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If nickel is added to copper sulfate solution, the nickel will replace the copper. write the balanced equation

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

(a) The given balanced ionic equation is,

Ni(NO_3)_2(aq)+Na_2S(aq)rightarrow NiS(s)+2NaNO_3(aq)

The ionic equation in separated aqueous solution will be,

Ni^{2+}(aq)+2NO_3^-(aq)+2Na^+(aq)+S^{2-}(aq)rightarrow NiS(s)+2Na^+(aq)+2NO_3^-(aq)

In this equation, Na^+text{ and }NO_3^- are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

Ni^{2+}(aq)+S^{2-}(aq)rightarrow NiS(s)

(b) The given balanced ionic equation is,

NaNO_3(aq)+KBr(aq)rightarrow KNO_3(s)+NaBr(aq)

The ionic equation in separated aqueous solution will be,

Na^+(aq)+NO_3^-(aq)+K^+(aq)+Br^-(aq)rightarrow KNO_3(s)+Na^+(aq)+Br^-(aq)

In this equation, Na^+text{ and }Br^- are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

NO_3^-(aq)+K^+(aq)rightarrow KNO_3(s)

(c) The given balanced ionic equation is,

Li_2SO_4(aq)+BaCl_2(aq)rightarrow BaSO_4(s)+2LiCl(aq)

The ionic equation in separated aqueous solution will be,

2Li^+(aq)+SO_4^{2-}(aq)+Ba^{2+}(aq)+2Cl^-(aq)rightarrow BaSO_4(s)+2Li^+(aq)+2Cl^-(aq)

In this equation, Li^+text{ and }Cl^- are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

SO_4^{2-}(aq)+Ba^{2+}(aq)rightarrow BaSO_4(s)

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