Question

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.

Answer 1

Answer: The mass of AgI in the precipitate is 9.55 g.

Explanation:

The chemical equation for this reaction is as follows.

 $Ag^{+}(aq) + Hg^{2+}(aq) + 3I^{-}(aq) \rightarrow AgI(s) + HgI_{2}(s)$

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 $I^{-}$

or,              x = $\frac{0.122}{3}$

                    = 0.0407 mol $I^{-}$

So,    $0.0407 moles I^{-} \times (\frac{126.9 g I^{-}}{1 mole I^{-}})$

                   = 5.16 g $I^{-}$

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

    0.0407 moles $I^{-} \times (\frac{1 mole Ag^{+}}{1 mole I^{-}})$  

                = 0.0407 moles $Ag^{+}$

   $0.0407 moles Ag^{+} \times (\frac{107.9 g Ag}{1 mole Ag})$

               = 4.39 g $Ag^{+}$

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

      Mass of AgI = mass of $Ag^{+}$ + mass of $I^{-}$  

                           = 5.16 + 4.39  

                           = 9.55 g AgI

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