Question

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 1

Answer:

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%