write a net ionic equation for the following: Copper (II) sulfate reacts with Ammonium hydroxide to produce Copper (II) oxide, Ammonium sulfate, and water.
CuSO4 + 2NH4OH → CuO + (NH4)2SO4 + H2O
To write the net ionic equation for the reaction between copper (II) sulfate and ammonium hydroxide, we first need to write the balanced molecular equation.
The balanced molecular equation is as follows:
CuSO4 + 2NH4OH -> Cu(OH)2 + (NH4)2SO4
Now, let's break down the equation into its ionic components. Only ions that undergo a change in the reaction are included in the net ionic equation.
The complete ionic equation would look like this:
Cu2+ + SO4^2- + 2NH4+ + 2OH- -> Cu(OH)2 + 2NH4+ + SO4^2-
Next, we cancel out the spectator ions, which are the ions that appear on both sides of the equation and have the same charge and number of atoms. In this case, the ammonium ions (NH4+) and sulfate ions (SO4^2-) are spectator ions, as they are present on both sides of the equation.
The remaining net ionic equation is:
Cu2+ + 2OH- -> Cu(OH)2
Therefore, the net ionic equation for the reaction is Cu2+ + 2OH- -> Cu(OH)2.
To write the net ionic equation for the given reaction, we need to first write the balanced molecular equation and then convert it into the net ionic equation.
The balanced molecular equation for the reaction is:
CuSO4 + 2NH4OH → CuO + (NH4)2SO4 + H2O
To write the net ionic equation, we need to break down all the aqueous ionic compounds into their individual ions and eliminate the spectator ions that remain unchanged on both sides of the equation.
Cu^2+ + SO4^2- + 2NH4+ + 2OH- → CuO + (NH4)2SO4 + 2H2O
The net ionic equation for the reaction is:
Cu^2+ + 2OH- → CuO + H2O