Question

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solve this one . for 50 points


9.0g water is added into oleum sample labelled as 112% H2SO4 then the amount of free SO3 remaining in the solution is *( StP = 1 atm and 273 k)

Answer is 3.73 L at STP

Answer 1

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112% oleum means 100g is the sample and 12g is the water required by sample to completely combine SO3 in the sample. 

And according to reaction 
SO3  +  H2O   =      H2SO4
80g       18g               98g

So, 12g water combines with 53.3g SO3 and hence in 100g sample 53.3g is SO3 and 46.7g is H2SO4.

Now, If another 9g of water is added to it, another 40g of SO3 is converted to H2SO4.
So, remaining free SO3 is 15.3g = 0.2 moles 
At STP, 1 mol has volume 22.4 L. SO, 0.2 moles has volume = 3.78 L.


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Answer 2

Answer:

Answer is 3.73 L at STP here is your answer enjoy