ZnCl2(aq)+Na2CO3(s)--> ZnCO3(s)+NaCl (aq) Balance the chemical equation and state solid, liq, gas, aq properly to each compound
Ok I have it balanced by 2 NaCl in the second part of the equation But now what is the total ionic and net ionic equations.
I think:
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
Help
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?? :
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
sorry on that first post.. I was mixing two problems But I also need help ith that first equation also the ZnCl2 one....
Thanks
To balance the chemical equation ZnCl2(aq) + Na2CO3(s) -> ZnCO3(s) + 2NaCl(aq), you need to ensure that the number of atoms of each element is equal on both sides of the equation. In this case, you balanced it correctly by adding a coefficient of 2 in front of NaCl to ensure equal numbers of Na, Cl, and O on both sides.
Now, to determine the state of each compound (solid, liquid, gas, or aqueous), it is important to know the solubility rules. ZnCO3 and NaCl are both soluble in water, so they are in the form of ions (aq) in the solution. Na2CO3 is a solid because it is not soluble in water (s). ZnCl2 is also soluble and exists as ions (aq) in the solution.
Next, let's write the total ionic equation. This equation shows all the ions in the solution as they exist, rather than as compounds.
Total Ionic Equation:
Zn^2+(aq) + 2Cl^-(aq) + 2Na^+(aq) + CO3^2-(s) -> ZnCO3(s) + 2Na^+(aq) + 2Cl^-(aq)
Lastly, to write the net ionic equation, we eliminate the spectator ions, which are the ions that appear on both sides of the equation. In this case, Na^+ and Cl^- ions are present on both sides, so they can be eliminated.
Net Ionic Equation:
Zn^2+(aq) + CO3^2-(s) -> ZnCO3(s)
In the net ionic equation, only the ions that actually undergo the reaction are shown, which are the Zn^2+ and CO3^2- ions forming ZnCO3 precipitate.
I hope this helps! Let me know if you have any further questions.