Homework Answers
Given
Concentration of PO4 3- = 25 mg/L = 25*10^-3 g / L
1 L of drinking water contains = 25 x 10^-3 g of PO43-
4.5 X 10^6 L of drinking water contains = 25 x 10^-3 x 4.5 x10^6 =
112500 g
remove 90 % of PO43-
PO43- removed = 0.90 x 112500 = 101250 g
moles of PO43- removed = mass / molecular weight
= 101250/95
= 1065.789 moles
The balanced chemical equation
5 Ca(OH2) + 3 PO4 3- = Ca5OH(PO4)3 +9 OH-
From the stoichiometry of the reaction
3 mole of PO43- = 5 moles of Ca(OH)2
1065.789 moles of PO43- = (5/3) x 1065.789 = 1776.316 moles of
Ca(OH)2
molecular mass of Ca(OH)2 = 74.1 g /mol
Mass of Ca(OH)2 recquired = moles x molecular weight
= 1776.316 moles x 74 g/mol
= 131447.368 g
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