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.

Step-by-step 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 =
`(0.122)/(3)`

= 0.0407 mol
`I^(-)`

So,
`0.0407 moles I^(-) * ((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^(-) * ((1 mole Ag^(+))/(1 mole I^(-)))`

= 0.0407 moles
`Ag^(+)`

`0.0407 moles Ag^(+) * ((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.