HI, can someone please help me. This is a Material Balance engineering question.

Question

HI, can someone please help me. This is a Material Balance engineering question. I am looking for the material balance for the follow. This is the 3rd time I have asked this question and this "so-called" expert "Anonymous" continues to answer it with coping and pasting an article from wikipedia?? WTH Chegg??. PLEASE someone else help me! I do NOT need an essay about Ethanol. I need help with this problem.

1000 kg of Ethanol (CH3CH2OH) is dehydrogenated in air over a catalyst and the following reactions take place:

CH3CH2OH --> CH3CHO +H2

2CH3CH2OH + 3O2 --> 4CO2 + 6H2

the product, CH3CHO (acetaldehyde), leaves as a liquid.

MW of ethanol = 46 g/mol MW of acetaldehyde= 44g/mol

Question:

1. If 8 kmol of CO2 leaves the reactor, how many kg acetaldehyde (CH3CHO) is produced per kg of ethanol fed?

2. If the product gas contains 10 kmol O2, how much air was fed?

3. What is the total moles and molar composition of the output gas?

Please let me know how to do the steps and what you are doing.. Please.

Explanation / Answer

CH3CH2OHCH3CHO+H2

2CH3CH2OH+3O24CO2+6H2

ADD UP

3CH3CH2OH+3O2 CH3CHO +4CO2+7H2 (OVERALL RXN)            

1) moles of ethanol fed=1000kg/46g/mol=21.7 kmol

4 moles of CO2 is produced when 3 moles of ethanol burns

8kmol of CO2 is produced when by ethanol=3/4*8 kmol=6kmol of ethanol

Relate moles of ethanol burned to moles of acetaldehyde produced

3 mol ethanol produces=1 mol acetaldehyde

6kmol ethanol produces=1/3*6kmol acetaldehyde=2k mol acetaldehyde.

In kg, 2 kmol acetaldehyde.=2kmol*44g/mol=88kg.

2) product gas=10kmol of O2 (excess)

Amount of o2 used up in burning 3 mol ethanol=3 mol

So amount o2 used up in burning 21.7k mol ethanol(total)=3/3*21.7 k mol =21.7kmol(equivalent amount)

So total O2 fed = 21.7 kmol+excess =21.7+10=31.7 kmol o2

[As by volume air=21% o2 and 79% N2

So 1 mol O2=79/21=3.76 mol N2

1 kmol o2=3.76kmol N2

Total air =1 kmol O2 +3.76 kmol N2=4.76 Kmol

Or , amount of air=4.76 Kmol per 1 kmol of O2]

Total air fed =31.7 kmol o2*4.76 kmol air/1 kmol o2=150.9kmol air

3) total air in output gas=150.9kmol air

Moles of O2 in output gas=31.7 kmol

Moles of CO2 in output gas=8kmol

%O2 =31.7/150.9*100=21.1%

%CO2=8/150.9*100=5.3%

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