# Question 4f432

Nov 19, 2017

("NH"_ 4)_ 2"S"_ ((aq)) + 2"AuNO"_ (3(aq)) -> "Au"_ 2"S" _ ((s)) darr + 2"NH"_ 4"NO"_ (3(aq))

#### Explanation:

You're dealing with a double-replacement reaction in which the gold(I) cations and the sulfide anions combine to form gold(I) sulfide, an insoluble ionic compound that precipitates out of the solution.

Both reactants are soluble in aqueous solution, so you can say that you have

("NH"_ 4)_ 2"S"_ ((aq)) -> 2"NH"_ (4(aq))^(+) + 2"S"_ ((aq))^(2-)

${\text{AuNO"_ (3(aq)) -> "Au"_ ((aq))^(+) + "NO}}_{3 \left(a q\right)}^{-}$

When you mix these two solutions, the gold(I) cations and the sulfide anions will combine to form the precipitate. The ammonium cations and the nitrate anions will form ammonium nitrate, a soluble ionic compound that exists as dissociated ions in the resulting solution.

("NH"_ 4)_ 2"S"_ ((aq)) + "AuNO"_ (3(aq)) -> "Au"_ 2"S" _ ((s)) darr + "NH"_ 4"NO"_ (3(aq))

Keep in mind that this chemical equation is not balanced! Next, write the complete ionic equation that matches the unbalanced chemical equation..

$2 {\text{NH"_ (4(aq))^(+) + "S"_ ((aq))^(2-) + "Au"_ ((aq))^(+) + "NO"_ (3(aq))^(-) -> "Au"_ 2"S"_ ((s)) darr + "NH"_ (4(aq))^(+) + "NO}}_{3 \left(a q\right)}^{-}$

Notice that you have two ammonium cations on the reactants' side and only one on the products' side. Multiply the ammonium cations present on the products' side by $\textcolor{b l u e}{2}$, but don't forget to do the same for the nitrate anions present on the products' side because these two ions form the soluble ammonium nitrate.

$2 {\text{NH"_ (4(aq))^(+) + "S"_ ((aq))^(2-) + "Au"_ ((aq))^(+) + "NO"_ (3(aq))^(-) -> "Au"_ 2"S"_ ((s)) darr + color(blue)(2)"NH"_ (4(aq))^(+) + color(blue)(2)"NO}}_{3 \left(a q\right)}^{-}$

To balance the nitrate anions, multiply the nitrate anions present on the reactants' side by $\textcolor{b l u e}{2}$, but don't forget to do the same with the gold(I) cations!

$2 {\text{NH"_ (4(aq))^(+) + "S"_ ((aq))^(2-) + color(blue)(2)"Au"_ ((aq))^(+) + color(blue)(2)"NO"_ (3(aq))^(-) -> "Au"_ 2"S"_ ((s)) darr + color(blue)(2)"NH"_ (4(aq))^(+) + color(blue)(2)"NO}}_{3 \left(a q\right)}^{-}$

Now everything is balanced out--notice that the gold atoms are balanced as well!

To get the net ionic equation, you need to eliminate the spectator ions, i.e. the ions that are present on both sides of the chemical equation.

$\textcolor{red}{\cancel{\textcolor{b l a c k}{2 {\text{NH"_ (4(aq))^(+)))) + "S"_ ((aq))^(2-) + color(blue)(2)"Au"_ ((aq))^(+) + color(red)(cancel(color(black)(color(blue)(2)"NO"_ (3(aq))^(-)))) -> "Au"_ 2"S"_ ((s)) darr + color(red)(cancel(color(black)(color(blue)(2)"NH"_ (4(aq))^(+)))) + color(red)(cancel(color(black)(color(blue)(2)"NO}}_{3 \left(a q\right)}^{-}}}}$

You will end up with

$\textcolor{b l u e}{2} {\text{Au"_ ((aq))^(+) + "S"_ ((aq))^(2-) -> "Au"_ 2"S}}_{\left(s\right)} \downarrow$

Therefore, you can say that the balanced chemical equation that describes this double-replacement reaction looks like this

("NH"_ 4)_ 2"S"_ ((aq)) + color(blue)(2)"AuNO"_ (3(aq)) -> "Au"_ 2"S" _ ((s)) darr + color(blue)(2)"NH"_ 4"NO"_ (3(aq))#