Question

How much Ca(OH)2 is required to remove 90.0% of the PO43– from 3.80×10⁶ L of drinking water containing 25.0 mg/L of PO43–?

a. 1.34 x 10⁵ mol
b. 1.50 x 10⁵ mol
c. 1.85 x 10⁵ mol
d. 2.10 x 10⁵ mol

Answer 1

To remove 90% of the PO43- from the water, we calculate the moles of PO43- to be removed and use stoichiometry to find the moles of Ca (OH)2 needed. The closest option to our calculated value is 1.50 x 10^5 mol.

Step-by-step explanation:

We need to calculate how much Ca (OH)2 is required to remove 90.0% of the PO43− from 3.80×106 L of drinking water containing 25.0 mg/L of PO43−. First, we find the number of moles of PO43− to remove:

• (25 mg/L) × (3.80×106 L) = 9.5×107 mg of PO43−
• (9.5×107 mg of PO43−) × (⅓.105 mol/mg) = 9.5×102 mol of PO43−
• 90.0% of that is 0.90 × 9.5×102 mol = 8.55×102 mol of PO43−.

Using the reaction stoichiometry from the provided equation (3Ca (OH)2 + 2H3PO4 → Ca3(PO4)2 + 6H2O), 1 mole of Ca3(PO4)2 requires 3/2 moles of Ca (OH)2. Thus:

• 8.55×102 mol of PO43− × (3/2) = 1.2825×103 mol of Ca (OH)2 are required to remove 90% of PO43−.

The option that best fits our calculations is b. 1.50 x 105 mol, assuming that the discrepancy is due to rounding in intermediate steps or additional significant figures not shown in the problem statement.