Question

What products result from the addition of aqueous solutions of `Cu(NO_3)_2` and `(NH_4)_2S`?

Answer 1

Solid copper(II) sulfide and aqueous ammonium nitrate.

Explanation:

Copper(II) nitrate, `"Cu"("NO"_3)_2`, and ammonium sulfide, `("NH"_4)_2"S"`, are both soluble ionic compounds, which means that the dissociate completely in aqueous solution and exist as cations and anions

`"Cu"("NO"_3)_text(2(aq]) -> "Cu"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-)`

`("NH"_4)_2"S"_text((aq]) -> 2"NH"_text(4(aq])^(+) + "S"_text((aq])^(2-)`

Now, when these two solutions are mixed, the copper(II) cations, `"Cu"^(2+)`, and the sulfide anions, `"S"^(2-)`, will react to form copper(II) sulfide, an insoluble solid that precipitates out of solution.

The other product of the reaction will be ammonium nitrate, `"NH"_4"NO"_3`, a soluble compound that will exist as cations and anions in solution.

This means that you can write

`"Cu"("NO"_3)_text(2(aq]) + ("NH"_4)_2"S"_text((aq]) -> "CuS"_text((s]) darr + 2"NH"_4"NO"_text(3(aq])`

The complete ionic equation will look like this

`"Cu"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-) + 2"NH"_text(4(aq])^(+) + "S"_text((aq])^(2-) -> "CuS"_text((s]) darr + 2"NH"_text(4(aq])^(+) + 2"NO"_text(3(aq])^(-)`

Notice that some ions are present on both sides of the equation - these ions are called spectator ions.

Removing these ions will give you the net ionic equation

`"Cu"_text((aq])^(2+) + color(red)(cancel(color(black)(2"NO"_text(3(aq])^(-)))) + color(red)(cancel(color(black)(2"NH"_text(4(aq])^(+)))) + "S"_text((aq])^(2-) -> "CuS"_text((s]) darr + color(red)(cancel(color(black)(2"NH"_text(4(aq])^(+)))) + color(red)(cancel(color(black)(2"NO"_text(3(aq])^(-))))`

which looks like this

`"Cu"_text((aq])^(2+) + "S"_text((aq])^(2-) -> "CuS"_text((s]) darr`