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KOH(aq) + CuSO4(aq) write the general and ionic equations for the reactions

Sagot :

Answer

The general equation for the reaction is:

[tex]2KOH\left(aq\right)+CuSO_4\left(aq\right)\rightarrow K_2SO_4(aq)+Cu(OH)_2(s)[/tex]

The ionic equation for the reaction is:

[tex]2K^+(aq)+2OH^-(aq)+Cu^{2+}(aq)+SO_4^{2-}\rightarrow2K^+(aq)+SO_4^{2-}+Cu(OH)_2(s)[/tex]

Explanation

Given unbalanced equation: KOH(aq) + CuSO4(aq)

The general equation for the reaction is:

[tex]2KOH\left(aq\right)+CuSO_4\left(aq\right)\rightarrow K_2SO_4(aq)+Cu(OH)_2(s)[/tex]

The reaction type is a double displacement reaction.

The ionic equation for the reaction is:

[tex]2K^+(aq)+2OH^-(aq)+Cu^{2+}(aq)+SO_4^{2-}\rightarrow2K^+(aq)+SO_4^{2-}+Cu(OH)_2(s)[/tex][tex]2K^+(aq)+2OH^-(aq)+Cu^{2+}(aq)+SO_4^{2-}\rightarrow2K^+(aq)+SO_4^{2-}+Cu(OH)_2(s)[/tex]

Cross out the spectators' ions on both sides of the complete ionic equation to get the net ionic equation as shown below:

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