Question

Amount of silver chloride precipitated, when
100 mL 0.1 M AgNO, was added to 200 mL 0.05 M
NaCl solution, is ?​

Answer 1

The reaction taking place is as follows:

      AgNO3 + NaCl → NaNO3 + AgCl

Number of moles of AgNO3 = M x V

                                              = 0.05mol / L x 0.1 L

                                              = 0.005 mol

Number of moles of NaCl     = M x V

                                              = 0.1 mol / L x 0.1 L

                                              = 0.01 mol

Thus,

     More AgNO3 is present than required.

Hence,

     NaCl is the limiting reagent.

Number of moles of AgCl obtained = 0.01 mol

Hence,

     Mass of AgCl formed = 0.01 x 143.5

                                         = 1.435 g AgCl