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.

Answer 1

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 = 2_x1 and total product O₂ = (10x₂) + y_1; so 2x₁ = 20 + y₁ ---(i)

similarly, total reactant H₂ = 22 and total product H₂ = 2y_1; 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