Answer:
14.33 g
Explanation:
Solve this problem based on the stoichiometry of the reaction.
To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and  finally calculate the mass of AgCl.
           2 AgNO₃  + CaCl₂   ⇒  Ca(NO₃)₂  + 2 AgCl
mass, g          6.97     6.39                   ?
MW ,g/mol     169.87    110.98                  143.32
mol =m/MW Â Â Â Â Â 0.10 Â Â Â Â 0.06 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.10
From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :
0.10 mol x 143.32 g/mol = 14.33 g
