Feed Ratio and Mole fraction outut

In summary: Then we could calculate the feed ratio of m3 air/kg oil by dividing the moles of air used by the moles of oil. In summary, the chemical analysis of the fuel shows that it contains 84.60 wt% C, 11.12 wt% H, 1.95 wt% S, and 2.33 wt% inerts. When the fuel is consumed in a steam generating furnace with 33% excess theoretical air, the sulfur and hydrogen are oxidized to SO2 and H2O while 95% of the C forms CO2 and the remaining 5% goes
  • #1
ScienceChem
7
0

Homework Statement


The chemical analysis of a fuel is determined to be: 84.60 wt% C, 11.12 wt% H, 1.95 wt % S and the balance, inerts. The fuel is consumed in a steam generating furnace with 33% excess theoretical air. The air is preheated to 190degreesC and enters the furnace at a gauge pressure of 225 mm Hg. The sulfur and hydrogen are oxidized to SO2 and H2O ; 95% of the C forms CO2 and the balance goes to CO. (Assume ideal gas)
Calculate the feed ratio of (m3 air)/ (kg oil) and the output on a mole fraction basis, dry and wet.


2. The attempt at a solution
Just going to give this a try as to how I should attempt this.
Have the fuel - 84.60% c
- 11.12% H
- 1.95% S
- 2.33% Inerts
33% excess air, thus we have 77% air consumed which is preheated to 190 degrees C at 255 mm HG
So we know the S + H -> SO2 and H2O while 95% C forms CO2 and 5% C forms CO

Could we say that we have a theoretical 100 moles which can then be divided down to 84.60 moles C and so forth.
We could then assume that we have 100 g of material and knowing that we have 77% of the air consumed work towards finding the ratio of this 77% to the 100g? I am slightly confused...
 
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  • #2
ScienceChem said:
Could we say that we have a theoretical 100 moles which can then be divided down to 84.60 moles C and so forth.
Is 84.60 wt% C the same as saying that for 100 moles of fuel 84.60 moles are C?

ScienceChem said:
We could then assume that we have 100 g of material and knowing that we have 77% of the air consumed work towards finding the ratio of this 77% to the 100g?
The feed ratio should be per kg fuel. Why not assume you have 1 kg of fuel? How much oxygen would that correspond to? And how much air in theory?

And you should probably start by writing down all the chemical reactions down to figure out the stoichiometry.
 
  • #3
Wait I just realized now that you pointed that out... we are dealing with wt% so it is the mass percent and would not be on the mole ratio right?
So if we assumed 1kg of fuel we know that 77% of the air is consumed in theory.
We don't know what the starting mixture is of the fuel we just know that it contains C, S, H, and inerts.
So to work out chemical equations how would that work?
Would we say S + H2 + 2O2 -> SO2 + 2H2O and 3C + 2O2 -> CO2 + 2CO
Could we look at it in terms of input and output from here knowing the %C formed.
So if we assume and input of 1kg of the fuel
Our input would then be
846g of C
111.2g of H
19.5g of S
23.3g of inerts
Our output would then be
SO2
H2O
CO2
CO
Of these we know that 95% of the C forms CO2 and 5% forms CO. Could we then say that 803.7g of CO2 is formed and 42.3g of CO is produced.
But then where would I go from here. Calculate moles and determine the amount of O2 originally reacted with the carbon?
 
  • #4
ScienceChem said:
Wait I just realized now that you pointed that out... we are dealing with wt% so it is the mass percent and would not be on the mole ratio right?
Right. There are many ways of attacking this kind of problem, and I think that I would start by converting these mass ratios into mole ratios. [Edit: or rather moles per kg, because of the presence of inerts)]

ScienceChem said:
So to work out chemical equations how would that work?
Would we say S + H2 + 2O2 -> SO2 + 2H2O and 3C + 2O2 -> CO2 + 2CO
I was a bit sloppy in my wording. I meant something like what you did there. You will need to know for each mole of each atom how many moles of O2 you need.

ScienceChem said:
Of these we know that 95% of the C forms CO2 and 5% forms CO. Could we then say that 803.7g of CO2 is formed and 42.3g of CO is produced.
But then where would I go from here. Calculate moles and determine the amount of O2 originally reacted with the carbon?
That's another way to proceed, instead of starting with mole ratios. But be careful: does having 846g of C and forming 95% CO2 mean that you form 803.7g of CO2?
 
  • #5
If we start with mole ratios I would then have:
846 g of C / 12.01 g/mol = 70.44 moles C
111.2g of H / 1.00 g/mol = 111.2 moles H
19.5g of S / 32.07 g/mol = 0.61 moles S
23.3g of inerts -> don't know moles because we don't know what the inerts contain.

the 846 g of C or 70.44 moles would mean that within the CO2 we have 803.7 g of C, but we don't know how much O2.

