Question

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

5Ca(OH)2(aq) + 3Po4^3-(aq)-->ca5(Po4)3(s) + 9OH^-(aq). How much Ca(OH)2 is required to remove 90% of the Po4^3- from 4.00 x 10^6 l of drinking water containing 25.0 mg / l of po4^3-?
a) ≈4.50×10^6 mol
b) ≈1.50×10^6 mol
c) ≈2.25×10^6 mol
d) ≈6.00×10^6 mol
e) ≈3.00×10^6 mol

Answer 1

First, we convert the total amount of PO4^3- in grams and then to moles. Next, we calculate 90% of that amount, and finally, use the stoichiometry from the balanced chemical equation to find the required moles of Ca(OH)2.

Step-by-step explanation:

To calculate how much Ca(OH)2 is required to remove 90% of the PO43- from 4.00 × 106 liters of drinking water containing 25.0 mg/L of PO43-, we must first determine the amount of PO43- in moles that needs to be removed. Given that 1 liter of water contains 25.0 mg of PO43-, 4.00 × 106 liters will contain 4.00 × 106 × 25.0 mg of PO43-, equal to 1.00 × 108 mg, or 1.00 × 105 grams since 1 gram is 1000 mg.