In some areas of the country, all the gasoline sold must contain at least 5 perc
ID: 1049805 • Letter: I
Question
In some areas of the country, all the gasoline sold must contain at least 5 percent by weight oxygen. The gasoline companies meet this requirement by blending into their ordinary gasoline with enough ethanol (C_2H_5OH) to meet the required 5 percent by weight oxygen. If we assume that their ordinary gasoline is the equivalent of C_8H_17 and that they plan to meet this oxygen requirement by blending enough ethanol with their ordinary gasoline, what weight percent ethanol must there be in the final blend? Assume that the densities of both ordinary gasoline and ethanol are the same, about 0.75 g/cm^3 What is the stoichiometric A/F ratio (lb/lb) for this mixture? Why is oxygen added to gasoline?Explanation / Answer
Ans:
Reaction of gasoline (C8H17) with O2 is as:
4 C8H17 + 49 O2 = 32 CO2 + 34 H2O
MW of C8H17 = 452 gm/mole
MW of O2= 32 gm/mole
Requirement of O2 per mole of C8H17 as per balance reaction is 49 mole or 49 x 32 = 1568 gm
Requirement of O2 for blending is 5% or = 5/100 x 1568 = 78.4 gm
MW of Ethanol= 46 gm/mole
46 gm of ethanol contains 16 gm of oxygen.
Therefore for 78.4 gm Oxygen requirement of ethanol = 46/16 x 78.4 gm = 225.4 gm.
Total molar weight of C8H17 and Oxygen = 1568+ 225.4 = 1646.4
Thus weight % of ethanol= 225.4 x100/1646.4= 13.69 %
Oxygen is usually employed as gasoline additives to reduce carbon monoxide and soot that is created during the burning of the fuel.
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