# How do you balance Na_2SO_4 + Ca(NO_3)_2 -> CaSO_4 + NaNO_3?

Dec 14, 2015

${\text{Na"_2"SO"_text(4(aq]) + "Ca"("NO"_3)_text(2(aq]) -> "CaSO"_text(4(s]) darr + 2"NaNO}}_{\textrm{3 \left(a q\right]}}$

#### Explanation:

Every time you're dealing with a double replacement reaction, you can make your life easier by looking at ions, rather than at atoms, when trying to balance the chemical equation.

A double replacement reaction is characterized by the reaction of the soluble ionic compounds and the formation of an insoluble solid that precipitates out of solution.

Sodium sulfate, ${\text{Na"_2"SO}}_{4}$, and calcium nitrate, "Ca"("NO"_3)_2, are both soluble ionic compounds, which means that they exist as ions in aqueous solution

${\text{Na"_2"SO"_text(4(aq]) -> color(red)(2)"Na"_text((aq])^(+) + "SO}}_{\textrm{4 \left(a q\right]}}^{2 -}$

${\text{Ca"("NO"_3)_text(2(aq]) -> "Ca"_text((aq])^(2+) + color(blue)(2)"NO}}_{\textrm{3 \left(a q\right]}}^{-}$

The reaction produces calcium sulfate, an insoluble solid that precipitates out of solution, and sodium nitrate, ${\text{NaNO}}_{3}$, another soluble ionic compound that exists as ions in solution

${\text{NaNO"_text(3(aq]) -> "Na"_text((aq])^(+) + "NO}}_{\textrm{3 \left(a q\right]}}^{-}$

Now, the idea here is that all the ions present on the reactants' side must be present, either as dissociated ions or as part of the solid, on the products' side.

Notice that the one calcium cation, ${\text{Ca}}^{2 +}$, and the one sulfate anion, ${\text{SO}}_{4}^{2 -}$, will group to form the insoluble calcium sulfate in a $1 : 1$ mole ratio, so you don't need to balance these two ions out.

On the other hand, notice that you have $\textcolor{red}{2}$ sodium cations on the reactants side, but only $1$ on the products' side. Likewise, you have $\textcolor{b l u e}{2}$ nitrate anions on the reactants' side, and once again only one on the products' side.

This tells you that will have to multiply the sodium nitrate by $\textcolor{g r e e n}{2}$ to balance these two ions out.

The complete chemical equation for this reaction will look like this

${\text{Na"_2"SO"_text(4(aq]) + "Ca"("NO"_3)_text(2(aq]) -> "CaSO"_text(4(s]) darr + color(green)(2)"NaNO}}_{\textrm{3 \left(a q\right]}}$

The net ionic equation, which doesn't include spectator ions, i.e. the ions that exist dissociated on both sides of the equation, will look like this

${\text{SO"_text(4(aq])^(2-) + "Ca"_text((aq])^(2+) -> "CaSO}}_{\textrm{4 \left(s\right]}} \downarrow$