A mixture of propane and butane is burned with pure oxygen. The combustion products contain 47.0 mole% H2O. After all the water is removed from the products, the residual gas contains 68.6 mole% CO2 and the balance O2 a. What is the mole percent of propane in the fuel? % b. It now turns out that the fuel mixture may contain not only propane and butane but also other hydrocarbons. The fuel does not contain oxygen. However, the dry combustion gases still contain 68.6% carbon dioxide. We wish to determine the elemental composition (carbon and hydrogen molar percentages) of the fuel feed. (Hint: Calculate the elemental compositions on an oxygen free basis). What is the mole% of carbon in the fuel?

Answer for part A is : = 11.11%

Answer for part B is: = 89.81%

Explanation:

Part a

The balanced reaction with unknown moles of O2

2C3H8 + 2C4H10 + XO2 = 18H2O + 14CO2 + YO2

Where, X = moles of O2 in reactants, Y = moles of O2 in products

moles of water formed = 47 mol

moles of carbon dioxide CO2 formed = 68.6 mol

Actual Mol ratio of CO2 : H2O = 68.6:47 = 1.459 : 1

From the stoichiometry of the reaction

moles of CO2 formed = 14 mol

Moles of water formed = 18 mol

Mol ratio of CO2 : H2O = 14: 18 = 0.777 : 1

The balanced combustion reaction of C3H8

C3H8 + 5O2 = 3CO2 + 4H2O

Stoichiometric ratio of CO2 : H2O = 3:4 = 0.75:1

The balanced combustion reaction of C4H10

2C4H10 + 14O2 = 8CO2 + 10H2O + O2

Stoichiometric ratio of CO2 : H2O = 8:10 = 0.8:1

The ratio of CO2 & H2O will be in between 0.75 & 0.80 with any propane and butane mixture

Actual Mol ratio of CO2 : H2O = 68.6:47 = 1.459 : 1

This is because of high C content to H combustion.

The balanced combustion reaction of C6H6

2C6H6 + 15O2 = 12CO2 + 6H2O

Stoichiometric ratio of CO2 : H2O = 12:6 = 2:1

The balanced combustion reaction of Naphthalene

C10H8 + 12O2 = 10CO2 + 4H2O

Stoichiometric ratio of CO2 : H2O = 10:4 = 2.5:1

Actual Mol ratio of CO2 : H2O = 68.6:47 = 1.459 : 1

Therefore the given gas mixture is impure.

Let us take a mixture for combustion

1 mol C3H8 + 1 mol C4H10 + 6 mol C6H6

The balanced combustion reaction

C3H8 + C4H10 + 6C6H6 + XO2 = 43CO2 + 27H2O + YO2

Where

X = moles of O2 in reactants

Y = moles of O2 in products

Stoichiometric ratio of CO2: H2O = 43: 27 = 1.59:1

From trial and error,

The combustion reaction we get

C3H8 + 2 C4H10 + 6 C6H6 + 63O2 = 47 CO2 + 32 H2O

Stoichiometric ratio of CO2: H2O = 47: 32 = 1.468:1

Actual Mol ratio of CO2 : H2O = 68.6:47 = 1.459 : 1

Mol % of propane = moles of propane / (moles of propane + moles of butane + moles of benzene)

= 1 / (1 + 2 + 6)

= 11.11%

Part b

Mass of C3H8 = moles x molecular weight

= 1 mol x 44 g/mol = 44 g

Mass of C4H10 = 2 mol x 58 g/mol = 116 g

Mass of benzene = 6 mol x 78 g/mol = 468 g

Total mass of mixture = 44 + 116 + 468 = 628 g

Mass % of C3H8 = mass of C3H8 x 100 / total mass

= 44*100/628 = 7.00%

Moles of carbon = moles of C in C3H8 + moles of C in C4H10 + moles of C in C6H6

= 3 + 8 + 36 = 47 mol

Mass of carbon = 47 mol x 12 g/mol = 564 g

% of C = mass of C x 100 / total mass

= 564*100/628 = 89.81%