Question

How would you find the net ionic equation of `"HCl" + "Zn" -> "H"_2 + "ZnCl"_2` ?

This information was given along with the question:

A piece of zinc metal is placed in a 1.0 M solution of hydrochloric acid at 25 C°.

Answer 1

`"Zn"_ ((s)) + 2"H"_ ((aq))^(+) -> "Zn"_ ((aq))^(2+) + "H"_ (2(g))`

Explanation:

For starters, make sure that you have a balanced equation to work with. To balance the equation given to you, multiply the hydrochloric acid by `2`

`"Zn"_ ((s)) + color(blue)(2)"HCl"_ ((aq)) -> "ZnCl"_ (2(aq)) + "H"_ (2(g))`

You know that hydrochloric acid is a strong acid, which means that it dissociates completely in aqueous solution to produce hydrogen ions, `"H"^(+)`, and chloride anions

`"HCl"_ ((aq)) -> "H"_ ((aq))^(+) + "Cl"_ ((aq))^(-)`

Now ,zinc chloride, `"ZnCl"_2`, is soluble in aqueous solution, which means that it too will exist as ions

`"ZnCl"_ (2(aq)) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)`

This means that you can write

`"Zn"_ ((s)) + color(blue)(2) xx ["H"_ ((aq))^(+) + "Cl"_ ((aq))^(-)] -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + "H"_ (2(g))`

This is equivalent to

`"Zn"_ ((s)) + 2"H"_ ((aq))^(+) + 2"Cl"_ ((aq))^(-) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + "H"_ (2(g))`

Now, the net ionic equation is obtained by removing the spectator ions, i.e. the ions that are present on both sides of the equation.

In this case, you have

`"Zn"_ ((s)) + 2"H"_ ((aq))^(+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) -> "Zn"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) + "H"_ (2(g))`

The net ionic equation that describes this single replacement reaction will thus be

`color(darkgreen)(ul(color(black)("Zn"_ ((s)) + 2"H"_ ((aq))^(+) -> "Zn"_ ((aq))^(2+) + "H"_ (2(g)))))`