.789g/mL*135mL=106.515g ~ 106.5g- the mass of ethanol.

Explanation:

Hope this helps :)

4 0

3 years ago