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Question:

At 300 K,Carbon powder and O2 are included in container of 10dm3.When the system is heated well,all the O2 converts into CO2.Then the system is cooled to the initial temperature. The volume of the Carbon powder is negligible. Find the percentage of the increasement of the density of the system.

My thoughts on the question:

C(s) + O2(g)--> CO2(g)
300K
ini: y x -

After (y-x) (x-x) (+x)
heating
(moles)

But what happens when the system is cooled again?Do all the CO2 cinvets into O2 so that the moles in the system is same as that of the 300K system?

So the increasement of density=(1/v) { [ (y*12)+(x*32)] - [(x*44)+( (y-x)*12 )] }
Here y cancels out but how do we find a numerical value for the percentage ,when x is there?

(1/v)= 1/volume of the container= 1/(10*(10)^-3)=100 m^(-3)

2 answers

Yes, mols CO2 is same as mols O2.
I see it this way.
Start with 10 dm3 O2 so end with 10 dm3 CO2.
d O2 to start = 32/22.4 = about 1.42 g/dm3 if at STP. At 300 K it will be 1.42 x 273/300 = ?

CO2 at STP will be 44/22.4 = about 1.96 g/dm3 at STP or 1.96 x 273/300 = ?

Then convert densities into percent change.
Thank you!