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26. 3CaCl2(ag) + 2N23PO4(ag) → Ca3(PO4)2(s) + 6NacI(ag) In a reaction mixture of 16 mL of 0.56 M CaCl2 and 3.4 mL of 1.7 M Na3PO4: The CaCl, can make moles of Ca3(PO4)2 The Na;PO4 can make moles of Ca3(PO4)2 The limiting reagent in this mixture is The theoretical yield of Ca3(PO4)2 in this reaction mixture is moles If 0.85 g of Ca3(PO4)2 (M = 310.18 g/mol) was isolated, the percent reaction yield is

26. 3CaCl2(ag) + 2N23PO4(ag) → Ca3(PO4)2(s) + 6NacI(ag) In a reaction mixture of 16 mL of 0.56 M CaCl2 and 3.4 mL of 1.7 M Na3PO4: The CaCl, can make moles of Ca3(PO4)2 The Na;PO4 can make moles of Ca3(PO4)2 The limiting reagent in this mixture is The theoretical yield of Ca3(PO4)2 in this reaction mixture is moles If 0.85 g of Ca3(PO4)2 (M = 310.18 g/mol) was isolated, the percent reaction yield is

Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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26. 3CaCl,(ag) + 2Na3PO4(ag) → Ca3(PO4)2(s) + 6NAC((ag) In a reaction mixture of 16 mL of 0.56 M CaCl2 and 3.4 mL of 1.7 M Na3PO4: The CaCl, can make moles of Ca3(PO4)2 The Na;PO4 can make moles of Ca3(PO4)2 The limiting reagent in this mixture is The theoretical yield of Ca3(PO4)2 in this reaction mixture is moles If 0.85 g of Ca3(PO4)2 (M = 310.18 g/mol) was isolated, the percent reaction yield is
Transcribed Image Text:26. 3CaCl,(ag) + 2Na3PO4(ag) → Ca3(PO4)2(s) + 6NAC((ag) In a reaction mixture of 16 mL of 0.56 M CaCl2 and 3.4 mL of 1.7 M Na3PO4: The CaCl, can make moles of Ca3(PO4)2 The Na;PO4 can make moles of Ca3(PO4)2 The limiting reagent in this mixture is The theoretical yield of Ca3(PO4)2 in this reaction mixture is moles If 0.85 g of Ca3(PO4)2 (M = 310.18 g/mol) was isolated, the percent reaction yield is
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