Question

what is the percentage of free SO3 in a sample of oleum labeled as 104.5%

Answer 1

Answer:

20% free SO3

Explanation:

104.5% labeled oleum means 100g oleum reacts with H2O to form 104.5g H2SO4,

therefore, mass of H2O = 4.5g

actually, SO3 + H2O forms  H2SO4

here,   1 mole SO3 reacts with 1 mole H2O to form 1 mole sulphuric acid

this means that 80g SO3 reacts with 18g H2O to form 98g H2SO4

Now, 80g SO3 reacts with 18g H2O

therefore 4.5g H2O reacts with 80/18 * 4.5g so = 20g SO3

hence % of free SO3 is 20% .