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ID: 815371 • Letter: I
Question
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A fuel oil is composed of C10H22 and C12H26 in an unknown ratio (Note, there are no other significant components in the fuel oil). When the fuel oil is burned with 20% excess air the exhaust is tested and the ratio of water to CO2 is found to be 1.088. Please note that the reaction proceeds to 100% completion (i.e. there is no detectable unburned fuel oil or carbon monoxide). Determine the composition of the fuel oil.
Explanation / Answer
20% excess air ensures that both fuels burn completely to give carbon dioxide and water.
C10H22 + x1O2 --> 10CO2 + y1H2O
since all C is converted to carbon dioxide, coefiicient of CO2 is 10.
total reactant O2 = 2x1 and total product O2 = (10x2) + y1; so 2x1 = 20 + y1 ---(i)
similarly, total reactant H2 = 22 and total product H2 = 2y1; so 22 = 2y1 ---(ii)
from (ii), y1 = 11. Using this value of y1 in (i), we get: 2x1 = 20+11 so x1 = 15.5
Hence, C10H22 + 15.5O2 --> 10CO2 + 11H2O
C12H26 + x2O2 --> 12CO2 + y2H2O
Similarly, we get: x2 = 18.5, y2 = 13 (using the same formula as above)
C12H26 + 18.5O2 --> 12CO2 + 13H2O
Let us assume that the ratio of C10H22 and C12H26 is a:b.
so CO2 from C10H22 would be 10a, and H2O from C10H22 would be 11a
and CO2 from C12H26 would be 12b, and H2O from C12H26 would be 13b.
Now, total H2O/total CO2 = (11a+13b)/(10a+12b) = 1.088
so 11a + 13b = 1.088(10a+12b)
11a + 13b = 10.88a + 13.056b
0.12a = 0.056b
a/b = 0.4666 Ans