Question

It is desirable to remove calcium ion from hard water to prevent the formation of precipitates known as boiler scale that reduce heating efficiency. The calcium ion is reacted with sodium phosphate to form solid calcium phosphate, which is easier to remove than boiler scale. What volume (in liters) of 0.478 M sodium phosphate is needed to react completely with 0.225 liter of 0.279 M calcium chloride

Answer 1

0.0876L of 0.478M Na₃PO₄ are needed

Step-by-step explanation:

The reaction of calcium chloride, CaCl₂, with sodium phosphate, Na₃PO₄ is:

3CaCl₂ + 2Na₃PO₄ → Ca₃(PO₄)₂ + 6NaCl

Where 3 moles of calcium chloride react with 2 moles of sodium phosphate to produce 1 mole of calcium phosphate.

To solve this question we need to find moles of CaCl₂ added. Using the reaction we can find the moles of Na₃PO₄ that are needed to react completely and the volume using its concentration:

Moles CaCl₂:

0.225L * (0.279mol / L) = 0.0628moles of CaCl₂

Moles Na₃PO₄:

0.0628moles of CaCl₂ * (2mol Na₃PO₄ / 3 mol CaCl₂) =

0.0419moles of Na₃PO₄

Volume 0.478M Na₃PO₄:

0.0419moles of Na₃PO₄ * (1L / 0.478mol) =

0.0876L of 0.478M Na₃PO₄ are needed