A military aircraft burns a hydrocarbon fuel called decane. Its chemical symbol is: C10 H 22 .This fuel is burned with air, so:C10 H 22 + a O 2 + a (79/21) N 2 -> b CO 2 + c H 2 O + d N 2 + e O 2a) First consider stoichiometric conditions, which means that there is not any excess O 2 orany excess fuel in the products (on the right side of the arrow). e = 0.Balance the atoms on the right and left side of the equation and compute a, b, c and d.For stoichiometric conditions (e = 0) compute fs , which is the stoichiometric fuel-airratio. Hint: the above equation is based on one mole of fuel. The mass of fuel is [12(10)+ 1(22)] grams, since a carbon atom C has a mass of 12 grams per mole and a hydrogenatom H has a mass of 1 gram per mole. The mass of air will be the mass of O 2 plus themass of N 2 . Use the mass of O 2 to be 32 grams/mole, and the number of moles of O 2 is a.b) Now consider fuel lean conditions (e is greater than 0). Replace a with (a/phi) where a isthe value of a computed in part (a). Phi (f) is the fuel-air equivalence ratio, which is[f / f s ]; f is the fuel-air ratio = mass of fuel to the mass of air, and f s is the value of r forstoichiometric conditions (e = 0). - Compute b, c, d and e in terms of phi.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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A military aircraft burns a hydrocarbon fuel called decane. Its chemical symbol is: C10 H 22 .
This fuel is burned with air, so:
C₁₀ H ₂₂ + a O ₂ + a (79/21) N _{2 ->}b CO 2 + c H 2 O + d N 2 + e O 2
a) First consider stoichiometric conditions, which means that there is not any excess O 2 or
any excess fuel in the products (on the right side of the arrow). e = 0.
Balance the atoms on the right and left side of the equation and compute a, b, c and d.
For stoichiometric conditions (e = 0) compute f_s, which is the stoichiometric fuel-air
ratio. Hint: the above equation is based on one mole of fuel. The mass of fuel is [12(10)
+ 1(22)] grams, since a carbon atom C has a mass of 12 grams per mole and a hydrogen
atom H has a mass of 1 gram per mole. The mass of air will be the mass of O ₂ plus the
mass of N₂. Use the mass of O ₂to be 32 grams/mole, and the number of moles of O₂ is a.
b) Now consider fuel lean conditions (e is greater than 0). Replace a with (a/phi) where a is
the value of a computed in part (a). Phi (f) is the fuel-air equivalence ratio, which is
[f / f _s ]; f is the fuel-air ratio = mass of fuel to the mass of air, and f_s is the value of r for
stoichiometric conditions (e = 0).

- Compute b, c, d and e in terms of phi.

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