Answer to Question #37239 in Physical Chemistry for radhika bauskar
3BaCl2 + 2Na3PO4 = Ba3(PO4)2 + 6NaCl
The mole ratio between BaCl2 and Na3PO4 is
The ratio of these compounds in the given mixture is 0.5/0.2 = 5:2 > 3:2.
Hence the BaCl2 is in excess and the Na3PO4 is the limiting reactant. The 1 mole of Ba3(PO4)2 occurs for each two moles of Na3PO4, hence the amount of Ba3(PO4)2
n[Ba3(PO4)2] = 1/2 * 0.2 = 0.1 mol.
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