One hundred fifty kmol of an aqueous phosphoric acid solution contains 5.00 mole

Question

One hundred fifty kmol of an aqueous phosphoric acid solution contains 5.00 mole % H3PO4. The solution is concentrated by adding pure phosphoric acid at a rate of 20.0 L/min
a) Write a differential mole balance on phosphoric acid and provide an initial condition. [Start by defining np (kmol) to be the total quantity of phosphoric acid in the tank at any time.]
b) Solve the balance to obtain an expression for np(t). Use the result to derive an expression for xp(t), the mole fraction of phosphoric acid in the solution.
c) How long will it take to concentrate the solution to 15 % H3PO4?

Explanation / Answer

kilo-moles of phosphoric acid initially present = 0.05*150 = 7.5 kmol = 7500 moles
kilo-moles of water initially present = 150-7.5 = 142.5 kmol = 142500 moles

assume density of pure phosphoric acid to be gms/L

then mass flow rate of phosphoric acid = 20* (gms/min)

molar mass of phosphoric acid = 98 gms/mol

so molar flow rate of phosphoric acid = (20/98) mol/min = (0.2041) mol/min

(a)

after time 't' minutes has passed,

total moles of phosphoric acid present in mixture = 7500 + (0.2041*t) moles

so, np(t) = 7500+0.2041t .....( put t in minutes to get np in moles)

(b)

we already have an expression for np(t) which is

np(t) = 7500+0.2041t

xp(t) = total moles of phosphoric acid at time 't'/total moles of solution at time 't'

total moles of phosphoric acid = np(t) = 7500+0.2041t

total moles of solution = total moles of phosphoric acid present at time 't' + moles of water initially present

so, total moles of solution = (np(t)+142500) = 7500+0.2041t + 142500 = (150000+0.2041t)

so,

xp(t) = (7500+0.2041t)/(150000+0.2041t)

(c)

put xp(t) = 0.15 in above equation and solve,

23112.245 = 0.173485t

so,

t = 23112.245/0.173485 = 133223.305/ minutes

in each answer put the value of in gms/L unit only to get final answer.

I looked up from the web and found that = 1885 gms/L

substituting that the answers for parts

a) np(t) = 7500+384.73t ... (put t in minutes to get np in moles)

b) xp(t) = (7500+384.73t)/(150000+384.73t) ...... (put t in minutes to get xp )

c) t = 70.68 minutes

however put the value of that you have in your textbook

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