Question

A mixture of propane and butane is burned with pure oxygen. The combustion products contain 46.4 mole% H2O. After all the water is removed from the products, the residual gas contains 69.2 mole% CO2 and the balance O2.

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 69.2% 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

The combustion products are H₂O and CO₂ and there is also unreacted O₂. Let number of moles of H₂O, CO₂ and O₂ ​be x, y and z respectively.

Mole percentage of H₂O = 46.4%; hence mole percentage of (CO₂ + O₂) = 100 - 46.4 = 53.6 %

Let total number of moles be 100, then 46.4 moles are H₂O and 53.6 moles are (CO₂ + O₂).

Given, dry gaseous mixture contains 69.2 mole% CO₂ ​ or {y / (y+z) } *100 = 69

when y+z = 53.6, {y / 53.6} * 100 = 69 or y = (69 / 100) * 53.6 = 36.98

So if total number of moles is 100, then number of moles of CO₂ is 36.98

and number of moles of H₂O = 46.4

Ratio of number of moles of H₂O to number of moles of CO₂ is (x / y) = (46.4/ 36.98)

Since number of moles of H per mol of H₂O is 2 and number of moles of C per mole of CO₂ is 1,

the ratio of number of moles of H, N_H ​ to number of moles of C, N_C ​ is,

N_H ​/ N_C   = 2x / y = 2 * (x / y) =  2 * (46.4/ 36.98) = 2.509

Hence, N_H ​/ N_C   = 2.509

N_H ​ =​ 2.509 * N_C

Hydrocarbons contain only C and H.

_​So mole percent of C is,      { N_C / (N_C+ N_H​) } * 100 = { N_C / [ N_C ​+ (2.509 * N_C )] } * 100

= { N_C / 3.509 * N_C } * 100 = 100 / 3.509 = 28.5

Hence, mole percent of C is 28.5% and mole percent of hydrogen is 100-28.5 = 71.5%

ANSWER: Mole % of carbon in fuel is 28.5% ; (mole % of hydrogen in fuel is 71.5%)