I suspect a balanced equation is the place to start...
3BaCl2 + 2K3PO4 >>> Ba3(PO4)2 + 6KCl
moles barium phosphate: 1/3 *moles barium chloride
calcuate the moles barium chloride, divide by three, then that is the moles of barium phosphate, convert that to grams of barium phosphate.
10.0 grams of barium chloride (208.3g/mol) is reacted with excess potassium phosphate and 7.22 of grams of barium phosphate (601.8 g/mol) is isolated. calculate the theoretical yield of barium phosphate
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