2.000 g of Na_3PO_4 middot 12H_2O is reacted with 2.000 g of SrCl_2 middot 6H_2O

Question

2.000 g of Na_3PO_4 middot 12H_2O is reacted with 2.000 g of SrCl_2 middot 6H_2O, calculate the theoretical yield of Sr_3(PO_4)_2 precipitate. a. Calculate the theoretical yield (mass) of Sr_3(PO_4)_2 produced from 2.000 g of Na_3PO_4 middot 12H_2O if we assume that the strontium chloride salt is in excess. b. Calculate the theoretical yield of Sr_3(PO_4)_2 produced from 2.000 g of SrCl_2 middot 6H_2O again assuming that the sodium phosphate is in excess c. Which reactant produces the smallest theoretical yield?

Explanation / Answer

Balanced equation:
2 Na3PO4.12H2O + 3 SrCl2.6H2O = Sr3(PO4)2 + 6 NaCl + 18H2O

2 gm of Na3PO4.12H2O = 2 / 379.94 = 0.00526 Mole

0.00526 Mole of Na3PO4.12H2O will produce (0.00526 / 2) 0.00263 Mole of Sr3(PO4)2

0.00263 Mole of Sr3(PO4)2 = 0.00263 x 452.802 = 1.191 gm

Theortical yield = 1.191 gm

Qustion 2

2 gm of SrCl2.6H2O = 2 /266.526 = 0.00750 Mole

0.00526 Mole of SrCl2.6H2O will produce ( 0.00750 / 3) 0.00250 Mole of Sr3(PO4)2

0.00250 Mole of Sr3(PO4)2 = 0.00263 x 452.802 = 1.132 gm

Theortical yield = 1.132 gm

Question c

SrCl2.6H2O produces less yield

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