BrO3⊖+5Br⊖⟶Br2+3H2OIf 50 mL 0.1 M BrO3⊖ is mixed with 30 mL of 0.5 M Br ⊖solution that contains excess of H⊕ ions, the moles of Br2 formed are

Balance the equation:                    BrO3⊖     +     5Br⊖          →   3Br2+3H2O                     1 mmol            5 mmol               1 mmol  Given:        50 x 0.1          30 x 0.5                     = 5 mmol       =15 mmolBr⊝ is the limiting reagent.5 mmol of Br⊝ gives ⇒3 mmol of Br215 mmol of Br⊝ gives ⇒3×155=9 mmol = 9 x 10-3 mol