What are the correct and accurate products of the following neutralization reaction. Then write the ionic form of the equation and the net ionic form.

KOH + H2SO4 ->

2KOH(aq) + H2SO4(aq) -> K2SO4(aq) + H2O(l)

Can you take it from here.
Convert the balanced molecular equation above to an ionic equation; e.g. KOH would be 2K^+*(aq) + OH^-(aq)
H2SO4 would be 2H^+(aq) + SO4^2-(aq)
For the net ionic equation, just cancel ions common to both sides.

To determine the correct and accurate products of a neutralization reaction, we need to identify the reactants and the products. In this case, the reactants are KOH (potassium hydroxide) and H2SO4 (sulfuric acid).

First, we can write the balanced equation for the reaction:

2KOH + H2SO4 -> K2SO4 + 2H2O

To write the ionic form of the equation, we need to separate the reactants and products into their individual ions (cation and anion).

KOH: K+ (cation) + OH- (anion)
H2SO4: 2H+ (cation) + SO4^2- (anion)
K2SO4: 2K+ (cation) + SO4^2- (anion)
2H2O: 4H+ (cation) + 4OH- (anion)

Therefore, the ionic form of the equation would be:

2K+ (aq) + 2OH- (aq) + 2H+ (aq) + SO4^2- (aq) -> 2K+ (aq) + SO4^2- (aq) + 4H2O (l)

Finally, to write the net ionic form of the equation, we need to remove the spectator ions, which are the ions that appear on both sides of the equation and do not participate in the reaction.

The spectator ions in this reaction are the potassium ions (K+), as they appear on both sides of the equation. Therefore, we can remove them from the equation.

The net ionic form of the equation would be:

2OH- (aq) + 2H+ (aq) + SO4^2- (aq) -> SO4^2- (aq) + 4H2O (l)

Note that the charges in the net ionic equation must balance, so we have 2 hydroxide ions (OH-) on the left side and 4 hydrogen ions (H+) on the right side, which balances the equation.