2naoh(aq) + zn(no3)2(aq) → zn(oh)2(s) + 2nano3(aq)the molecular equation you determined in part b is shown above for your convenience. examine each of the chemical species involved to determine the ions that would be present in solution.be sure to consider both the coefficients and subscripts of the molecular equation, and then write this precipitation reaction in the form of a balanced complete ionic equation.express your answer as a chemical equation including phases. type an underscore (_) or a carat (^) to add subscripts and superscripts more quickly.

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AsiaWoerner AsiaWoerner · Answer 1 · 2018-09-30T16:52:46+02:00

The molecular equation is

2NaOH(aq) + Zn(NO_3)_2(aq) → Zn(OH)^2(s) + 2NaNO_3(aq)

The net ionic equation will not include the spectator ions it will only includes the other ions

So first let us consider the balanced ionic reaction

2Na^+ + 2OH^-(aq) + Zn^+2 + 2NO3^-1 (aq) → Zn(OH)_2(s) + 2Na^+ + 2NO3-(aq)

spectator ions are

Na+ and NO3^-

So net ionic equation will be

2OH^- (aq) + Zn^+2 → Zn(OH)_2(s)

IlaMends IlaMends · Answer 2 · 2019-03-26T08:55:54+01:00

Explanation:

The given molecular equation:

[tex]2NaOH(aq)+Zn(NO_3)_2(aq)\rightarrow Zn(OH)_2(s)+2NaNO_3(aq)[/tex]

The sodium hydroxide will give sodium ion and hydroxide in their aqueous solution.

[tex]NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)[/tex]

The zinc nitrate will zinc ion and nitrate ions in their aqueous solution.

[tex]Zn(NO_3)_2(aq)\rightarrow Zn^{2+}(aq)+2NO_{3}^{-}(aq)[/tex]

So the molecular equation can be rewrite as:

[tex]2Na^+(aq)+2OH^-(aq)+Zn^{2+}(aq)+2NO_{3}^{-}(aq)\rightarrow Zn(OH)_2(s)+2Na^+(aq)+2NO_{3}^{-}(aq)[/tex]

Spectator ions will get cancelled out from the the sides and the net ionic equation comes out to be:

[tex]Zn^{2+}(aq)+2OH{-}(aq)\rightarrow Zn(OH)_2(s)[/tex]