Question e8eae

Jun 19, 2017

Explanation:

Assuming that you're dealing with lead(II) nitrate, "Pb"("NO"_3)_2, you can say that this reaction will produce aqueous potassium nitrate and lead(II) iodide, an insoluble solid that precipitates out of solution.

The balanced chemical equation that describes this double replacement reaction looks like this

${\text{Pb"("NO"_ 3)_ (2(aq)) + 2"KI"_ ((aq)) -> "PbI"_ (2(s)) darr + 2"KNO}}_{3 \left(a q\right)}$

The complete ionic equation looks like this

${\text{Pb"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + 2"K"_ ((aq))^(+) + 2"I"_ ((aq))^(-) -> "PbI"_ (2(s)) darr + 2"K"_ ((aq))^(+) + 2"NO}}_{3 \left(a q\right)}^{-}$

The net ionic equation, which doesn't include the spectator ions, i.e. the ions that are present on both sides of the equation

"Pb"_ ((aq))^(2+) + color(red)(cancel(color(Black)(2"NO"_ (3(aq))^(-)))) + color(red)(cancel(color(black)(2"K"_ ((aq))^(+)))) + 2"I"_ ((aq))^(-) -> "PbI"_ (2(s)) darr + color(red)(cancel(color(black)(2"K"_ ((aq))^(+)))) + color(red)(cancel(color(Black)(2"NO"_ (3(aq))^(-))))#

will look like this

${\text{Pb"_ ((aq))^(2+) + 2"I"_ ((aq))^(-) -> "PbI}}_{2 \left(s\right)} \downarrow$

You can thus say that the reaction will produce lead(II) iodide, a yellow insoluble solid that will precipitate out of the solution. 