Question

A 1.00 mL solution containing 0.400g BaCl2-2H20 is mixed with 50.00 mL of a solution containing 0.300 g NalO3. Determine the mass of Ba(IO3)2 formed. Ksp(Ba(IO3)2) = 1.6x10-9

Answer 1

Given weight of Ball 2420 = 0.4oog Molar mass of Bach. 2420 = 244.269 l mal mol of Bach, PH2O = given weight Molar mass mol = = 0.00164 moj 0.yg 244.26gmoln NAIO3 = 0.33 0.29 1 = 0.00152 mol 199.89 9 moi Reaction between Bacle 8 Nq 103 Bauz (oup.) +mW9103 cap.) → Ba(IO3), + anece Imola Baca react with 2 mol Na IO₃ 0.00164 mol Bacle react with 0.00328 Mol N 9103 But mol g Na I03 available = 0.00152 mol so NqI03 is limting reagent and consume completely in rachion. and reaction proceed according to Na 103 2 mol of N9I03 give 1 mol Ba (103) 0.00152 mol N9103 give 0.00076 mol 84(103) molg B2 (103)ą produced = 0.00076 mol mang Ba(103), = mol X molar mas = 0.00076 Dzot X 487013 à = 0.3709 mans of Precipitate = 0.370g Answer