So if we look at the equations then
S + H2 + 2O2 -> SO2 + 2H2O
3C + 2O2 -> CO2 + 2CO

Could we say that we need 4O2 molecules to react in order to get the products described. I am not sure where to proceed from here though. We know the S and H in the first equation so we could determine using a mole ratio the amount of moles of O2 on a 1:2 ratio. Same with the 2nd equation on a 3:2 ratio from C. Then we could determine the total moles of O2 reacted and use that to find the volume of air? Could we then use PV=nRT since we know the temp is 190degreesC and enters the furnace at a gauge pressure of 225 mm Hg. Now that we have moles we solve for volume? And since we initally started with a basis of 1kg of fuel, that would represent the amount of O2 in 1 kg oil?
Then the outputs we could again use ratios to determine the amount produced from the inputs?
 
  • #6
ScienceChem said:
If we start with mole ratios I would then have:
846 g of C / 12.01 g/mol = 70.44 moles C
111.2g of H / 1.00 g/mol = 111.2 moles H
19.5g of S / 32.07 g/mol = 0.61 moles S
23.3g of inerts -> don't know moles because we don't know what the inerts contain.
Good.

ScienceChem said:
the 846 g of C or 70.44 moles would mean that within the CO2 we have 803.7 g of C, but we don't know how much O2.
I don't understand what you are trying to do here. I suggest you keep everything in moles for now.

ScienceChem said:
So if we look at the equations then
S + H2 + 2O2 -> SO2 + 2H2O
3C + 2O2 -> CO2 + 2CO
Separate all the "reactions":
S + O2 → SO2
2H + 1/2 O2 → H2O
C + O2 →CO2
C + O →CO

Can you figure out how many moles of O2 are needed per kg fuel?

ScienceChem said:
Then we could determine the total moles of O2 reacted and use that to find the volume of air? Could we then use PV=nRT since we know the temp is 190degreesC and enters the furnace at a gauge pressure of 225 mm Hg. Now that we have moles we solve for volume? And since we initally started with a basis of 1kg of fuel, that would represent the amount of O2 in 1 kg oil?
Then the outputs we could again use ratios to determine the amount produced from the inputs?
Did you mean "that would represent the amount of O2 needed for 1 kg oil"? Otherwise, that sounds good! And don't forget that you are given the gauge pressure and that there is excess air.
 
  • #7
So if we have:
S + O2 → SO2 S=0.61 moles thus 1:1 ratio so 0.61 moles of O2
2H + 1/2 O2 → H2O H=111.2 moles thus 2:1/2 ratio so 27.8 moles O2
C + O2 →CO2 here we know 95% of the C went to CO2 and 1:1 ratio so 66.92 moles O2
C + O →CO here 5% of C went to CO so 1:1 ratio so 3.52 moles O2
So total we know there is 0.61 + 27.8 + 66.92 + 3.52 =98.85 total moles of O2 per kg of fuel

PV=nRT
V=nRT/P
V=(98.85moles)*(0.0821 liter·atm/mol·K)*(463K)/ (0.296 atm)
V=12694.3 L
V=12.69m^3 O2 per kg of oil

Does that seem right?
 
  • #8
ScienceChem said:
C + O →CO here 5% of C went to CO so 1:1 ratio so 3.52 moles O2
You made an error here.

ScienceChem said:
V=(98.85moles)*(0.0821 liter·atm/mol·K)*(463K)/ (0.296 atm)
What does "gauge pressure" mean?
 
  • #9
You forgot to account for the nitrogen in the air, and you forgot to multiply by 4/3 to account for the excess air.

Chet
 

Related to Feed Ratio and Mole fraction outut

1. What is feed ratio and how is it calculated?

Feed ratio is a measure of the relative amounts of two substances in a mixture. It is calculated by dividing the mass or volume of one substance by the mass or volume of the other substance in the mixture.

2. How does feed ratio affect the output of a chemical reaction?

The feed ratio can significantly impact the output of a chemical reaction. A higher feed ratio can result in a higher yield of the desired product, while a lower feed ratio can result in a lower yield. An optimal feed ratio is often necessary for the most efficient and effective reaction.

3. What is mole fraction output and how is it related to feed ratio?

Mole fraction output is the ratio of the moles of a specific substance in the output to the total moles of all substances in the output. It is directly related to feed ratio, as the feed ratio determines the initial amount of each substance in the mixture, which in turn affects the mole fraction output.

4. How can feed ratio and mole fraction output be used to optimize a chemical reaction?

By carefully adjusting the feed ratio and monitoring the resulting mole fraction output, scientists and engineers can optimize a chemical reaction to achieve the desired yield of the desired product. This involves finding the optimal balance between the amounts of reactants and products in the reaction mixture.

5. Can feed ratio and mole fraction output be used to determine the purity of a substance?

Yes, feed ratio and mole fraction output can be used to determine the purity of a substance. By comparing the expected mole fraction output of a pure substance to the actual mole fraction output of a sample, scientists can calculate the purity of the sample. A higher purity will result in a closer match between the expected and actual mole fraction output.

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