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Grade: 11

                        

what is the % labelling of oleum if mole fraction of so3 in the oleum is 0.5

6 months ago

Answers : (2)

Arun
24742 Points
							
Calculation of free SO3 in 118% H2SO4 labelled oleum
H2O + SO3 → H2SO4
As 18g of water combines with 80g of SO3
100g of oleum contains 80g of SO3 or 80% free SO3
6 months ago
Vikas TU
12276 Points
							
Oleum is mixture of H2SO4 and SO3 gas. When we add water to oleum, the SO3 in it reacts with water to from H2SO4. Labeling of oleum tells us about the total amount of H2SO4 that would be there if water is added to it.
SO3+H2O———−>H2SO4
Okay, so lets say we have x gm of SO3 and (100-x) gm of H2SO4 in 100gm of Oleum.
The x gm of SO3 would react with H2O to form some amount of H2SO4.
Molecular mass of SO3 is 80. So the moles of SO3 taking part in the reaction would be x/80. ( Moles= Given Mass/ Molecular Mass )
Analyzing the Stoichiometric coefficients of the reaction of SO3 in Water.
1 mole of SO3 reacts with 1 mole of H2O to form 1 mole H2SO4.
So x/80 moles of SO3 would reacts with x/80 moles of H2O to produce x/80 mole of H2SO4.
Now, as we know the moles of H2SO4 formed, we can calculate the mass of H2SO4 so formed ( Molecular mass of H2SO4 is 98)
GivenMass=Moles∗MolecularMass
So, Mass of H2SO4 so formed is 98x/80.
We know that the total mass of H2SO4 that would be formed when water is added to 109% oleum is 109 gms and we assumed the mass of H2SO4 to be (100-x) gm. Which gives us the equation:
Mass of H2SO4 formed+Mass of H2SO4 already present in Oleum = Total mass of oleum
98x/80+(100-x) = 109
Hence, x=40
Percentage of SO3, (40/100)∗100 = 40%
6 months ago
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