2Na(s) + 2H2O(l) ==> 2Na^+(aq) + 2OH^-(aq) + H2(g)

What is the the net ionic equation because it doesn't look like their is any spectator ions.

That is the net ionic equation.

To write the net ionic equation, we need to show only the species that participate in the reaction and exclude any spectator ions.

In the given reaction:

2Na(s) + 2H2O(l) --> 2Na^+(aq) + 2OH^-(aq) + H2(g)

The sodium (Na) and hydroxide (OH^-) ions are involved in the reaction. However, the sodium ion (Na^+) is a spectator ion, meaning it does not change during the reaction.

Therefore, the net ionic equation is:

2OH^-(aq) + H2O(l) --> 2OH^-(aq) + H2(g)

To determine the net ionic equation, we need to first write the balanced molecular equation and then break down the equation into its ionic forms. After that, we can eliminate the spectator ions to get the net ionic equation.

The balanced molecular equation for the given reaction is:

2Na(s) + 2H2O(l) → 2Na^+(aq) + 2OH^-(aq) + H2(g)

Breaking down the equation into its ionic forms, we have:

2Na(s) → 2Na^+(aq) + 2e^-
2H2O(l) → 2H^+(aq) + 2OH^-(aq)
H2(g) → 2H^+(aq) + 2e^-

To eliminate the spectator ions, we can cancel out the common ions that appear on both sides of the equation. In this case, we see that 2Na^+(aq) and 2OH^-(aq) appear on both sides of the equation and can be canceled out. The resulting net ionic equation is:

2H2O(l) → 2H^+(aq) + H2(g)

Therefore, the net ionic equation for the given reaction is 2H2O(l) → 2H^+(aq) + H2(g).