In a polluted river, a chemical treatment is used to remove excess phosphate (PO4^3-) ions, causing them to precipitate as calcium phosphate (Ca3(PO4)2). Calculate the amount of calcium chloride (CaCl2) required to treat 1,000 liters of river water with a phosphate concentration of 2.0 mg/L.
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The equation for the reaction between calcium chloride and phosphate ions is:
3CaCl2 + 2Na3PO4 → Ca3(PO4)2 + 6NaCl
From this equation, we can see that 2 moles of calcium phosphate are produced for every 3 moles of calcium chloride. Therefore, the mole ratio between calcium chloride and calcium phosphate is 3:2.
To calculate the amount of calcium chloride required, we need to determine the number of moles of phosphate ions in 1,000 liters of river water.
First, convert the phosphate concentration from mg/L to moles/L:
2.0 mg/L * (1 g/1000 mg) * (1 mol PO4^3-/95.99 g) = 0.02083 mol/L
Next, calculate the total number of moles of phosphate ions in 1,000 liters of river water:
0.02083 mol/L * 1000 L = 20.83 mol
Since the mole ratio between calcium chloride and calcium phosphate is 3:2, we can calculate the moles of calcium chloride required as follows:
(20.83 mol PO4^3-) * (3 mol CaCl2/2 mol Ca3(PO4)2) = 31.25 mol CaCl2
Finally, convert the moles of calcium chloride to grams:
31.25 mol CaCl2 * (110.98 g CaCl2/mol) = 3470.375 g
Therefore, approximately 3,470 grams (or 3.47 kilograms) of calcium chloride are required to treat 1,000 liters of river water with a phosphate concentration of 2.0 mg/L.
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