Question

In: Chemistry

When solutions of excess barium chloride and 300 mL of sodium phosphate are mixed, 2.50 g...

When solutions of excess barium chloride and 300 mL of sodium phosphate are mixed, 2.50 g of a solid are produced.

A) Write the formula of the solid produced.

B) What is the original concentration of the sodium phosphate?

Solutions

Expert Solution

When solutions of excess barium chloride and 300 mL of sodium phosphate are mixed, 2.50 g of a solid are produced.

  1. Write the formula of the solid produced.

Given: Volume of Sodium phosphate = 300 mL = 0.300 L

Mass of solid = 2.50 g

Lets write the reaction :

3BaCl2 + 2Na3PO4 (aq) --> Ba3(PO4)2 (s) +6 NaCl

Ba3(PO4)2 (s) is the formula for the solid produced.

  1. What is the original concentration of the sodium phosphate?

Lets calculate moles of Ba3(PO4)2 (Barium phosphate)

Mol Barium phosphate = Mass of Barium phosphate / Molar mass of Barium phosphate

= 2.50 g / 601.93 g per mol   ( molar mass of barium phosphate = 601.93 g )

= 0.00415 mol of Ba3(PO4)2

From the reaction stoichiometry mol ratio of Ba3(PO4)2 to sodium phosphate is 1 : 2

We use that as a conversion factor to calculate moles of sodium phosphate

= 0.00415 mol of Ba3(PO4)2 * 2 mol sodium phosphate / 1mol Ba3(PO4)2

= 0.00831 mol sodium phosphate

[sodium phosphate ]= #mol sodium phosphate / volume in L

= 0.00831 mol / 0.300 L

= 0.0277 M

[Sodium phosphate] = 0.0277 M


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