Answer:
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
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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