4 replies to this topic

[aquaboy]

SO2 is oxidized to SO3 in a reactor. SO2 and 80% excess air are fed to the reactor. The reaction proceeds to a 45% SO2 conversion, and the production rate of SO3 is 100 lb/min. Air can be taken to be composed of 79 mol% N2 and 21 mol% O2.

 

a) Write the balanced chemical reaction equation.

 

b) Determine the extent of the reaction.

 

c) Calculate the amount of air fed to the process.

[breizh]

aquaboy ,

 what is the scope ? To do your homework ?

 let us know your difficulty and we will help .

Hope you understand

Breizh

Edited by breizh, 01 April 2015 - 10:04 PM.

[aquaboy]

Sorry, I'm new to this forum. The only part I'm having difficulty understanding is part C. If the O2 is supplied at 80% excess, does that mean, that it is supplied at 80% excess of the stoichiometric feed?

 

a)  2SO₂+ O₂ ---> 2SO₃

 

 

b) Mass balance on SO₃

 

N_SO3,OUT = 2*E

where E = extent

Solving yields,

E = 0.625 lbmol/min

 

c)  Find the amount of SO2 in feed then use stoichiometry to find O2 feed which then can be used to find excess air feed

f_{conversion, SO2} = 0.45 = N_SO3,CONSUMED / N_SO3,FEED = (2*E) / N_SO3,FEED

Solving the above gives,

N_SO3,FEED = 2.78 lbmol/min

 

N_{O2, FEED, NON-EXCESS} = 2.78lbmol/min *(1/2) = 1.39 lbmol/min

N_{AIR, FEED, NON-EXCESS} = 1.39/0.21 = 6.62 lbmol/min

N_{AIR, FEED, EXCESS} = (1.8) * 6.62 lbmol/min = 11.91 lbmol/min

 

Is the bolded answer above correct?

 

[aquaboy]

More importantly, is my method correct?

[breizh]

Consider my table , results  seems to be similar to yours

 

Breizh