Question #98443
1 Answer
Here's what I got.
Explanation:
Nickel(II) chloride,
The idea here is that both reactants are soluble ionic compounds, as shown by the solubility rules listed below
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Nickel sulfide is an insoluble solid that precipitates out of solution, while ammonium chloride is a soluble ionic compound that will exist as ions in the solution, i.e. the reaction produces aqueous ammonium chloride.
You will thus have
"NiCl"_ (2(aq)) + ("NH"_ 4)_ 2"S"_ ((aq)) -> "NiS"_ ((s)) darr + 2"NH"_ 4"Cl"_ ((aq))NiCl2(aq)+(NH4)2S(aq)→NiS(s)⏐↓+2NH4Cl(aq)
To get the net ionic equation, start by writing out the complete ionic equation
"Ni"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + 2"NH" _ (4(aq))^(+) + "S"_ ((aq))^(2-) -> "NiS"_ ((s)) darr + 2"NH"_ (4(aq))^(+) + 2"Cl"_ ((aq))^(-)Ni2+(aq)+2Cl−(aq)+2NH+4(aq)+S2−(aq)→NiS(s)⏐⏐↓+2NH+4(aq)+2Cl−(aq)
Eliminate the spectator ions, i.e. the ions that are present on both sides of the equation
"Ni"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) + color(red)(cancel(color(black)(2"NH"_ (4(aq))^(+)))) + "S"_ ((aq))^(2-) -> "NiS"_ ((s)) darr + color(red)(cancel(color(black)(2"NH"_ (4(aq))^(+)))) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-))))
to get
"Ni"_ ((aq))^(2+) + "S"_ ((aq))^(2-) -> "NiS"_ ((s)) darr
