# 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°.

Dec 20, 2016

${\text{Zn"_ ((s)) + 2"H"_ ((aq))^(+) -> "Zn"_ ((aq))^(2+) + "H}}_{2 \left(g\right)}$

#### 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 $2$

${\text{Zn"_ ((s)) + color(blue)(2)"HCl"_ ((aq)) -> "ZnCl"_ (2(aq)) + "H}}_{2 \left(g\right)}$

You know that hydrochloric acid is a strong acid, which means that it dissociates completely in aqueous solution to produce hydrogen ions, ${\text{H}}^{+}$, and chloride anions

${\text{HCl"_ ((aq)) -> "H"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

Now ,zinc chloride, ${\text{ZnCl}}_{2}$, is soluble in aqueous solution, which means that it too will exist as ions

${\text{ZnCl"_ (2(aq)) -> "Zn"_ ((aq))^(2+) + 2"Cl}}_{\left(a q\right)}^{-}$

This means that you can write

${\text{Zn"_ ((s)) + color(blue)(2) xx ["H"_ ((aq))^(+) + "Cl"_ ((aq))^(-)] -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + "H}}_{2 \left(g\right)}$

This is equivalent to

${\text{Zn"_ ((s)) + 2"H"_ ((aq))^(+) + 2"Cl"_ ((aq))^(-) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + "H}}_{2 \left(g\right)}$

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

${\text{Zn"_ ((s)) + 2"H"_ ((aq))^(+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) -> "Zn"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) + "H}}_{2 \left(g\right)}$

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

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{Zn"_ ((s)) + 2"H"_ ((aq))^(+) -> "Zn"_ ((aq))^(2+) + "H}}_{2 \left(g\right)}}}}$