Chemistry Chemical Process Analysis - Mole Flow Rate

[nobodyuknow]

Homework Statement

120 mol/min of Propane (C3H8) is burned in the presence of air (21% O2 and 79% N2) in a furnace, two reactions occur:
Complete Combustion: 67%, propane is burned to CO2 and H2O
Incomplete Combustion: 18%, propane is burned to CO and H2O

Oxygen is supplied at 70% excess.

Homework Equations

%Excess = (In - Req'd *100%)/Req'd

Complete Combustion: 2C3H8 + 10O2 -> 6O2 + 8H2O
Incomplete Combustion: 2C3H8 + 7O2 -> 6O + 8H2O

The Attempt at a Solution

So for the complete combustion...
120 Moles C3H8 * 5 Required Moles of O2 * 67% Required for Complete Combustion = 402 Moles
So for the incomplete combustion...
120 Moles C3H8 * 3.5 Required Moles of O2 * 18% Required for Complete Combustion = 75.6 Moles

Total of 477.6 Moles required
Using %Excess Formula, you get, In = 477.6 * 0.7 + 477.6 = 811.92 Moles...

HOWEVER, my lecturer simply had:

120moles/min * 1.7 * 5/1 = 1020 moles of O2.

I'm extremely clueless to as how he got that.

Thanks

 

[mallinath]

1. State what is "Kauzmann Paradox" and provide your own explanation (or solution) for the paradox and derive a unique explanation using different approaches.

2. If you observe a phase transition phenomenon in your own experiment, how to determine the whether the transition is binodal or spinodal?

3. Explain the origin of the friction when an elastic rubber ball is bouncing on a ground made of a crystalline material using an entropy analysis.

Thanks in Advance
Mallinath

 

[ldc3]

Homework Statement

120 mol/min of Propane (C3H8) is burned in the presence of air (21% O2 and 79% N2) in a furnace, two reactions occur:
Complete Combustion: 67%, propane is burned to CO2 and H2O
Incomplete Combustion: 18%, propane is burned to CO and H2O

Oxygen is supplied at 70% excess.

Homework Equations

%Excess = (In - Req'd *100%)/Req'd

Complete Combustion: 2C3H8 + 10O2 -> 6O2 + 8H2O
Incomplete Combustion: 2C3H8 + 7O2 -> 6O + 8H2O

The Attempt at a Solution

So for the complete combustion...
120 Moles C3H8 * 5 Required Moles of O2 * 67% Required for Complete Combustion = 402 Moles
So for the incomplete combustion...
120 Moles C3H8 * 3.5 Required Moles of O2 * 18% Required for Complete Combustion = 75.6 Moles

Total of 477.6 Moles required
Using %Excess Formula, you get, In = 477.6 * 0.7 + 477.6 = 811.92 Moles...

HOWEVER, my lecturer simply had:

120moles/min * 1.7 * 5/1 = 1020 moles of O2.

I'm extremely clueless to as how he got that.

Thanks

I think that you missed a very important bit of information.

Complete Combustion: 67%, propane is burned to CO2 and H2O
Incomplete Combustion: 18%, propane is burned to CO and H2O

This only accounts for 85% of the propane, so your calculations exclude the oxygen used in 15% of the reactions.

The professor took the flow rate of the propane, multiplied by the 70% excess (1.7) and then by the molar ratio (from the balanced chemical equation for complete combustion) (5 moles of O₂ over 1 mole of propane).

The answer should be 1020 moles of O₂ per minute.