D) 150,000 mol of aqueous phosphoric acid (H3PO4) solution contains 5.00% H3PO4

Question

D) 150,000 mol of aqueous phosphoric acid (H3PO4) solution contains 5.00% H3PO4 by moles. The solution is concentrated by adding pure H3PO4 at a rate of 20.0 L/min.

D.1) Write a differential mole balance on H3PO4 and write the initial condition. (Hint: Start by defining np to be the total quantity of H3PO4 in the tank at any given time.

D.2) Solve to find np(t). Then derive an expression for the mole fraction of H3PO4 in the solution, xp(t).

D.3) How long will it take to concentrate the solution to 15% H3PO4?

Explanation / Answer

1) If np is total quantity of H3PO4 in the tank at any given time.

by mole balance,

rate of moles of H3PO4 in - rate of moles of H3PO4 out = d(np)/dt

weight of 20 L H3PO4 = 1.885*20 = 37.7 kg

moles of H3 PO4 in 20 L = 0.38469 Kmoles

rate of moles in of H3PO4 = 384.69 moles/min

rate of moles out =0

so, 384.69 -0 = d(np)/dt

2)np = 384.69 t +constant

at time, t=0, np = 0.05*150000 = 7500 moles

no.of moles of water initially = 150000*0.95 = 142500 moles

so, np = 384.69 t +7500

xp(t) = np/(np+142500)

xp(t) = (384.69 t +7500)/(384.69 t+150,000)

3)for it to become 15% H3PO4

xp(t)=0.15

0.15 = (384.69 t +7500)/(384.69 t+150,000)

solving, t= 45.87 minutes

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