Al2(SO4)3 + 3Ca(OH)2 2Al(OH)3 + 3Ca+2 + 3SO4-2 1. Assuming that the concentratio

Question

Al2(SO4)3 + 3Ca(OH)2 2Al(OH)3 + 3Ca+2 + 3SO4-2

1. Assuming that the concentration of calcium in the water is 80 mg/L (as Ca), how many mg/L of quick lime (CaO) would be needed to reduce the calcium concentration to 35 mg/L. Also assume that the reaction (conventional lime softening) is stoichiometric, that there is no excess CO2 present that needs to be removed (alum is not added as a coagulant), and that no magnesium will be removed. (Hint: all of the calcium hardness is carbonate hardness.) Atomic weights: Ca=40; O=16; C=12; H=1

2. What is the annual cost of chlorine when the plant is operated at the design flow rate of 20 MGD? Assume that the chlorine is added as a solution of sodium hypochlorite (NaOCl) having a strength of 120 g/L as Cl2, that the dosage is 5 mg/L as Cl2, and that the cost of the solution is $1.75 per gallon.

Explanation / Answer

Al2(SO4)3 + 3Ca(OH)2 ------> 2Al(OH)3 + 3Ca2+ + 3SO42-

Molar mass of Ca = 40 g/mole

Thus, millimoles of Ca in per litre of water = 80/40 = 2

millimoles of Ca required per litre of water = 35/40 = 0.875

Thus, millimoles of Ca required to be removed per litre of water = 2-0.875 = 1.125

Thus, millimoles of CaO required to be removed = millimoles of Ca required to be removed = 1.125

Molar mass of CaO = 56 g/mole

Thus, mass of CaO required to be removed = 1.125*56 = 63 mg/litre of water

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