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

Question

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°.

Noah Aldrich · Accepted Answer

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

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 implies that you are capable of writing.

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

This is comparable to

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

The spectator ions, or the ions that are present on both sides of the equation, are now removed in order to obtain the net ionic equation.

Here, you've

#"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))#

Consequently, the net ionic equation for this one replacement reaction will be

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

Kai Bowman · Answer

undefined The balanced molecular equation is written first, then the complete ionic equation is written, dissecting all soluble ionic compounds into their individual ions: H⁺(aq) + Cl⁻(aq) + Zn(s) -> H₂(g) + Zn²⁺(aq) + 2Cl⁻(aq). Lastly, spectator ions (ions that appear on both sides of the equation without changing) are eliminated from the complete ionic equation to obtain the net ionic equation: Zn(s) -> Zn²⁺(aq) + 2e

How would you find the net ionic equation of #"HCl" + "Zn" -&gt; "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°.

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°.