What is the total ionic equation and net ionic equation of ammonium sulphide solution added to gold nitrate solution?
Molecular equation is
3(NH4)2S + 2Au(NO3)3 ==> Au2S3 + 6NH4NO3
Net ionic equation is
3S^2-(aq) + 2Au^3+(aq) ==> Au2S3(s)
You should be able to do the total ionic equation from what I've written above.
To determine the total ionic equation and net ionic equation of ammonium sulfide solution added to gold nitrate solution, we first need to identify the compounds involved and their respective ions.
The compounds involved are:
1. Ammonium sulfide (NH4)2S, which dissociates in water into ammonium ions (NH4+) and sulfide ions (S2-).
2. Gold nitrate Au(NO3)3, which dissociates in water into gold ions (Au3+) and nitrate ions (NO3-).
Now, let's write the complete balanced chemical equation:
(NH4)2S + Au(NO3)3 → 2NH4NO3 + AuS
In the complete ionic equation, we separate all the dissolved ionic compounds into their respective ions:
2NH4+(aq) + S2-(aq) + Au3+(aq) + 3NO3-(aq) → 2NH4+(aq) + 2NO3-(aq) + AuS(s)
By canceling out the spectator ions that appear on both sides of the equation (NH4+ and NO3-), we can write the net ionic equation:
S2-(aq) + Au3+(aq) → AuS(s)
Therefore, the total ionic equation is 2NH4+(aq) + S2-(aq) + Au3+(aq) + 3NO3-(aq) → 2NH4+(aq) + 2NO3-(aq) + AuS(s), and the net ionic equation is S2-(aq) + Au3+(aq) → AuS(s).