What mass of free SO3 is present in 50g oleum sample which is labelled as 108.1%? 1) 24g. 2)36g. 3)18g. 4)12g
Answers
The correct answer is 18 grams
GIVEN
Mass of Oleum = 50 gram
Percentage of SO₃ present = 108.1%
TO FIND
Mass of free SO₃ present
SOLUTION
We can simply solve the above problem as follows,
Chemical formula of Oleum = H₂SO₄.SO₃
108.1% oleum means 100 grams of oleum requires 8.1 grams of H₂O to form 108.1 grams of H₂SO₄
SO₃ + H₂O ----------> H₂SO₄
1 moles SO₃ requires 1 mole of H₂O
Here,
Mass of H₂O given = 8.1 grams
Moles of H₂O = 8.1/18 = 0.45 mole
Moles of SO₃ required by 0.45 moles of H₂O = 0.45 × 80 = 36 grams
This means, 100 grams of oleum has 36 grams of free SO₃
So, 50 grams of Oleum has (36/100)×50 = 18 grams .
Hence, The correct answer is 18 grams
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ANSWER :
3)18 grams.
50 grams of Oleum has
(36/100)×50 = 18 grams.
Explanation:
Oleum is also known as fuming sulfuric acid in chemistry.
Sulfur trioxide and sulfuric acid are the main ingredients.
Olem is a sulphur trioxide solution in anhydrous sulfuric acid that is heavy, greasy, and very corrosive.
Oleum is made up of H2S2O7.
How can you figure out the percentage of oleuml?
If X grammes of water must be added to a 100 gramme sample of (SO3+H2SO4) in order for all of the So3 to completely react with the water to generate H2SO4, the percentage labelling of the oil will be (100+X)%.
Oleum is utilised as a fundamental ingredient in the sulfonation process, which involves chemically adding sulphate, in the creation of nylon, colours, nitrating reactions, and hydrofluoric acid (HF).
Sulfur trioxide can be made into oleum by mixing it with strong sulfuric acid.
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