Question

A mixture, weighing 6.000 grams, contains 41.50 % Na3PO4 and 58.50 % BaCl2 of the mixture, by weight. When dissolved in water, a precipitate forms via the double displacement reaction:

BaCl2(aq) + Na3PO4(aq) NaCl(aq) + Ba3(PO4)2(s)

If the reaction is quantitative, the yield of the solid barium phosphate should be

Answer 1

Answer on Question #51826, Chemistry, Inorganic Chemistry

Question: A mixture, weighing 6.000 grams, contains 41.50 % Na3PO4 and 58.50 % BaCl2 of the mixture, by weight. When dissolved in water, a precipitate forms via the double displacement reaction:

$$3 \mathrm{BaCl_2(aq)} + 2 \mathrm{Na_3PO_4(aq)} \rightarrow 6 \mathrm{NaCl(aq)} + \mathrm{Ba_3(PO_4)_2(s)}$$

If the reaction is quantitative, the yield of the solid barium phosphate should be

Answer:

The mass of Na3PO4 in the mixture: 6.000*0.4150 = 2.49g

The number of moles of Na3PO4: 2.49/164 = 0.0152 moles.

The mass of BaCl2: 6 – 2.49 = 3.51g.

The number of moles of BaCl2: 3.51/208 = 0.0169

From the equation and our previous calculation it can be determined that Na3PO4 is a limiting reactant. The number of moles of Ba3(PO4)2 is 2 times smaller than the number of moles of Na3PO4, so the number of moles of Ba3(PO4)2 will be: 0.0152/2 = 0.0076 moles.

Yield of Ba3(PO4)2 is 0.0076*601 = 4.568g.

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