Question

Question #d3b6a

Answer 1

Here's what I got.

Explanation:

Barium nitrate, `"Ba"("NO"_3)_2`, and sodium fluoride, `"NaF"`, are soluble ionic compounds, which means that they exist as ions in aqueoues solutions.

`"Ba"("NO"_ 3)_ (2(aq)) -> "Ba"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-)`

`"NaF"_ ((aq)) -> "Na"_ ((aq))^(+) + "F"_ ((aq))^(-)`

When two aqueous solutions of barium nitrate and sodium fluoride are mixed, two ionic compounds are formed

• barium fluoride, `"BaF"_2`, an insoluble ionic compound that precipitates out of solution
• sodium nitrate, `"NaNO"_3`, a soluble ionic compound that exists as ions in the resulting solution

You can write the unbalanced chemical equation that describes this double replacement reaction like this

`"Ba"("NO"_ 3)_ (2(aq)) + "NaF"_ ((aq)) -> "BaF"_ (2(s)) darr + "NaNO"_ (3 (aq))`

To balance this chemical equation, add a coefficient of `2` to sodium fluoride

`color(darkgreen)(ul(color(black)("Ba"("NO"_ 3)_ (2(aq)) + 2"NaF"_ ((aq)) -> "BaF"_ (2(s)) darr + "NaNO"_ (3 (aq)))))`

Now, you can start from the balanced chemical equation and write the complete ionic equation

`"Ba"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + 2 xx ["Na"_ ((aq))^(+) + "F"_ ((aq))^(-)] -> "BaF"_ (2(s)) darr + 2 xx ["Na"_ ((aq))^(+) + "NO"_ (3(aq))^(-)]`

This is equivalent to

`"Ba"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + 2"Na"_ ((aq))^(+) + 2"F"_ ((aq))^(-) -> "BaF"_ (2(s)) darr + 2"Na"_ ((aq))^(+) + 2"NO"_ (3(aq))^(-)`

In order to get the net ionic equation, you must remove the spectator ions, i.e. the ions that are present on both sides of the equation.

`"Ba"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-)))) + color(red)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + 2"F"_ ((aq))^(-) -> "BaF"_ (2(s)) darr + color(red)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-))))`

This will get you

`color(darkgreen)(ul(color(black)("Ba"_ ((aq))^(2+) + 2"F"_ ((aq))^(-) -> "BaF"_ (2(s)) darr)))`