Question

calculate the mass in grams of AgCl that can be formed fron the reaction of 85.0 mL of 0.110 M MgCl2 solution with 65.0 mL of a 0.150 M AgC2H3O2 solution.
MgCl2(aq) + AgC2H3O2 (aq) -> AgCl(s) + Mg(C2H3O2)2 (aq)

Answer 1

Reaction taking place; MgCl2 +2CH3Coog -29C1 +(Chy ces), mg Molar , mass. 143.329/mol 85 mL of o. 110mAgCl, 85m2x o llomol 9.35 mmol of Mg42

65mL of 0.15M Ag CH₂EDO 65 my x 0.15 mol & 9.75 mind of CH3 CooAg

2AgCl +(CH₂ (oo, My MgCh + 2 CH₂ Coong 9.35 9.75 mmol mmol It is clear that CHz CooAg is the limiting reagent and decides yield of Aga 2 mol of Chy Coodg → 2mol of Agc1 : 9.75 mmol 1 giver 9.75 monol Aga mass = moles x molar mass ... = 9.75 X 143.32 mmol 9 = 1402.245 mg I mass=1.402g] . 1.402 g of Agl. can be produced from the reaction