Question #f9544

You're dealing with a double replacement reaction in which ammonium carbonate, `("NH"_4)_2"CO"_3`, and cobalt(II) chloride, `"CoCl"_2`, react to form cobalt(II) carbonate, `"CoCO"_3`, and insoluble solid, and aqueous ammonium chloride, `"NH"_4"Cl"`.

Both ammonium carbonate and cobalt(II) chloride are soluble in aqueous solution, which means that you can write them as

`("NH"_ 4)_ 2 "CO"_ (3(aq)) -> 2"NH"_ (4(aq))^(+) + "CO"_ (3(aq))^(2-)`

and

`"CoCl"_ (2(aq)) -> "Co"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)`

When the two solutions are mixed, the cobalt(II) cations, `"Co"^(2+)`, will combine with the carbonate anions ,`"CO"_3^(2-)`, to form the insoluble solid which then precipitates out of solution.

The complete ionic equation for this reaction will look like this

`2"NH"_ (4(aq))^(+) + "CO"_ (3(aq))^(2-) + "Co"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) -> "CoCO"_ (3(s)) darr + 2"NH"_ (4(aq))^(+) + 2"Cl"_ (2(aq))^(-)`

As you can see, some of the ions are present on both sides of the equation. These ions are called spectator ions and can be removed from the equation

`color(red)(cancel(color(black)(2"NH"_ (4(aq))^(+)))) + "CO"_ (3(aq))^(2-) + "Co"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) -> "CoCO"_ (3(s)) darr + color(red)(cancel(color(black)(2"NH"_ (4(aq))^(+)))) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-))))`

to get the net ionic equation

`"Co"_ ((aq))^(2+) + "CO"_ (3(aq))^(2-) -> "CoCO"_ (3(s)) darr`

It's worth noting that cobalt(II) carbonate is light purple in color.

![https://www.flickr.com/photos/37388341@N00/1508161448]()

Therefore, you will have

`("NH"_ 4)_ 2 "CO"_ (3(aq)) + "CoCl"_ (2(aq)) -> "CoCO"_ (3(s)) darr + 2"NH"_ 4"Cl"_ ((aq))`