[iratus]

Hello

I am Erasmus student in a Chemical university in Prague and a teacher give us subjects that we are not able to solve (because we are not really students of chemistry)

This week he gave us this problem


what is the mixing ratio of sulfur dioxide in flue gas (% volume) ?
Fuel is heating oil with 2,5% (by mass) of sulfur.In the fuel to eath atom of carbon there are two atoms of hydrogen

Expect

The air excess λ =1,2
All carbon from the fuel is burnt to carbon dioxide
All hydrogen from fuel is burnt to water
All sulfur from the fuel is burnt to sulfur dioxide The water formed by combustion stays in gas state
3% of total sulfur is chemically bound in ash
the air to burner contains 20,9% oxygen ang 80,1% inert gasses (nitrogen)

so2= ppm



Can somebody help me

my first problem is what is the equation from the left side

thank you

[ARGOS++]

Dear iratus;

Welcome!

For your fist step  -  simply translate the words:

           In the fuel to eath (each) atom of carbon there are two atoms of hydrogen   

   C_nH_2n   +   O₂    ------->       CO₂       +       H₂O

Now you have to balance the equation and I hope you know how to do.

I hope to have been of help to you.
Good Luck!
                    ARGOS⁺⁺

[iratus]

And what about the sulfur

how can i find it's mass if i don't really know about the carbon 

[ARGOS++]

Dear iratus;

Do the sulphur separately in a second part and make finally the mass balance!

For the stoichiometry the n is not important, meaning that you can select the simplest n ( n = 1) as a representative/"surrogate".
Think about that the fuel oil is anyway a mixture of different n.

Good Luck!
                    ARGOS⁺⁺

[iratus]

sorry again but what do you mean

Do the sulphur separately in a second part

I think that is possible to find the mass of CH2 and then find the 2,5% of it's mass that is the sulfur's mass.

Is this correct

[ARGOS++]

Dear iratus;

Do it simpler!,  -  burn each separately and count the required oxygen together.
(That makes the calculation much simpler.)

For example:
Burn  1.00 mole CH₂ and declare it to be 97.5%; then you know how much Sulphur you have additionally to burn and build the sum.

Good Luck!
                    ARGOS⁺⁺

[iratus]

And what should i do with the air excess that is 1,2

Is this makes a different in my calculations ?

[iratus]

Do you know where i can find a solved problem similar to this so i can understand how should i do this one ??

[ARGOS++]

Dear iratus;

The excess of 20% air is very important!

Sorry I don’t know a similar example, not on the Forums, and not elsewhere.

I did the whole calculation and got a result in the neighbourhood of 0.1% v/v.

The recipe to find the mixture of the exhausting gas goes as follow:
     A.)    Find the required amount of oxygen for CH₂.
     B.)    Find the required amount of oxygen for Sulphur.
     C.)    Sum A.) and B.)  and convert it to required amount of air.
     D.)    Multiply C.) for the excess of 20%.
     E.)    Calculate from D.)  the total amount of exhausting gas.
     F.)    Calculate how much % v/v that B.) is from E.)

I hope to have been of help to you.
Good Luck!
                    ARGOS⁺⁺

[iratus]

Hello again and thank you for you for helping me so much.

I have some questions

I made A and B and the i multiple the oxygen from A with 97,5% and the oxygen from b with 2,5% and after i sum the total grams. Is this correct ?? Because if i only sum the oxygen what is the pupose of 2,5% of sulfur mass

Also in your steps in step E should i make a new reaction with the excess of air

[ARGOS++]

Dear iratus;

Let’s make it much simpler and more convenient for calculation!

Think first that you can use the Ideal Gas Law (IGL) by constant pressure:
From the IGL with p = const. it follows that volumes are proportional to number of moles!
That means that you can do the whole calculation in moles.

Start with A.) and give me a balanced reaction equation and then start with 1.00 mole CH₂ as I told you before. What we also need is the molecular weight of CH₂ and of sulphur for the calculation of B.).

Good Luck!
                    ARGOS⁺⁺

[iratus]

i made this

2CH2 +3O2 -> 2CO2 + 2H20

S2 + 2O2 -> 2SO2

WEIGHT OF O2 IN FIRST=28GR

WEIGHT OF O2 IN SECOND IS 64 GR

is this already correct ?

[ARGOS++]

Dear iratus;

Both reaction equations (RxN) are ok!

But the molecular weight of CH₂ is 14g/mole and that of Sulphur is 32 g/mole.

Let now start the calculation:
A.)    Assume that 97.5% fuel are 1.00 mole CH₂. How many mole Oxygen is required for it and how many mole total exhaust (CO₂ & H₂O) is produced?

Good Luck!
                    ARGOS⁺⁺

[iratus]

one mol on CH4 needs (3/2) mol oxygen  and produce 1 mol of CO2 one mol of H20

nOW

[ARGOS++]

Dear iratus;

Correct!  -  But why you are now dealing with CH₄?

So the results for A.) are:
     A.)   We need 1.50 mole Oxygen and produce 2.00 mole exhausts.

Now for B.):
   If  14.0g (= 1.0 mole) CH₂  correspond to 97.5%, how many gram Sulphur we have to burn too?
   And how many mole Sulphur would that be?

Good Luck!
                    ARGOS⁺⁺