Question in the Description

Question

Phosphate can be removed from drinking water supplies by treating the water with Ca(OH)2.

How much Ca(OH)2(s) is required to remove 90% of the PO4 3− from 4.5 × 106L of drinking water containing 25 mg/L PO4 ^3− ?

5 Ca(OH)2 (aq) + 3 PO4 3− (aq)  Ca5(PO4)3OH (s) + 9 OH− (aq)

Answer 1

Let's look at the balanced equation for the reaction taking place...

5Ca(OH)₂ (aq) + 3PO₄³⁻ (aq) ==> Ca₅(PO₄)₃OH (s) + 9OH⁻ (aq)

How many moles of PO₄^3- are present?

25 mg PO₄^3-/L and molar mass ofPO₄^3- is 94.97 g/mol, so what do we have...?

25 mg x 1 g/1000 mg x 1 mol/94.97 g = 2.63x10^-4 moles PO₄^3-

90% of this 0.9 x 2.63x10^-4 moles = 2.37x10^-4 moles PO₄^3-

How many moles of Ca(OH)2 do we need to react with this many moles of PO₄^3-?

2.37x10^-4 moles PO₄^3- x 5 moles Ca(OH)₂ / 3 moles PO₄^3- = 3.95x10-4 moles Ca(OH)₂ needed

So, this would be the answer in moles. The question is somewhat ambiguous since it doesn't ask for moles or mass. It simply asks "how much". So this would be a correct answer.