Question

An acetylene -oxygen mixture, which contains 25 mol % acetylene (C2H2) and 75% mol oxygen is fed at the rate of 1g-mol/min to a reactor where acetylene reacts completely with oxygen to form CO2 and water vapor according to the stoichiometry C2H2+2.5 O2 ----> 2CO2 +H2O.

Determine the molar flow rates and mole fractions of the various species in the reactor effluent.

Answer 1

The balanced reaction

C2H2 + 2.5 O2 ----> 2CO2 + H2O

At input

Moles of C2H2 = 1 x 0.25 = 0.25 mol/min

Moles of O2 = 1 x 0.75 = 0.75 mol/min

From the stoichiometry of the reaction

1 Mol C2H2 required = 2.5 mol O2

0.25 mol C2H2 required = 2.5 x 0.25 = 0.625 mol O2

But we have more moles of O2 than required

Limiting reactant = C2H2

Excess reactant = O2

At outlet

Moles of CO2 = 2 x 0.25 = 0.50 mol/min

Moles of H2O = 0.25 mol/min

Moles of O2 = Initial - reacted = 0.75 - 0.625 = 0.125 mol/min

Total Moles = 0.50 + 0.25 + 0.125 = 0.875 mol/min

Mol fraction of CO2 = Moles of CO2 / total moles

= 0.50/0.875 = 0.571

Mol fraction of H2O = 0.25/0.875 = 0.286

Mol fraction of O2 = 0.125/0.875 = 0.143