Sulfur trioxide (SO₃) dissolves in and reacts with water to form an aqueous solution of sulfuric acid (H₂SO₄). The vapor in equilibrium with the solution contains both SO and H₂O. If enough SO₃ is added, all of the water reacts and the solution becomes pure H2S04. If still more SO₃ is added, it dissolves to form a solution of SO₃ in H2SO₄, called oleum or fuming sulfuric acid. The vapor in equilibrium with oleum is pure SO₃. A 20% oleum by definition contains 20 kg of dissolved SO₃ and 80 kg of H2SO₄ per hundred kilograms of solution. Alternatively, the oleum composition can be expressed as % SO₃ by mass, with the constituents of the oleum considered to be SO, and H₂O.

(a) Prove that a 15.0% ole urn contains 84.4% SO₃.

(b) Suppose a gas stream at 40°C and 1.2 atm containing 90 mole% SO₃ and 10% N₂ contacts a liquid stream of 98 wt% H2SO₃ (aq), producing 15% oleum at the tower outlet. Tabulated equilibrium data indicate that the partial pressure of SO₃ in equilibrium with this oleum is 1.15 mm Hg. Calculate (i) the mole fraction of SO, in the outlet gas if this gas is in equilibrium with the liquid product at 40°C and 1 atm. and (ii) the ratio (m’ gas feed)/(kg liquid feed).