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°.
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°.
1 Answer
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
#"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,
#"HCl"_ ((aq)) -> "H"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#
Now ,zinc chloride,
#"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)))))#