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Chlorobenzene (C H Cl), an important solvent and intermediate in the production of many other chemicals, is produced by bubbling chlorine gas through liquid benzene in the presence of ferric chloride catalyst. In an undesired side reaction, the product is further chlorinated to dichlorobenzene, and in a third reaction the dichlorobenzene is chlorinated to trichlorobenzene. The feed to a chlorination reactor consists of essentially pure benzene and a technical grade of chlorine gas (98 wt% Cl , the balance gaseous impurities with an average molecular weight of 25.0). The liquid output from the reactor contains 65.0 wt% C H , 32.0 wt% C H Cl, 2.5 wt% C H Cl , and 0.5 wt% C H Cl . The gaseous output contains only HCl and the impurities that entered with the chlorine.

You wish to determine:
a. the percentage by which benzene is fed in excess
b. the fractional conversion of benzene
c. the fractional yield of monochlorobenzene
d. the mass ratio

Sagot :

Answer:

a) % of excess =  236 %

b) Conversion of benzene = 0.27

c) Yield = 83.8%

d) ratio = 0.265

Explanation:

Step 1: Data given

The liquid output from the reactor contains:

65.0 wt% C6H6

32.0 wt% C6H5Cl

2.5 wt% C6H4Cl2

0.5 wt% C6H3Cl3

Molar mass of C6H6 = 78.11 g/mol

Molar mass of C6H5Cl = 112.56 g/mol

Molar mass of C6H4Cl2 = 147.01 g/mol

Molar massof C6HCl3 = 181.45 g/mol

Step 2: The equations

General form: A + B → C +D

C6H6 + Cl2 → C6H5Cl + HCl

C6H5Cl + Cl2 → C6H4Cl2 + HCl

C6H4Cl2 + HCl → C6H3Cl3 + HCl

Step 3: Calculate number of moles

Lets suppose the total mass in the reactor is 100 grams

Number of moles =mass / molar mass

Number of moles C6H6 = 65.0 grams / 78.11 g/mol

Number of moles C6H6 = 0.832 moles C6H6

Number of moles C6H5Cl = 32.0 grams / 112.56 g/mol

Number of moles C6H5Cl = 0.284 moles C6H5Cl

Number of moles C6H4Cl2 = 2.5 grams / 147.01 g/mol

Number of moles C6H4Cl2 = 0.017 moles C6H4Cl2

Number of moles C6H3Cl3 = 0.50 grams / 181.45 g/mol

Number of moles C6H3Cl3 = 0.0028 moles C6H3Cl3

Total number of moles = moles C6H6 + moles C6H5Cl + moles C6H4Cl2 + moles C6H3Cl3

Total number of moles = 0.832 + 0.284 + 0.017 + 0.0028

Total number of moles = 1.14 moles

Liquid part: C6H6 + Cl2 → C6H5Cl + HCl

n1 → n3 + n4 ( HCl)

Gas part: n2 (Cl2) + n3 → nTotal (1.14 moles)

Moles of carbon = 6*n1 = 1.14 moles of A

moles of Hydrogen = 6*n1 = 6* moles of C6H6 + 5*moles of C6H5Cl + 4* moles of C6H4Cl2 + 3* moles of C6H3Cl3 + moles of HCl ( = n4)

moles of Hydrogen =6*n1 = 6*0.832 + 5*0.284 + 4*0.017 + 3*0.0028 + n4

n4 = 0.352 moles of D (HCl)

Moles of Chlorine: 2*n2 = 1* moles of HCl + 1* moles of C6H5Cl + 2* moles of C6H4Cl2 + 3* moles C6H3Cl3

Moles of Chlorine: 2*n2 = n4 + 1*0.284 moles + 2* 0.017 moles + 3* 0.0028 moles

Moles of chlorine = 0.339 moles B

Step 4: Calculate the excess of benzene

For 1 mol benzene we need 1 mol Cl2

For 0.339 moles Cl2 we have 0.339 moles benzene

% of excess = (total amount of moles - moles of benzene) / moles of benzene

% of excess = ((1.14 - 0.339)/ 0.339) * 100 % = 236 %

Step 5: Calculate the conversion of benzene

Conversion of benzene = (1.14 moles - 0.832 moles ) / 1.14 moles

Conversion of benzene = 0.27

Step 6: the fractional yield of monochlorobenzene

Yield = ( moles of monochlorobenzene produced / total moles) * 100%

Yield = (0.284 moles / 0.339 moles ) * 100 %

Yield = 83.8%

Step 7: Calculate mass of gass

Mass of gass (Cl2) = (Moles of Cl2 * molar mass of Cl2) * 98%

Mass of gass (Cl2) = (0.339 moles*70.9 g/mol ) * 0.98

MAss of gass = 23.6 grams

Step 8: Calculate mass of liquid

Mass of liquid = total amount of moles * molar mass of C6H6

Mass of liquid = 1.14 moles * 78.11 g/mol

Mass of liquid = 89.0 grams

Step 9: Calculate the mass ratio

Mass ratio = mass gass / mass liquid

Mass ratio = 23.6 grams / 89.0 grams

Mass ratio = 0.265

lhmarianateixeira

In this exercise we have to use the knowledge of chemistry to calculate the compositions of benzene, in this way we find that

a) 236 %

b) 0.27

c)  83.8%

d) 0.265

First, let's punctuate the information given in the text, which can be summarized in:

