AlCl3(aq)+ NH4OH(aq)-> Al(OH)3(s) + NH4Cl(aq)
AlCl3(aq)+ 3NH4OH(aq)-> Al(OH)3(s) + 3 NH4Cl(aq)... is balanced-- Now total ioic?? Is this right?? :
Al +3Cl+3NH4 +OH --> Al+ 3OH+ 3NH4+ 3Cl
Crossing out on both sides I end up with OH --> 3OH
I do not know what to do or how it works and what each compound ends up as solid, liq, gas or aq
See your other post. If you still need assistance, explain exactly what your trouble is. The first thing see that may be causing trouble is that you have displayed ions as atoms.
Please reply and your answer is wrong
To determine the total ionic equation, you need to break down all the compounds into their ions.
Starting with the balanced equation:
AlCl3(aq) + 3NH4OH(aq) → Al(OH)3(s) + 3NH4Cl(aq)
Breaking down the compounds into their ions:
AlCl3(aq) → Al3+(aq) + 3Cl-(aq)
3NH4OH(aq) → 3NH4+(aq) + 3OH-(aq)
Al(OH)3(s) does not break down as it is already a solid.
3NH4Cl(aq) → 3NH4+(aq) + 3Cl-(aq)
Now let's write the total ionic equation using only the ions that are involved in the reaction:
Al3+(aq) + 3Cl-(aq) + 3NH4+(aq) + 3OH-(aq) → Al(OH)3(s) + 3NH4+(aq) + 3Cl-(aq)
Notice that there are 3 NH4+(aq) ions and 3 Cl-(aq) ions on both the reactant and product sides. They cancel each other out. Also, the OH- ions on both sides cancel each other out as well.
Simplifying the equation:
Al3+(aq) + 3OH-(aq) → Al(OH)3(s)
This is the simplified total ionic equation for the reaction. Note that the solid Al(OH)3 does not get broken down into its ions because it does not exist in the aqueous solution.
To determine the total ionic equation for the reaction between AlCl3(aq) and NH4OH(aq), you need to break down the reactants and products into their individual ions.
First, let's write out the complete ionic equation:
Al^3+ + 3Cl^- + NH4+ + OH^- → Al(OH)3(s) + NH4+ + Cl^-
In this equation, AlCl3(aq) dissociates into Al^3+ and 3 Cl^- ions, while NH4OH(aq) dissociates into NH4+ and OH^- ions.
Next, you can eliminate the spectator ions, which are the ions that appear on both the reactant and product sides without undergoing any change. In this case, the spectator ions are NH4+ and Cl^-. So, we can remove them from both sides of the equation:
Al^3+ + OH^- → Al(OH)3(s)
This is the net ionic equation.
Regarding the states of the compounds:
- AlCl3(aq) and NH4OH(aq) are both in aqueous form, which means they are dissolved in water.
- Al(OH)3 is a solid, denoted as (s), indicating that it precipitates out of the solution.
- NH4Cl(aq) is also in aqueous form.
The net ionic equation above represents the reaction between Al^3+ and OH^- ions to form solid Al(OH)3. The NH4+ and Cl^- ions remain in the solution as NH4Cl(aq).