# Question #4f73e

Mar 2, 2016

${\text{BaCl"_(color(red)(2)text((aq])) + "Na"_color(red)(2)"SO"_text(4(aq]) -> color(red)(2)"NaCl"_text((aq]) + "BaSO}}_{\textrm{4 \left(s\right]}} \downarrow$

#### Explanation:

Your reactants, barium chloride, ${\text{BaCl}}_{2}$, and sodium sulfate, ${\text{Na"_2"SO}}_{4}$, are soluble ionic compounds that exist as ions in aqueous solution.

On the other hand, only one of the two products, sodium chloride, $\text{NaCl}$, is soluble in aqueous solution. The other product, barium sulfate, ${\text{BaSO}}_{4}$, is insoluble and will precipitate out of solution as the reaction progresses.

You can thus say that you're dealing with a double replacement reaction in which two soluble ionic compounds react to form an insoluble solid.

The unbalanced chemical equation looks like this

${\text{BaCl"_(color(red)(2)text((aq])) + "Na"_color(red)(2)"SO"_text(4(aq]) -> "NaCl"_text((aq]) + "BaSO}}_{\textrm{4 \left(s\right]}} \downarrow$

Right from the start, you can say that since the reactants' side contains $\textcolor{red}{2}$ chloride anions and sodium cations, multiplying the sodium chloride, $\text{NaCl}$, by $\textcolor{red}{2}$ will balance these ions out.

$\textcolor{g r e e n}{{\text{BaCl"_(color(red)(2)text((aq])) + "Na"_color(red)(2)"SO"_text(4(aq]) -> color(red)(2)"NaCl"_text((aq]) + "BaSO}}_{\textrm{4 \left(s\right]}} \downarrow}$

This will be the overall balanced chemical equation.

It could be easier for you to write the unbalanced complete ionic equation first. Use the fact that three out of the four compounds involved in the reaction are soluble

${\text{Ba"_text((aq])^(2+) + color(red)(2)"Cl"_text((aq])^(-) + color(red)(2)"Na"_text((aq])^(+) + "SO"_text(4(aq])^(2-) -> "Na"_text((aq])^(+) + "Cl"_text((aq])^(-) + "BaSO}}_{\textrm{4 \left(s\right]}} \downarrow$

You can once again see that the reactants' side contains $\textcolor{red}{2}$ chloride anions, ${\text{Cl}}^{-}$, and $\textcolor{red}{2}$ sodium cations, ${\text{Na}}^{+}$, but that the products' side contains only $1$ of these two ions.

To balance these out, multiply the compound that contains both ions on the products' side, sodium chloride, by $\textcolor{red}{2}$.

${\text{Ba"_text((aq])^(2+) + color(red)(2)"Cl"_text((aq])^(-) + color(red)(2)"Na"_text((aq])^(+) + "SO"_text(4(aq])^(2-) -> color(red)(2)overbrace(("Na"_text((aq])^(+) + "Cl"_text((aq])^(-)))^(color(purple)("sodium chloride")) + "BaSO}}_{\textrm{4 \left(s\right]}}$ $\downarrow$

This will once again get you

$\textcolor{g r e e n}{{\text{BaCl"_(color(red)(2)text((aq])) + "Na"_color(red)(2)"SO"_text(4(aq]) -> color(red)(2)"NaCl"_text((aq]) + "BaSO}}_{\textrm{4 \left(s\right]}} \downarrow}$

The net ionic equation, which doesn't show spectator ions, will look like this

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

Barium sulfate is a white insoluble solid that precipitates out of solution.