US Hethane and oxygen react in the presence of a catalyst to form formaldehyde.

Question

US Hethane and oxygen react in the presence of a catalyst to form formaldehyde. In a parallel reaction, methane is oxidized to carbon dioxide and water The feed to the reactor contains equimolar-mounts of methane and oxygen. Assume a bass of 100.? mal feed/s. ?. How many depees of heedom reman, for the overall process? b. T fractional conversion of methane is 0.900 and the fractional yield of maldehyde is 0.870 mal HCHOmol?4fed. what is the compostion arths att stream, mens co, C. What is the selectivity (mol HCHOmal ??) ofthe process? mol HCHO/mol CO? Question Attempts: 0 of 3 used SAVE FOR LATER

Explanation / Answer

Part a

Degrees of freedom = number of unknowns + number of independent reactions - number of molecular species

Number of unknowns in the output = CH4, O2, HCHO, H2O, CO2 = 5

number of independent Reactions = 2

number of molecular species = 5

DOF = 5 + 2 - 5 = 2

2 degrees of freedom remain for the overall process

Part b

Moles of Feed = 100 mol/s

Moles of CH4 + moles of O2 = 100

Stoichiometric ratio of CH4 and O2 = 1 : 1

Initial Moles of CH4 = Initial moles of O2 = 50 mol

Fractional conversion of CH4 = 0.900

(Initial mol of CH4 - Final mol of CH4)/(Initial mol CH4) = 0.9

(50 - Final mol CH4) / 50 = 0.9

Final mol of CH4 = 5 mol/s

Theoretically Maximum moles of HCHO produced = 50 mol/s

Yield of HCHO = 0.870

Actual mol of HCHO produced / maximum mol HCHO = 0.87

Actual mol HCHO produced = 0.87 x 50 = 43.5 mol/s

For the first reaction

Extent of reaction = e1 = Actual mol HCHO produced

e1 = 43.5 mol/s

Final mol CH4 = Initial mol CH4 - e1 - e2

5 = 50 - 43.5 - e2

e2 = 1.5 mol/s

Final mol O2 = 50 - e1 - 2e2

= 50 - 43.5 - 2*1.5

= 3.5 mol/s

Final mol H2O = e1 + 2e2

= 43.5 + 2*1.5 = 46.5 mol/s

Final mol CO2 = e2 = 1.5 mol/s

Total Moles = 100 mol/s

Mol% O2 = 3.5%

Mol% HCHO = 43.5%

Mol% H2O = 46.5 %

Mol% CO2 = 1.5%

mol% CH4 = 5%

PART C

Mol HCHO / mol CO2 = 43.5/1.5 = 29

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