Question

6. A 1.500 g sample of a mixture of solid Na PO, 12H.O mixed with NaCl, was dissolved in water, reacted with aqueous 1 M SrCl, and 0.520 g Sr.(PO), precipitate was collected and dried to a constant mass. If the reagent tests demonstrated that Na PO:12H.O was the limiting reactant in the precipitation reaction, calculate the mass percent of Na PO, 12H 0 in the original sample. a. How many moles of Sr,(PO), were produced? b. What is the balanced chemical reaction? c. How many moles of Na,PO, were present in the aqueous solution? d. If the moles of Na PO, are equivalent to the moles of Na PO, 12H,O, what is the theoretical mass of Na, PO, 12H,0 in the solid mixture? What is the mass percent of sodium phosphate dodecahydrate in the solid mixture?

Answer 1

Sol Geren 1.5g of sample of a mixture of solid Naz Poy. 12H20 and Natl. When the said mixture il dissolved in water and reacted with 1M solde 0.52g of S63 (Pou), was precipitated it wasion is tiniting reagent, Total Na Pou is reacted with Sort in sich 2 Nag Pou +358 – 6Na (d & Srg (Poule Molecular weight of Sr Poul = 3x88+ 2 (3H642 M 454 a) Number of moles of StaPou) precipitated is 0.52 0.001145 454 From balanced equation number of moles of Na, pou reacted - 2xn, = 2x 0.001145 = 0.00229 b) giove, Rom + 2po2 +358 +2 Sog(Pou) c) No. of moles of Nag Pou is 0.00229

d) Mass of Naz Pon 12 H2O = 0.00229x( 380 = 0.8702g. percent of Naz Pon 12H20 in solid minture . 0-8702 x 100 = 58.0133% 105 e Masse