what is the net ionic equation for

AlBr3(aq) + 3LiOH(aq)

Al^3+(aq) + 3OH^-(aq) ==> Al(OH)3(s)

To find the net ionic equation for the reaction between AlBr3(aq) and 3LiOH(aq), we need to first write the balanced molecular equation and then convert it to the ionic equation.

Step 1: Write the balanced molecular equation
The balanced molecular equation for the reaction is:

AlBr3(aq) + 3LiOH(aq) → Al(OH)3(s) + 3LiBr(aq)

Step 2: Split the compounds into their ionic forms (if applicable)
AlBr3(aq) can be broken down into Al3+ and 3Br- ions:
Al3+ (aq) + 3Br- (aq)

LiOH(aq) can be broken down into Li+ and OH- ions:
3Li+ (aq) + 3OH- (aq)

Al(OH)3(s) remains unchanged since it is not ionic.

LiBr(aq) can be broken down into Li+ and Br- ions:
3Li+ (aq) + 3Br- (aq)

Step 3: Write the complete ionic equation
Now that we have the ions, we can write the complete ionic equation by representing all the dissolved ionic compounds as their separate ions:

Al3+ (aq) + 3Br- (aq) + 3Li+ (aq) + 3OH- (aq) → Al(OH)3(s) + 3Li+ (aq) + 3Br- (aq)

Step 4: Cancel out the spectator ions
In the net ionic equation, we remove the spectator ions that appear on both sides of the equation. In this case, the spectator ions are Li+ and Br-.

The net ionic equation is:

Al3+ (aq) + 3OH- (aq) → Al(OH)3(s)

So, the net ionic equation for the reaction is Al3+ (aq) + 3OH- (aq) → Al(OH)3(s).