Chemistry, asked by shwetasingh29112003, 3 months ago

A1.88 g sample mixture of BaCl2.2H20 (molar mass =
244.3 g/mol) and NagPO4.12H,0 (molar mass 380.2
g/mol) was dissolved in 250 mL water, and after
filtration and drying of the resulting barium phosphate
precipitate Ba3(PO4)2, (molar mass 601.9 g/mol),
weighed 0.45 g. Ifa drop of barium chloride solution
added to the filtrate yielded a precipitate, then the
mass % of the excessS reactant in the sample mixture
is equal to 2 Na3 PO4)+3 BaCl2) Ba3(PO4)0) 6 NaCles)​

Answers

Answered by avinahkr2298
0

Answer:

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Explanation:

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