Question

If 45.0 mL of 0.250 M CaCl₂ is added to 17.5 mL of 1.00 M AgNO₃, what is the mass of the AgCl precipitate?

Answer 1

- 2 AgNO3 (42) + Caldelas) y 2 Ag (18+ Ca(NO3)2 (ag) - No of moles= Molarity & Volume o No of moles of Calle= 0.250 ml x 45.0 mL = 0.250 melle x 4.5.0 Xtól (le m= 163) = 0.01125 mol I No of moles of AgNO₃ = 1.00 mallLy 17.503 L = 0.0175 mol - I E I I In the balanced oxh, 1 male Calle requires 2 mole AgNO₃ to react Completely 8 0.0125 mol Calle should require 0.01125x2= 0.0225 mole AgNO3. But as no. of moles of AgNO3 available are less so AgNO₃ is the limiting Reagent and thus Agnez gives the amount of product Cie. precipitate of Ari) that can be formed g moke AgNO3 gives 2 mole Agci t No of moles of Age formed = 0.0175 mal i Mass of AgCl precipitated= No of moles x Molar mass = 0.0175 mal* 143.32 g ml E = = gesig