Question

Experiment 9 Data, Results and Discussion Name Beaker Sample 1 1. a. 4.412 g Mass of SrCl2-6H20 and container Mass of container - 2.901 g Mass of SrCl2-6H20 added to Beaker I 1.511 grams Moles of SrCl2-6H20 added to Beaker I 1.511g/266.64 g/mole = 0.00567 moles Theoretical moles of Sr3(PO4)2 0.00567 mole Srdlz6H20 x 1mole Srs(PO4)2/3 mole Srl determined from moles of SrCl2-6H20 0.00189 2. a. Mass of Na3PO4-12H20 with container 3.586 g - 2.583 g Mass of container 1.003 grams Mass of Na3PO4-12H20 added to Beaker I Moles of Na3PO4:12H20 added to beaker I d. 1.003 grams/380.18 grams/mole Theoretical moles of Sr3(PO4)2 determined from moles of Na3PO4-12H20 Theoretical moles of Sr3(PO4)2 in Beaker I determined from the limiting reagent #moles Sr3(PO4)2 452.80 grams 4. Theoretical yield of Sra(PO4)2 in grams 5. a. Observation of SrCl2 test reagent, added to filtrate from Beaker I Cloudy - precipitate b. Observation of Na3PO4 test reagent added to filtrate No change - clear c. Formula of limiting reactant d. Formula of excess reactant Page 1

Answer 1

This question is basically based on stochiometry of reaction. Simple mole concept is applied in this question.
at Mass of Srila, 6H₂O added to Boke I : 1511 g Moles of Sau, . 66,0 added to Beaker I. 1.511 96664 = 0.00567 males Theoretical from moles of Srl, . 6120 Sng (Pon) formed and Nazpon 12120 - 0.00567 rowes 3 0.00189 moles Basically . Following Chemical Reaction is happening 3 Sau, 6H2O + ana, pon. 12H20 - Sng (Pon) + 6 Nace + 142 M₂0 Mass of Naz Pon 12 H₂O added to beakes I = 1.003 g Moles of Naz Pon 12 H₂O = 1.003 . -= 380.18 0.00264 mal 2(e) Theoretical moles of Siz (pon), delamined from moles of Nazpon - 19 H₂O to According Balanced Chemical equation Moles of Siz (Pon), = 0.00064 0.00132 moles - g

= Detcimination of Limiting Reagent 3 moles of Soci, 6H2O Na, pon lang required a moles of 0.00567 moles of Srl 6H₂O required 9x 0.00567 male of Nagpon moles 2x0.00567 modes Theoretically required of Nazpon 12 H2O : 0.00378 modes But available moles of Nazpon 121₂0 =0.00364 mol The face Limiting Reagent = Naz pon. 121120 Excess Reagent = Se U 6H2O of Theoretical from moles Limiting Sez(pon) Reagent 0.00864 - = 0.00132 mol ų Theoretical Yield of Se, (POM), in gm = 0.00132 x 459.89 : 0.5977 g