Propylene (C3H6), ammonia (NH3), and oxygen (O2) react in a continuous, steady s

Question

Propylene (C3H6), ammonia (NH3), and oxygen (O2) react in a continuous, steady state reactor to form acrylonitrile (C3H3N) and water (H2O), as shown below. The feed to the reactor contains 150 lbmol/hr of C3H6, 150 lbmol/hr of NH3, and 100 lbmol/hr of O2 a. b. c. What is the limiting reactant? What is the percent excess of the two non-limiting reactants? If the conversion of the limiting reactant is 60% (mol), how much O2 comes out in the product stream? d. If the conversion of the limiting reactant is 60% (mol), what is the extent of e. If the conversion of the limiting reactant is 60% (mol), how much NH3 comes out f. If the conversion of the limiting reactant is 60% (mol), how much C3H3N exits in g. If the conversion of the limiting reactant is 60% (mol), how much C3H6 exits in h. If the conversion of the limiting reactant is 60% (mol), how much water is in the reaction? in the product stream? the product stream? the product stream? product stream?

Explanation / Answer

To produce acrylonitrile,

The balanced equation is

2 C3H6 + 2 NH3 + 3 O2 = 2 C3H3N + 6 H2O

Part a

From the stoichiometry of the reaction

Moles of C3H6 required to react with 150lbmol/hr NH3

= 2 lbmol C3H6 x 150 lbmol/hr NH3 / 2 lbmol NH3

= 150 lbmol/hr C3H6

Moles of O2 required to react with 150lbmol/hr NH3

= 3 lbmol O2 x 150 lbmol/hr NH3 / 2 lbmol NH3

= 225 lbmol/hr O2

But we have only 100 lbmol/hr of O2

It means O2 is limiting reactant.

Part b

Excess reactants are NH3 and C3H6

Moles of C3H6 required to react with 100 lbmol/hr O2

= 2 lbmol C3H6 x 100 lbmol/hr O2 / 3 lbmol O2

= 66.67 lbmol/hr C3H6

Similarly

Moles of NH3 required to react with 100 lbmol/hr O2

= 66.67 lbmol/hr NH3

% excess of C3H6 = % excess of NH3

= (Initial moles - reacted moles) *100/(initial moles)

=(150 - 66.67)*100/150

= 55.55%

Part C

If conversion of O2 = 60%

O2 reacted = 100*0.60 = 60 lbmol/hr

O2 in product stream = initial moles - reacted moles

= 100 - 60 = 40 lbmol/hr

Part d

Mol balance of O2

O2 in outlet stream = O2 in inlet stream - 3*Extent of reaction

40 = 100 - 3*Extent of reaction

Extent of reaction = (40-100)/(-3)

= 20 lbmol/hr

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