Asked Nov 11, 2019

a 1.500 g sample of a mixture of solid Na3PO4 *12H2O and BaCL2* 2H2O was dissolved in water and formed 0.520g of Ba3PO4 precipitate . If reagent tests demonstrated that Na3PO4* 12H2O was the limitng reactant in the precipitaion reaction, calculate the mass percent of each reactant salt in the original sample. 


Expert Answer

Step 1

The reaction proceeds according to the following equation


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Baз (РОд)2 + 6NaCl + 30H,о ЗВаСl), 2H,0 2Na;POд 12H-0

Step 2

Since Na3PO4.H2O is the limiting reagent


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2 moles of Na,PO4.12H,0 produces 1 mole of Ba,(PO4)2 therefore when 0.520 g of Ba3 (PO4),is produced given mass no of moles of Ba^ (PO4)2- molar mass 0.520g 602g 0.00086mol no of moles of Na^PO4.12H,0= 2x 0.00086 =0.00172mol

Step 3

Mass of Na3PO4.H2O


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mass of Na,PO4.12H20 =moles x molar mass 0.00172 x380g = 0.6536g


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