A 0.515 g sample of CaCl2 reacts with aqueous sodium phosphate to give 0.484 g Ca3(PO4)2....
A 0.515 g sample of CaCl2 reacts with aqueous sodium phosphate to give 0.484 g Ca3(PO4)2. Sow the theorectical yield of Ca3(PO4)2.
Homework Answers
3 CaCl2 + 2 Na3PO4 → Ca3(PO4)2 + 6 NaCl
Supposing there is excess sodium phosphate:
(0.515 g CaCl2) / (110.9840 g CaCl2/mol) x (1 mol Ca3(PO4)2 / 3 mol
CaCl2) x (310.1767 g Ca3(PO4)2/mol) =
0.4797 g Ca3(PO4)2 in theory
(0.484 g) / (0.4797 g) = 100.8 = 100.8 % yield
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A 0.515 g sample of CaCl2 reacts with aqueous sodium phosphate
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