Write the net ionic equation for the following reaction :
Na3SO4(aq) + Ba(NO3)2(aq) —> BaSO4 (s) + 2 NaNO3(aq)
Ba++ + SO4-- >> BaSO4 (s)
You made an error; it's Na2SO4.
2Na^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2NO3^-(aq) ==> BaSO4(s) + 2Na^+(aq) + 2NO3^-(aq)
I have written the complete ionic equation above. To make it into a net ionic equation, cancel those ions that appear on both sides. The net ionic equation is what is left.For example, I see 2Na^+(aq) on both sides so cancel those. Etc. I'll check your final answer is you will post it.
Yes my bad Na2SO4. So would the answer be Ba(NO3)2 (aq) -> BaSO4 (s)
To write the net ionic equation for a reaction, we first need to write the balanced equation.
The given balanced equation is:
Na3SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2 NaNO3(aq)
The next step is to separate the aqueous substances into their respective ions. We use the solubility rules to determine if a compound dissociates into ions in water. In this case, Na3SO4 and Ba(NO3)2 will dissociate into ions.
Na3SO4(aq) dissociates into Na+(aq) + 3 SO4^2-(aq)
Ba(NO3)2(aq) dissociates into Ba^2+(aq) + 2 NO3^-(aq)
Now we can write the complete ionic equation by including the dissociated ions for the reactants and products:
3 Na+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2 NO3^-(aq) —> BaSO4(s) + 2 Na+(aq) + 2 NO3^-(aq)
Next, we can cancel out the ions that appear on both sides of the equation. In this case, Na+ and NO3- appear on both sides:
3 Na+(aq) + SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s) + 2 Na+(aq)
Finally, we obtain the net ionic equation by removing the spectator ions (ions that appear on both sides of the equation but do not participate in the reaction) from the complete ionic equation:
SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)
So, the net ionic equation for the given reaction is:
SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)