What is the molecular, total ionic, and net ionic equation for the reaction of Bi(NO3)3 + BaS?

First, you must recognize that this is a .

Next you must write the for the products.

You know that ##"NO"_3## has an ionic charge of -1 and ##"S"## has a charge of -2. So the ionic charge on ##"B"i## is +3, while on ##"Ba"## it is -2.

After the cations have changed partners, the formulas of the products are ##"Bi"_2"S"_3## and ##"Ba"("NO"_3)_2##.

Next, you use the solubility rules to determine if there are any precipitates.

In this case, the important rules are:

1. All nitrates are soluble.
2. All sulfides are insoluble except those of Groups 1 and 2.

So

• ##"Ba"("NO"_3)_2## is soluble.

• ##"BaS"## is soluble, because ##"Ba"## is in Group 2.

• ##"B"_2"S"_3## is insoluble, because ##"Bi"## is in group 15.

The unbalanced molecular equation is

##"Bi"("NO"_3)_3"(aq)" + "BaS (aq") → "Bi"_2"S"_3"(s)" + "Ba"("NO"_3)_2"(aq)"##

The balanced molecular equation is

##"2Bi"("NO"_3)_3"(aq)" + "3BaS(aq)" → "Bi"_2"S"_3"(s)" + "3Ba"("NO"_3)_2(aq)##

The total ionic equation is

##"2Bi"^(3+)"(aq)" + "6NO"_3^(-)"(aq)" + "3Ba"^(2+)"(aq)" + "3S"^(2-)"(aq)" → "Bi"_2"S"_3"(s)" + "3Ba"^(2+)"(aq)"+ "6NO"_3^(-)"(aq)"##

To get the net ionic equation, we cancel the spectator ions.

##"2Bi"^(3+)"(aq)" + color(red)(cancel(color(black)("6NO"_3^(-)"(aq)"))) + color(red)(cancel(color(black)("3Ba"^(2+)"(aq)"))) + "3S"^(2-)"(aq)" → "Bi"_2"S"_3"(s)" + color(red)(cancel(color(black)("3Ba"^(2+)"(aq)"))) + color(red)(cancel(color(black)("6NO"_3^(-)"(aq)")))##

##"2Bi"^(3+)"(aq)" + "3S"^(2-)"(aq)" → "Bi"_2"S"_3"(s)"##

Here's a good video on solubility rules and net ionic equations.