question archive ##SO_(4(aq))^(2-) + Pb_((aq))^(2+) -> PbSO_(4(s))## The balanced equation for this reaction is ##MgSO_(4(aq)) + Pb(NO_3)_(2(aq)) ->Mg(NO_3)_2(aq) + PbSO_4(s)## The complete ionic equation is ##Mg_((aq))^(2+) + SO_(4(aq))^(2-) + Pb_((aq))^(2+) + 2NO_(3(aq))^(-) -> PbSO_(4(s)) + Mg_((aq))^(2+) + + 2NO_(3(aq))^(-)## If you eliminate spectator ions, which are the ions thatcan be found both on the reactants' and on the products' side, you'll get the net ionic equation ##SO_(4(aq))^(2-) + Pb_((aq))^(2+) -> PbSO_(4(s))## Lead (II) sulfate is insoluble in aqueous solution and will form a precipitate
##SO_(4(aq))^(2-) + Pb_((aq))^(2+) -> PbSO_(4(s))## The balanced equation for this reaction is ##MgSO_(4(aq)) + Pb(NO_3)_(2(aq)) ->Mg(NO_3)_2(aq) + PbSO_4(s)## The complete ionic equation is ##Mg_((aq))^(2+) + SO_(4(aq))^(2-) + Pb_((aq))^(2+) + 2NO_(3(aq))^(-) -> PbSO_(4(s)) + Mg_((aq))^(2+) + + 2NO_(3(aq))^(-)## If you eliminate spectator ions, which are the ions thatcan be found both on the reactants' and on the products' side, you'll get the net ionic equation ##SO_(4(aq))^(2-) + Pb_((aq))^(2+) -> PbSO_(4(s))## Lead (II) sulfate is insoluble in aqueous solution and will form a precipitate
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##SO_(4(aq))^(2-) + Pb_((aq))^(2+) -> PbSO_(4(s))##
The balanced equation for this reaction is
##MgSO_(4(aq)) + Pb(NO_3)_(2(aq)) ->Mg(NO_3)_2(aq) + PbSO_4(s)##
The complete ionic equation is
##Mg_((aq))^(2+) + SO_(4(aq))^(2-) + Pb_((aq))^(2+) + 2NO_(3(aq))^(-) -> PbSO_(4(s)) + Mg_((aq))^(2+) + + 2NO_(3(aq))^(-)##
If you eliminate spectator ions, which are the ions thatcan be found both on the reactants' and on the products' side, you'll get the net ionic equation
##SO_(4(aq))^(2-) + Pb_((aq))^(2+) -> PbSO_(4(s))##
Lead (II) sulfate is insoluble in aqueous solution and will form a precipitate.
Here is another example which illustrates how to write a net ionic equation. video from: Noel Pauller
