Question

Identify the spectator ions for the reaction that occurs when aqueous solutions of lithium sulfide and copper(ii) nitrate are mixed.
a.cu2+ and no3-
b.li+ and cu2+
c.li+ and no3-
d.s2- and cu2+
e.no spectator ions are present.

Answer 1

The answer: c) Li⁺ and NO₃⁻

Firstly, we have to show molecular equation (all compounds are showed as neutral molecules) for the reaction:

Li₂S(aq) + Cu(NO₃)₂(aq)  ⇒ CuS(s) + 2LiNO₃(aq)

Then, the reaction has to be written as an ionic equation:

2Li⁺(aq) + S²⁻(aq) + Cu²⁺(aq) + 2NO₃⁻(aq) → CuS(s) + 2Li⁺(aq) + 2NO₃⁻(aq)

Soluble ionic compounds Li₂S and Cu(NO₃)₂ are shown as dissociated ions. As you can see on both side of the equation we have Li⁺ and NO₃⁻ ions. They do not participate in the reaction, and they are called spectator ions. All the above is an ionic equation.

Finally, when we cancel out spectator ions, it remains just ions and molecules involved in the reaction and we get the net ionic equation:

S²⁻(aq) + Cu²⁺(aq) → CuS(s)

Answer 2

The spectator ions for the reaction between lithium sulfide and copper nitrate are $\boxed{{\text{c}}{\text{. L}}{{\text{i}}^ + }{\text{ and NO}}_3^ - }$.

Further Explanation:

Spectator ions:

These are the ions that exist in the same form on both sides of the reaction. These ions do not affect the equilibrium of any reaction.

The three types of equations that are used to represent the chemical reaction are as follows:

1. Molecular equation

2. Total ionic equation

3. Net ionic equation

The reactants and products remain in undissociated form in the molecular equation. In the case of the total ionic equation, all the ions that are dissociated and present in the reaction mixture are represented while in the case of the overall or net ionic equation only the useful ions that participate in the reaction are represented.

The steps to write the net ionic reaction are as follows:

Step 1: Write the molecular equation for the reaction with the phases in the bracket.

In the reaction, ${\text{L}}{{\text{i}}_2}{\text{S}}$ reacts with ${\text{Cu}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}$ to form CuS and ${\text{LiN}}{{\text{O}}_3}$. The balanced molecular equation of the reaction is as follows:

${\text{L}}{{\text{i}}_{\text{2}}}{\text{S}}\left( {aq} \right) + {\text{Cu}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}\left( {aq} \right) \to {\text{CuS}}\left( s \right) + 2{\text{LiN}}{{\text{O}}_3}\left( {aq} \right)$

Step 2: Dissociate all the compounds with the aqueous phase to write the total ionic equation. The compounds with solid and liquid phases remain the same. The total ionic equation is as follows:

$2{\text{L}}{{\text{i}}^ + }\left( {aq} \right) + {{\text{S}}^{2-}}\left({aq}\right)+{\text{C}}{{\text{u}}^{2+}}\left( {aq} \right) + 2{\text{NO}}_3^-\left({aq}\right)\to {\text{CuS}}\left(s\right)+2{\text{L}}{{\text{i}}^+}\left({aq}\right)+2{\text{NO}}_3^-\left( {aq}\right)$

Step 3: The common or spectator ions on both sides of the reaction get cancelled out to get the net ionic equation.

$\boxed{2{\text{L}}{{\text{i}}^ + }\left( {aq} \right)} + {{\text{S}}^{2 - }}\left( {aq} \right) + {\text{C}}{{\text{u}}^{2 + }}\left( {aq} \right) + \boxed{2{\text{NO}}_3^ - \left( {aq} \right)} \to {\text{CuS}}\left( s \right) + \boxed{2{\text{L}}{{\text{i}}^ + }\left( {aq} \right)} + \boxed{2{\text{NO}}_3^ - \left( {aq} \right)}$

${\text{L}}{{\text{i}}^ + }$ and ${\text{NO}}_3^ -$ ions are present in the same form on both the reactant and the product side and therefore are known as spectator ions.

Therefore, the net ionic equation is as follows:

${{\text{S}}^{2 - }}\left( {aq} \right) + {\text{C}}{{\text{u}}^{2 + }}\left( {aq} \right) \to {\text{CuS}}\left( s \right)$

Learn more:

1. Balanced chemical equation: https://brainly.com/question/1405182

2. Oxidation and reduction reaction: https://brainly.com/question/2973661

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: net ionic equation, spectator ions, CuS, Cu2+, Li+, NO3-, S2-, 2 Li+, 2 NO3-.