One method of removing phosphate from wastewater effluents is to precipitate it

Question

One method of removing phosphate from wastewater effluents is to precipitate it with aluminum sulfate. A plausible stoichiometry (but not exact because aluminum and phosphate can form many different chemical materials) is:

2 PO43- + Al2(SO4)3 ? 2 AlPO4 + 3 SO42-
If the concentration of phosphate (PO43-) is 30 mg/L, how many kg of aluminum sulfate must be purchased annually to treat 40 L/sec of wastewater? How many kg of precipitate will be formed as sludge if all of the phosphate is precipitated as AlPO4?

Explanation / Answer

annually waste water to treat=40*3600*24*365=1.26*10^9L

total amount of  (PO43-)=30/1000*1.26*10^9L=3.78*10^7g

moles of (PO43-)=3.78*10^7g/95=3.98*10^5 moles

for 1 mole of  (PO43-) is required 1/2 mole of  aluminum sulfate

moles of  aluminum sulfate = 1/2*3.98*10^5 moles =1.99*10^5 moles

weight of  aluminum sulfate=1.99*10^5 mole*342=6.81*10^4 kg

1 mole of (PO43-) is reqiured to 1 mole of  AlPO4

kg of precipitate will be formed as sludge if all of the phosphate is precipitated as AlPO4

=3.98*10^5*122g=4.86*10^4kg

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