Oleum contains H2SO4 and SO3 . 118% oleum is given which means 100g oleum will be diluted with 18gH2O . This H2O reacts with SO3. ∴ Moles of H2O= Moles of SO3 Or ‌
‌ Mass of ‌H2O
‌ Molar mass of ‌H2O
=‌
‌ Mass of ‌SO3
‌ Molar mass of ‌SO3
Or ‌
18
18
=‌
‌ Mass of ‌SO3
80
Or Mass of SO3=80g It means mass of H2SO4= Mass of oleum - Mass of SO3 ‌=100−80 ‌=20g Moles of H2SO4=‌
‌ Mass of ‌H2SO4
‌ Molar mass of ‌H2SO4
=‌
20
98
Moles of SO3=‌
‌ Mass of ‌SO3
‌ Molar mass of ‌SO3
=‌
80
80
=1 The reaction of NaOH takes place with both H2SO4 and SO3 as follows H2SO4+2‌NaOH⟶Na2SO4+H2O According to calculations, moles of H2SO4 available are 20/98. So, moles of NaOH=2×‌
20
98
SO3+2‌NaOH⟶Na2SO4+H2O Moles of NaOH=2 ∴ Total moles of NaOH=2×‌