Question

0.5g of fuming H₂SO₄ oleum is diluted with 100ml of water. This solution is completely neutralised by 26.7ml of 0.4N NaOH.

Find mass of SO₃ and percentage of free SO₃ .

Answer 1

H2SO4 + SO3 + H2O -> 2(H2SO4) .

==> SO3 + H2SO4 -----> H2SO4

( Equivalent = Eq )

∴ Eq of SO3 = 1/2 of H​2SO4

==> Eq of H2SO4 + Eq of SO3 = Eq of total H2SO4

Eq of total H2SO4 = (Normality x volume (ml) )/1000 = (0.4 x 26.7) / 1000 = 10.68/1000


Let mass of SO2 =x
mass of H2SO4 = (0.5-x)

= (0.5 - x)/49 + x/40 = 10.68/1000

Solving.... We get
x = 0.103g

mass of SO3 = 0.103 g

Percentage of free SO3 = =100x/ .5 = (0.1036 x 100) /0.5 = 20.6 %