Question

What mass of silver chloride can be precipitated from a silver nitrate solution by 200mL of a solution of 0.50 M CaCl2?

Answer 1

reaction is

2AgNO₃ + CaCl₂    2AgCl + Ca(NO₃)₂

calculate mole of CaCl₂

no. of mole = molarity volume of solution in liter

no. of mole of CaCl₂  = 0.50 M 0.200 L = 0.1 mole

According to reaction 1 mole of CaCl₂ produce 2 mole of AgCl then 0.1 mole of CaCl₂ produce 0.2 mole of AgCl

molar mass of AgCl = 143.32 gm/mole that mean 1 mole of AgCl = 143.32 gm then 0.2 mole of AgCl =

143.32 0.2 = 28.664 gm

28.664 gm mass of silver chloride can be precipitated from a silver nitrate solution by 200mL of a solution of 0.50 M CaCl2