Question

How much Ca(OH)2 is required to remove 90.0% of the PO43– from 2.70×10⁶ L of drinking water containing 25.0 mg/L of PO43–?

a. 1.34×10⁵ mol
b. 2.45×10⁵ mol
c. 3.12×10⁵ mol
d. 4.02×10⁵ mol

Answer 1

To remove 90.0% of the PO43– from the drinking water, approximately 1.42x10^3 mol of Ca(OH)2 is required.

Step-by-step explanation:

To remove 90.0% of the PO43–, we can determine the initial concentration of PO43– in the water:

[PO43–] = 25.0 mg/L * 2.70x10^6 L = 67.5x10^6 mg = 67.5 kg

To convert to moles, we can use the molar mass of PO43–:

Molar mass of PO43– = 3*16.0 + 4*1.0 = 95.0 g/mol

[PO43–] = 67,500 g / 95.0 g/mol = 710.5 mol

Since Ca(OH)2 reacts with 2 moles of PO43– to remove 2 moles of PO43–, we need twice the amount of Ca(OH)2:

Amount of Ca(OH)2 required = 2 *710.5 mol = 1421 mol

Therefore, the amount of Ca(OH)2 required to remove 90.0% of the PO43– is approximately 1.42x10^3 mol.