If you added an excess of sodium phosphate dodecahydrate to 4.512 g barium chloride dihydrate, how many grams of barium phosphate could you synthesize

The mass of barium phosphate synthesized is 3.612 g.

The equation of the reaction is;

3 BaCl2.2H2O + 2 Na3PO4.12 H2O → 6 NaCl + Ba3(PO4)2 + 14H2O

Molar mass of barium chloride dihydrate = 244.26 g/mol

The number of moles of barium chloride dihydrate =  4.512 g/244.26 g/mol

= 0.018 moles

From the balanced reaction equation shown;

If 3 moles of barium chloride dihydrate yields 1 mole of  barium phosphate

0.018 moles of  barium chloride dihydrate yields 0.018 moles × 1 mole/3 moles

= 0.006 mole of barium phosphate

Mass of barium phosphate = 602 g/mol × 0.006 mole

= 3.612 g

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3.35g of Ba3(PO4)2

Explanation:

The balanced chemical equation is given as;

3 BaCl2 + 2 Na3PO4 → 6 NaCl + Ba3(PO4)2

Fro the reaction;

3 mol of BaCl2 (624g) reacts wth 2 mol of Na3PO4 (327.88g) to form 1 mol of Ba3(PO4)2 (601.93g)

Since Na3PO4 is in excess, the limiting reactant is; BaCl2. It would determine how much of Ba3(PO4)2 is formed.

624g of BaCl2 = 601.93g of Ba3(PO4)2

4.512g of BaCl2= xg of Ba3(PO4)2

x = 4.512g * 601.93g / 624g

x = 3.35g of Ba3(PO4)2

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