# How do you write the balanced chemical, complete ionic and net ionic equation for the reaction between aqueous solutions of sodium carbonate and calcium chloride?

Jan 27, 2016

Here's what I got.

#### Explanation:

You're dealing with a double replacement reaction in which two soluble ionic compounds react in aqueous solution to form an Insoluble solid.

As you know, soluble ionic compounds exist as ions in solution. In your case, a solution of sodium carbonate, ${\text{Na"_2"CO}}_{3}$, will contain

${\text{Na"_2"CO"_text(3(aq]) -> 2"Na"_text((aq])^(+) + "CO}}_{\textrm{3 \left(a q\right]}}^{2 -}$

Likewise, a solution of calcium chloride, ${\text{CaCl}}_{2}$, will contain

${\text{CaCl"_text(2(aq]) -> "Ca"_text((aq])^(2+) + 2"Cl}}_{\textrm{\left(a q\right]}}^{-}$

When these two solutions are mixed, the calcium cations, ${\text{Ca}}^{2 +}$, will apir up with the carbonate anions, ${\text{CO}}_{3}^{2 -}$, and form the Insoluble calcium carbonate, ${\text{CaCO}}_{3}$, which precipitates out of solution.

The complete ionic equation, which features all the ions that are present in solution, will look like this

$2 {\text{Na"_text((aq])^(+) + "CO"_text(3(aq])^(2-) + "Ca"_text((aq])^(2+) + 2"Cl"_text((aq])^(-) -> "CaCO"_text(3(s]) darr + 2"Na"_text((aq])^(+) + 2"Cl}}_{\textrm{\left(a q\right]}}^{-}$

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will look like this

${\text{CO"_text(3(aq])^(2-) + "Ca"_text((aq])^(2+) -> "CaCO}}_{\textrm{3 \left(s\right]}} \downarrow$

To get the overall balanced chemical equation, simply look at he complete ionic equation and group the ions back to an ionic compound

overbrace(2"Na"_text((aq])^(+) + "CO"_text(3(aq])^(2-))^(color(red)("Na"_2"CO"_3)) + overbrace("Ca"_text((aq])^(2+) + 2"Cl"_text((aq])^(-))^(color(blue)("CaCl"_2)) -> "CaCO"_text(3(s]) + overbrace(2"Na"_text((aq])^(+) + 2"Cl"_text((aq])^(-))^(color(green)(2"NaCl"))

Notice that you have two sodium cations and two chloride anions on the products' side, which is why you have $2 \text{NaCl}$.

${\text{Na"_2"CO"_text(3(aq]) + "CaCl"_text(2(aq]) -> "CaCO"_text(3(s]) darr + 2"NaCl}}_{\textrm{\left(a q\right]}}$