Aluminum sulfate reacts with alkalinity according to the following reaction in t

Question

Aluminum sulfate reacts with alkalinity according to the following reaction in the coagulation tank.

Al2(SO4)3.14H2O + HCO3- ==> Al(OH)3 (s) + CO2 + H2O + SO4+ ......1

a. Balance this stoichiometric equation

b. Montebello water treatment plant processess 250 million gallons per day of water from loch raven resevior with Al2(SO4)3.14H2O at a dose of 6.25 g/L. Aluminuma sulafate coagulation is used to remove particulate matters,reduce the concentration of organic matter and reduce the alkalinity of the water according tho equation 1. If the organic matter concentration is reduced from 8mg/L to 3mg/L, determine the total mass of alkalinity consumed and the total mass of dry solids removed per day. (note Alkalinity is the [HCO3-]) (Al=27; S =32, O =16, H =1; C=12;)

Explanation / Answer

Al2(SO4)3.14H2O + 6 HCO3- ==> 2Al(OH)3 (s) + 6CO2 + 14H2O + 3SO42-

As per this equation 1 mole of Aluminium sulfate uses 6 moles of bicarbonate

6.25 g/L x 250 million gallons x 3.78 litres/gallon = 5906250 Kg Aluminium sulfate is used which is 5906250 x 103/594 = 9943 x 103 moles

So we need alkalinity if 9943 x 103 x 6 = 59659090 moles of HCO3- which is 59659090 moles x 61 = 3639204 Kg of bicarbonate.

1 mole of aluminium sulfate gives 2 moles of Al(OH)3 so 9943 x 103 moles of Aluminium sulfate will give

9943 x 103 molesx2 = 19886000 moles of Al(OH)3 which is 19886000 moles x 78 = 1551108 Kg of dry solid/day

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