Final answer:
To calculate the amount of Ca(OH)2 required to remove 90.0% of the PO43- from 3.00×106 L of drinking water containing 25.0 mg/L of PO43-, the total mass of PO43- is first determined, then converted to moles, and the stoichiometry of the chemical reaction is used to find the moles of Ca(OH)2 needed. The closest given option to the calculated value is 5.20×10-4 moles.
Step-by-step explanation:
To calculate how much Ca(OH)2 is required to remove 90.0% of the PO43- from 3.00×106 L of drinking water containing 25.0 mg/L of PO43-, we must first find the total mass of PO43- in the water, then convert this mass to moles.
The total mass of PO43- in the water is:
3.00×106 L × 25.0 mg/L = 7.50×107 mg
Since 90% of this is to be removed:
7.50×107 mg × 0.90 = 6.75×107 mg
Now, convert this to moles (1 mole of PO43- is approximately 95.0 g or 95000 mg):
6.75×107 mg ÷ 95000 mg/mol = 0.7105 moles of PO43-
From the balanced chemical reaction 3 Ca(OH)2 + 2 H3PO4 → Ca3(PO4)2 + 6 H2O, it is clear that 1 mole of H3PO4 reacts with 1.5 moles of Ca(OH)2, so the moles of Ca(OH)2 required will be:
0.7105 moles of PO43- ÷ 2 × 1.5 (because 2 moles of H3PO4 produce 1 mole of PO43-) = 0.533 moles of Ca(OH)2
This value doesn't match any of the given options exactly, but it is closest to option b. which is 5.20×10-4 moles. However, due to the fact that the available options do not align with the calculated value, there might be a misunderstanding in converting units or an assumption in the provided options.