[cseil]

Hi everyone, here's my problem.

A gas mixture contained 51.3% of H2S, 48.74% of CO2. 1750mL of this mixture (21°C, 760mmHg) passed through a tubular oven at 350°C and then they were cooled down rapidly. The gas outlet passed through a tube containing CaCl2 and the weight of CaCl2 increased of 34.7mg. Calculate the Kp at 350°C for the reaction:

H2S(g) + CO2(g) = COS + H2O

Firstly I calculated the volume of H2S and CO2 in the 1750mL taken for the reaction.
Easily: 1075*0.513=551.47mL (H2S) and 523.52mL (CO2).

I know that [tex]n=\frac{pV}{RT}[/tex] so I can calculate the moles of H2S and CO2 at the beginning (I know that T=294K, 21°C and p=1atm).
So: n(H2S)=0.0228 mol, n(CO2)=0.0217 mol.

So at the equilibrium I have that:
H2S = 0.0228-x
CO2= 0.0217-x
H2O=COS=x

I know the x, because it's the weight increase of CaCl2 caused by the water adsorbed so 0.0347/18=1.92x10-3 mol.

The moles at equilibrium are: (0.0228-x)+(0.0217-x)+2x = .04454 mol.
So I can calculate the molar fraction of every component.

x(H2S)=(0.0228-x)/ntot
etc etc

x(H2S)=0.469
x(CO2)=0.444
x(H2O)=0.0431
x(COS)=0.0431

[tex]Kp = P^{v_i} \times \frac{x_{COS}x_{H2O}}{x_{H2S}x_{CO2}} = 0.00892.[/tex]
The member P^vi is 1.


After all of this.. the answer is not correct. It should be 0.00315.
And my question is.. WHY? 

[mjc123]

It looks as if the percentages of H₂S and CO₂ are mass percentages, not volume or molar. They should specify.

[cseil]

It looks as if the percentages of H₂S and CO₂ are mass percentages, not volume or molar. They should specify.

They are volume percentages. I didn't write it, sorry!

[mjc123]

Well, the given answer is right if they are mass percentages; your answer is right if they are volume percentages!

[cseil]

I think that my book wants to see me suffer
I didn't think about that because it is clearly written that it's a volume percentage.

Thank you anyway for your time!