Concentration of Aqueous Phosphoric Acid Solution

What is the mole fraction of phosphoric acid in the tank at 25 minutes?

The mole fraction of phosphoric acid in the tank at 25 minutes can be calculated based on the given data:

Calculation of Mole Fraction of Phosphoric Acid

Given data:
Initial moles of phosphoric acid in the tank = 150 kmol
Initial mole fraction of H3PO4 = 5.00% = 0.05
Rate of addition of pure phosphoric acid = 30.0 L/min
Molecular weight of phosphoric acid = 98
Density of phosphoric acid = 1.834 kg/L

Calculation:
- Moles of phosphoric acid added in 25 minutes = 30 L/min * 25 min = 750 L
- Mass of phosphoric acid added = Density * Volume added = 1.834 kg/L * 750 L = 1375.5 kg = 1375.5/98 = 14.036 kmol
- Moles of phosphoric acid in the tank after addition = Initial moles + Moles added = 150 kmol * 0.05 + 14.036 kmol = 21.536 kmol
- Moles of water in the tank after addition = Initial moles * (1 - Initial mole fraction) = 150 kmol * 0.95 = 142.5 kmol

Now, calculate the mole fraction of phosphoric acid in the tank:
Mole fraction = Moles of phosphoric acid / (Moles of phosphoric acid + Moles of water)
Mole fraction = 21.536 kmol / (21.536 kmol + 142.5 kmol) = 0.1313

Therefore, the mole fraction of phosphoric acid in the tank at 25 minutes is 0.1313.

← The power of nuclear fission chain reactions Gas effusion understanding the relative amounts of helium and argon →

More Posts: