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 (alum). A plausible stoichiometry (but not exact because aluminum and phosphate can form many different chemical materials) is: 2 POUND^3-_4 + Alpha_2(SO_4)_3 rightarrow 2 AlPO_4 + 3 SO^2-_4 If the concentration of phosphate (POUND^3-_4) is 30 mg/L, how many kg of alum must be purchased annually to treat 40 L/s of wastewater? How many kg/yr of solid precipitate will be formed (dry basis) if all of the phosphate is precipitated as AlPO_4?

Explanation / Answer

Mass of PO4(3-) per second = 30 mg/L * 40L/s = 1200 mg = 1.2 g/seconds

Number of seconds in a year = 3.154 * 10^7

Mass of Po4(3-) = 1.2 * 3.154 * 10^7 grams/year = 3.7848 * 10^7 grams

2 moles of PO4(3-) requires one mole of alum

Number of moles of PO4(3-) = 95 grams/mol

number of moles of PO4(3-) = 3.7848 * 10^7/95 = 3.984 * 10^5 moles

Number of moles of Al2(SO4)3 required = 3.984/2 * 10^5 moles = 1.992 * 10^5 moles

Molar mass of Al2(SO4)3 = 342.15 gm/mol

mass of al2(so4)3 = 1.992 * 10^5 * 342.15 = 6.8156 * 10^4 Kg = 68156.28 Kg

number of moles of AlPO4 formed = 3.984 * 10^5 moles

molar mass of AlPO4 = 121.95 gm/mol

Mass of AlPO4 formed = 48584.88 kg

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