NalO3 is the limiting reagent in this synthesis. Since 2 mol Nalo, =1 mol Ca(103), (from Equation 1) 1 mol Ca(IO, ), 6 HO Theoretical yield of Ca(10), 6 H, 0 = mol NaO x - xMola mess of Ca(10),6 7,0 2 mol Nalo Mass of Ca(10), 6 H,0 produced 2. Percent yield of Ca(10,), 6 H,0 = Theoretical yield of Cal 103), 6 H,0 2100 Table 1. Experimental Data Mass of CaCl, 2H20 used (g) 0.8359 Molar mass of CaCl, 2H,01 g/mol) Mass of Nalo, used (g) 2.0059 Molar mass of Nalo; (g/mol) Mass of Ca(10,), 6 H,0 produced (g) 2.003g Molar mass of CaI0), 6H,0 (gr mol ) Table 2. Calculated Values Theoretical yield of Ca(103), 6 H,0 (9) Percent vield of Ca(TO. 16 H.O