Question

Both Ca?and Biº form insoluble ionic compounds with the phosphate ion (PO), Sodium phosphate a convenient source of this anion and we will use it to precipitate both of the cations we are trying to quantify 3Ca2+(aq) + 2Na3PO,(aq) Ca3(PO) (s) + 6Na+ B. Calt in Calcium Carbonate Antacids 1.1999 1. Mass of the Calcium carbonate tablet 2. Mass of Calcium in the antacid 204mg Ca 3. Theoretical Yield of Ca3(PO4)2 Show Calculations 4. Mass of the filter paper 0.305 22,775 5. Mass of the watch glass 6. Combined mass of the Ca3(PO4)2 filter paper and watch glass after drying. 7. Mass of the Ca3(PO4)2 Obtain from the class enerare curos 0.9606g 0.73009 0.8795 yours 8. Average Mass of the Ca3(PO4)2 9. Percent Yield of Ca,(PO4)2 actual yield percent yield - theoretical yield Tetical yield * 100 Show Calculations 10. Mass of Calcium in the Ca.(POA)

Most importantly, want to find theoretical yield of Ca_3 (PO_4)_2

Answer 1

Mass of calcium in antacid = 204 mg = 0.204 grams mass of I mole cate = 40.078 g/mol molar mass of calcium Now, we have to convert 0.204 grams into moles of Cath →

i mol cath = 40.078 g/mol x mol Cat2 = 0.204 g : 0.204 g x Imol Cath = 0.00509 mol. . 40.078 8 /mol - 0.204 g of Cat2 = 0.00509 mol of Cat2 3 Cath + 2Nq (104) Cą (PO4), (s) + 6 Na I caq) - (aq) (aq) 3 moles of Cath is giving Imole ca, (pow. solid (According to above equation) 0.00509 moles of Cath gives 0.00509 mol of C (pow.), solid & number of moles of caz (py), formed = 0.00509 = 0.0016967 moles. 1 3 ui

molar mass of caz (PO4) = 310.18 g/mol 0.0016967 mol of caz (Pou), = 0.0016967 mol x 310.18 g/mol = 0.5263 grams. Theoretical Yield of Coz (PO4) = 0.5263 grams

8) Average mass of Caz (PO4)2 = 0.9606 + 0-7300 + 0-879 = 0.8565 g. 9 percent yield = 0.8565 g x 100 0.5263 g = 162.74%