4. (2 points) One method of removing phosphate from wastewater effluent is to pr

Question

4. (2 points) One method of removing phosphate from wastewater effluent is to precipitate it W1 with aluminium sulfate (alum). A possible stoichiometry (but not exact because aluminium and phosphate can form many different chemical materials) is: 2 PO43. + Al2(SO4)3 2 AlPO4 + 3 SO42. If the concentration of phosphate (POs3) is 30 mg/L, how many kg of alum must be purchased annually to treat 40L/s of wastewater? How many kg/yr of solid precipitate will be formed (dry basis) i if all of the phosphate is precipitated as AIPO?

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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