How much Ca(OH)2 is required to remove 90.0% of PO4^-3 from 7.5 times 10^6 L of drinking water containing 25 mg/l of PO4^-3?

Answer 1

How much PO4^3- must be removed? That is
0.025 g/L x 7.5E6 L = 187,500 g.

What's 90% of that number?
That is 0.90 x 187,500 = 168,750 g.

How many moles is that?
mol PO4^3- = grams/molar mass PO4^3-

3Ca(OH)2 + 2PO4^3- ==> Ca3(PO4)2 + + 6OH^-
The equation tells you that moles Ca(OH)2 is 3/2 times PO4^3- moles.

I assume the problem wants grams although that isn't shown.
grams Ca(OH)2 = mol Ca(OH)2 x molar mass Ca(OH)2
Check my numbers; you best confirm them.