  • [tex]65.0 \ wt\% \ C6H6[/tex]
  • MM: [tex]C6H6 = 78.11 g/mol[/tex]

  • [tex]32.0 \ wt\% \ C6H5Cl[/tex]
  • MM:  [tex]C6H5Cl = 112.56 g/mol[/tex]

  • [tex]2.5 \ wt\% \ C6H4Cl2[/tex]
  • MM: [tex]C6H4Cl2 = 147.01 g/mol[/tex]

  • [tex]0.5 \ wt\% \ C6H3Cl3[/tex]
  • MM:  [tex]C6HCl3 = 181.45 g/mol[/tex]

Now writing the equations of the compositions given above, we can find that, general form:

[tex]A + B \rightarrow C +D[/tex]

equations of the compositions form:

  • [tex]C6H6 + Cl2 \rightarrow C6H5Cl + HCl[/tex]
  • [tex]C6H5Cl + Cl2 \rightarrow C6H4Cl2 + HCl[/tex]
  • [tex]C6H4Cl2 + HCl \rightarrow C6H3Cl3 + HCl[/tex]

from the informed data and equations we can now calculate the number of moles in each of the forms, so lets suppose the total mass in the reactor is 100 grams, in that case we have:

[tex]Number of moles = mass / molar\ mass[/tex]

  • C6H6: [tex]65.0 / 78.11 = 0.832 moles[/tex]
  • C6H5Cl:  [tex]32.0 / 112.56 = 0.284 moles[/tex]
  • C6H4Cl2: [tex]2.5 / 147.01 = 0.017 moles[/tex]
  • C6H3Cl3: [tex]0.50 / 181.45 = 0.0028 moles[/tex]
  • Total number of moles : [tex]0.832 + 0.284 + 0.017 + 0.0028 = 1.14 moles[/tex]

The liquid part can be rewritten as:

[tex]C6H6 + Cl2 \rightarrow C6H5Cl + HCl\\n1 \rightarrow n3 + n4 ( HCl)[/tex]

The gas part can be rewritten as:

[tex]n2 (Cl2) + n3 \rightarrow 1.14 moles[/tex]

  • Moles of Carbon:  [tex]1.14 moles[/tex]
  • Moles of Hydrogen: [tex]6*0.832 + 5*0.284 + 4*0.017 + 3*0.0028 + n4[/tex]
  • Moles of Chlorine: [tex]0.339 moles[/tex]

Now we can calculate the excess benzene from the equations described above, thus we find:

  • For 1 mol benzene we need 1 mol Cl2
  • For 0.339 moles Cl2 we have 0.339 moles benzene

[tex]\% \ of \ excess = (total \ amount \ of \ moles - moles \ of \ benzene) / moles \ of \ benzene[/tex]

[tex]\% \ of \ excess = ((1.14 - 0.339)/ 0.339) * 100 = 236 \%[/tex]

Calculate the conversion of benzene, we have:

[tex]Conversion \ of \ benzene = (1.14 moles - 0.832 moles ) / 1.14 moles = 0.27[/tex]

The fractional yield of monochlorobenzene, that can be say:

[tex]Yield = ( moles\ monochlorobenzene / total\ moles) * 100\\Yield = (0.284 moles / 0.339 moles ) * 100 \\Yield = 83.8%[/tex]

Calculate mass of gass, we have:

[tex]Mass of gass (Cl2) = (Moles of Cl2 * molar mass of Cl2) * 98\\Mass of gass (Cl2) = (0.339 moles*70.9 g/mol ) * 0.98\\Mass of gass = 23.6 grams[/tex]

Calculate mass of liquid, we have that:

[tex]Mass \ of \ liquid = total \ amount \ of \ moles * molar \ mass \ of \ C6H6\\Mass \ of \ liquid = 1.14 moles * 78.11 g/mol\\Mass \ of\ liquid = 89.0 grams[/tex]

Calculate the mass ratio, we can say that:

[tex]Mass \ ratio = mass \ gass / mass \ liquid\\Mass \ ratio = 23.6 grams / 89.0 grams\\Mass \ ratio = 0.265[/tex]

See more about chemistry at brainly.com/question/1882888

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