Answer:
14.6 g of barium phosphate
3.35 g of sodium metal
Explanation:
2Na3PO4(aq) + 3Ba(s) -------> Ba3(PO4)2(aq) + 6Na(s)
The first step in any such reaction is to but down the balanced reaction equation according to the stoichiometry of the reaction.
The two products formed are barium phosphate and sodium metal.
Number of moles of barium corresponding to 10.0g of barium = mass of barium/ molar mass of barium
Molar mass of barium = 137.327 g
Number of moles of barium = 10/137.327
Number of moles of barium = 0.0728 moles
For barium phosphate;
3 moles of barium yields 1 mole of barium phosphate
0.0728 moles yields 0.0728 moles × 1/3 = 0.0243 moles of barium phosphate
Molar mass of barium phosphate = 601.93 g/mol
Therefore mass of barium phosphate = 0.0243 moles × 601.93 g/mol = 14.6 g of barium phosphate
For sodium metal
3 moles of barium yields 6 moles of sodium metal
0.0728 moles of barium yields 0.0728 × 6 / 3 = 0.1456 moles of sodium
Molar mass of sodium metal= 23 gmol-1
Mass of sodium metal= 0.1456g × 23 gmol-1 = 3.35 g of sodium metal
