Question #98443

Nickel(II) chloride, #"NiCl"_2#, will react with ammonium sulfide, #("NH"_4)_2"S"#, to form nickel sulfide, #"NiS"#, and ammonium chloride, #"NH"_4"Cl"#.

The idea here is that both reactants are soluble ionic compounds, as shown by the solubility rules listed below

![http://highered.mheducation.com/olcweb/cgi/

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))#

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))^(-)#

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#