What mass, in g, of agcl is formed from the reaction of 75.0 ml of a 0.078 m agc2h3o2 solution with 55.0 ml of 0.109 m mgcl2 solution? 2 agc2h3o2 (aq) + mgcl2 (aq) â 2 agcl (s) + mg (c2h3o2) 2 (aq) ?

Given is 75.0 ml of a 0.078 m agc2h3o2 solution with 55.0 ml of 0.109 m mgcl2 solution. AgC2H3O2 = 75.0 mL x 0.078 M = 5.85 mmol MgCl2 = 55.0 mL x 0.109 M = 5.995 mmol since 5.995 mmol MgCl2 contains 11.99 mmol Cl - 5.85 mmol Ag + reacts with 5.85 mmol Cl and gives MgCl2 is in excess That is 5.85 mmol AgCl = > 0.00585 mol AgCl Now mass of AgCl = 0.00585 mol AgCl x 143.34 g/mol = 0.839 g of AgCl