Combustion of Toluene with 30% excess air (Himmelblau)

[Bernardo32Rey]

Homework Statement

Toluene, C7H8, is burned with 30% excess air. A bad burner cause 15% of the carbon to
form soot (pure C) deposited on the walls of the furnace, what is the Orsat analysis of the
gases leaving the furnace?

Homework Equations

C7H8 + 9 O2 => 7 CO2 + 4 H2O
30% excess of air

What would be the theoretical toluene on which the excess air is based?

The Attempt at a Solution

I get lost with the "A bad burner cause 15% of the carbon to
form soot (pure C) deposited on the walls of the furnace".
If it weren't for that, I could solve this problem easily!

I need help with the 15% carbon to soot.

 

[Borek]

Of each mole of carbon present in the toluene, how many moles will reach the apparatus?

 

[Bernardo32Rey]

Of each mole of carbon present in the toluene, how many moles will reach the apparatus?

Hi, Borek.

Ok, supposing there are an initial 100 kg-mol of toluene, C7H8.
Each mole has 7 moles of carbon, C.
Then we have 700 kg-mol of C.
The 15% of that is 105 kg-mol.

That means that 85% of toluene will react with oxygen.
Buuuuttt... what happens with the other hydrogen in toluene, C7H8, since Carbon formed soot (pure C), does the hydrogen form H2?

Is the 30% of excess air based on the 85 kg-mol of toluene that do react with oxygen or the original amount?
That's my doubt! Thanks for your time.

 

[Borek]

You are over complicating it. You are not told anything about hydrogen so simply assume it is all converted to H₂O.

Excess air doesn't matter - your analysis detects only gaseous combustion products.

 

[Bernardo32Rey]

You are over complicating it. You are not told anything about hydrogen so simply assume it is all converted to H₂O.

Excess air doesn't matter - your analysis detects only gaseous combustion products.

Wow, wow, wow!
I never thought of that!

Yeah, obviously, the rest of hydrogen forms water with the oxygen of air.

I need to do the calculations.

Thanks!