# Question f9544

Jun 8, 2016

Yes, the reaction is possible.

#### Explanation:

You're dealing with a double replacement reaction in which ammonium carbonate, ("NH"_4)_2"CO"_3, and cobalt(II) chloride, ${\text{CoCl}}_{2}$, react to form cobalt(II) carbonate, ${\text{CoCO}}_{3}$, and insoluble solid, and aqueous ammonium chloride, $\text{NH"_4"Cl}$.

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

${\text{CoCl"_ (2(aq)) -> "Co"_ ((aq))^(2+) + 2"Cl}}_{\left(a q\right)}^{-}$

When the two solutions are mixed, the cobalt(II) cations, ${\text{Co}}^{2 +}$, will combine with the carbonate anions ,${\text{CO}}_{3}^{2 -}$, to form the insoluble solid which then precipitates out of solution.

The complete ionic equation for this reaction will look like this

$2 {\text{NH"_ (4(aq))^(+) + "CO"_ (3(aq))^(2-) + "Co"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) -> "CoCO"_ (3(s)) darr + 2"NH"_ (4(aq))^(+) + 2"Cl}}_{2 \left(a q\right)}^{-}$

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

$\textcolor{red}{\cancel{\textcolor{b l a c k}{2 {\text{NH"_ (4(aq))^(+)))) + "CO"_ (3(aq))^(2-) + "Co"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) -> "CoCO"_ (3(s)) darr + color(red)(cancel(color(black)(2"NH"_ (4(aq))^(+)))) + color(red)(cancel(color(black)(2"Cl}}_{\left(a q\right)}^{-}}}}$

to get the net ionic equation

${\text{Co"_ ((aq))^(2+) + "CO"_ (3(aq))^(2-) -> "CoCO}}_{3 \left(s\right)} \downarrow$

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

Therefore, you will have

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