Question

A stream of 400 mol/min gases containing N₂, O₂, 1.85% H₂O and 13.56% CH₄ (all % mole basis) is combusted with 450 mol/min air containing N₂, O₂, 1.67% Ar and 0.72%  H₂O. The exiting gaseous product contains CO₂, CO, N₂, Ar, H₂O and O₂ (only). After the water is completely removed from the exiting gas, the composition is found to be 1.01% Ar and 4.39% CO.

i. What is the molar flowrate of H₂O is the product?

ii. What is the molar flowrate of CO in the product?

iii. What is the molar flowrate of CO₂ in the product?

Answer 1

Combustion of CH₄ will take place -

Moles of CH₄ entering = 0.1356 x 400 = 54.24 mol/min

Moles of O₂ required for complete combustion = 2 x 54.24 = 108.48 mol/min

Moles of O₂ required for partial combustion = 1.5 x 54.24 = 81.36 mol/min

Moles of H₂O entering = (0.0185 x 400) + (0.0072 x 450) = 10.64 mol/min

Let mole fraction of N₂ be x and y in two streams, then

Moles of O₂ entering = (0.8459-x) x 400 + (0.9761-y) x 450

Moles of H₂O produced by complete combustion = 2 x 54.24 = 108.48 mol/min

Moles of H₂O produced by partial combustion = 2 x 54.24 = 108.48 mol/min

Molar flowrate of H₂O in the product = 108.48 + 10.64 = 119.12 mol/min

Molar flowrate of CO in the product = 108.48 mol/min

Molar flowrate of CO₂ in the product = 108.48 mol/min