Selective oxidation of hydrocarbons is a known method to produce alcohols. Howev

Question

Selective oxidation of hydrocarbons is a known method to produce alcohols. However, the alcohols react with the oxygen to produce aldehydes, and the latter react with oxygen to produce organic acids. A 60.0 mol/s stream consisting of 90.0% ethane and 10.0% nitrogen is mixed with a 39.0 mol/s air stream and fed into a catalytic reactor. The following reactions take place in the reactor:

The oxygen conversion is 80.0%, and the concentration of the ethanol in the product stream is three times that of the aldehyde and four times that of the acetic acid.

What is the fractional conversion of ethane?

What is the production rate of ethanol?

Explanation / Answer

let calculate for 1s:
mol of ethane = 0.9*60 = 54 mol
mol of O2 = 39 mol
mol of O2 reacted = 0.8*39 = 31.2 mol

Let number of moles of acetic acid formed be x mol
number of moles of aldehyde formed be 4/3x mol
number of moles of ethanol formed be 4x mol

Mol of O2 for x mol of acetic acid formation from C2H6= x mol
Mol of O2 for 4/3x mol of aldehyde formation from C2H6= 4/3 x mol
Mol of O2 for 4x mol of ethanol formation from C2H6= 4x/2 = 2x mol

x + 4/3 x + 2x = 31.2
x= 7.43 mol

Total mol of ethanol formed initailly= x+4/3 x + 4 = 33.2 mol
mol of ethane reacted = 33.2 mol
fractional conversion of ethane = 33.2/ 54 =0.615

Production arte of ethanol = 4xmol/s = 29.72 mol/s

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