Question

Propane (C₃H₈) is burned in an oxygen atmosphere with 10% deficit oxygen with respect to the stoichiometric requirement. Assuming no hydrocarbons in the products, the volume percentage of CO in the products is _______.

Answer 1

Stoichiometric Requirement:

\({C_3}{H_8} + a\;\left( {{O_2} + 3.76\;{N_2}} \right) \to bC{O_2} + c{H_2}O + 3.76a\;{N_2}\)

On comparing both side

H: 2c = 8 ⇒ c = 4

C: b = 3

O: 2a = 2b + c ⇒ a = 5

\({C_3}{H_8} + 5\;\left( {{O_2} + 3.76\;{N_2}} \right) \to 3C{O_2} + 4{H_2}O + 18.8\;{N_2}\)

Assuming no hydrocarbons in the products:

\({C_3}{H_8} + 5\;{O_2} \to 3C{O_2} + 4{H_2}O\)

∵ Now 10% deficit oxygen is used:

Oxygen = 0.9 × 5 = 4.5

\({C_3}{H_8} + 4.5\;{O_2} \to aC{O_2} + bCO + c{H_2}O\)

On comparing

H: 2c = 8 ⇒ c = 4

O: 2a + b + c = 9 ⇒ 2a + b = 5    ---(I)

C: a + b = 3     ---(II)

Eq. (I) and (II): a = 2, b = 1

\(\therefore {C_3}{H_8} + 4.5\;{O_2} \to 2C{O_2} + 1\;CO + 4{H_2}O\)

Volume % of CO in product:

\({V_{CO}} = \frac{b}{{a + b + c}} = \frac{1}{{2\; + \;1\; + \;4}} \times 100\;\% = 14.28\;\% \)