A solution contains Ag and Hg2 ions. The addition of 0.100 L of 1.71 M NaI solution is just enough to precipitate all the ions as AgI and HgI2. The total mass of the precipitate is 39.6 g . Find the mass of AgI in the precipitate. Express your answer to two significant figures and include the appropriate units.

The mass of AgI in the precipitate is 9.55 g.

Explanation:

The chemical equation for this reaction is as follows.

So, we will calculate the moles of  added as follows.

     Moles = Molarity × Volume  

                = 1.71 \times 0.100

                = 0.171 moles  added

Let us assume that  consumes x moles of  and  consumes 2x moles of

               3x = 0.122 mol

or,              x =

                    = 0.0407 mol

So,    

                   = 5.16 g

Hence, the formula AgI the mole ratio of  to  is 1:1.

    0.0407 moles  

                = 0.0407 moles

               = 4.39 g

Therefore, we will calculate the mass of AgI as follows.

      Mass of AgI = mass of + mass of  

                           = 5.16 + 4.39  

                           = 9.55 g AgI

Therefore, we can conclude that the mass of AgI in the precipitate is 9.55 g.

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the correct answer is:

a. formation of precipitation