Question

A tank contains one hundred fifty kmol of an aqueous phosphoric acid solution with 5.00 mole% H3PO4. The solution is concentrated by adding pure phosphoric acid at a rate of 30.0 L/min. Molecular weight of phosphoric acid is 98. Density of phosphoric acid is 1.834 kg/L. Determine the mole fraction of phosphoric acid in the tank at 25 minutes.

Answer 1

Mole fraction = 0.1313

Step-by-step explanation:

given data

Addition rate of pure phosphoric acid = 30L/min

Phosphoric acid added in 25 min = 750 L

solution

mass is express as here

mass = density × volume and number of moles = mass ÷ molar mass

So, In terms of moles

phosphoric acid added = (750 × 1.834) ÷ 98

phosphoric acid added = 14.036 kmol

and

moles of phosphoric acid in tank = 0.05 × 150 + 14.036

moles of phosphoric acid in tank = 21.536 kmol

and

Moles of water in tank = 0.95 × 150

Moles of water in tank = 142.5 kmol

So that Mole fraction of phosphoric acid in tank

Mole fraction = 21.536 ÷ (21.536+142.5)

Mole fraction = 0.1313