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

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    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 ##Bi## is +3 while on ##Ba## it is -2.
    After the cations have changed partners the formulas of the products are ##Bi_2S_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:
    So
    ##Ba(NO_3)_2## is soluble.
    ##BaS## is soluble because ##Ba## is in Group 2.
    ##B_2S_3## is insoluble because ##Bi## is in group 15.
    The unbalanced molecular equation is
    ##Bi(NO_3)_3(aq) + BaS (aq) Bi_2S_3(s) + Ba(NO_3)_2(aq)##
    The balanced molecular equation is
    ##2Bi(NO_3)_3(aq) + 3BaS(aq) Bi_2S_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_2S_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_2S_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_2S_3(s)##
    Here’s a good video on solubility rules and net ionic equations.

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