Question

What is the molecular, total ionic, and net ionic equations of these reactions? $AuB{{r}_{3}}+{{H}_{2}}C{{O}_{3}}$ And $Bi{{(N{{O}_{3}})}_{3}}+BaS$

Answer 1

Hint: Molecular equations, total ionic and net ionic equations are possible when a reaction occurring, is a double displacement reaction. The places of ions are displaced in these kinds of reactions.

Complete answer: A molecular equation is the equation that is expressed in terms of the whole molecules involved in a chemical reaction. The molecular equation does not have the constituent atoms of the molecules, rather the atoms represent a whole molecule.
The total ionic equations of any compounds are written in the form of their respective ions. The ions that are not involved in the reaction, called spectator ions, are also involved. The ions are without their molecular form.
Net ionic equations only involve the ions that are taking part in the chemical reaction to form the products, the extra spectator ions are not involved in these types of equations.
For the reaction $AuB{{r}_{3}}+{{H}_{2}}C{{O}_{3}}$,
Molecular equation is, $2AuB{{r}_{3}}(aq)+{{H}_{2}}C{{O}_{3}}(aq)\to A{{u}_{2}}{{(C{{O}_{3}})}_{3}}\downarrow +6HBr(aq)$
Total ionic equation is, $2A{{u}^{3+}}(aq)+6B{{r}^{-}}(aq)+6{{H}^{+}}+C{{O}_{3}}^{2-}(aq)\to A{{u}_{2}}{{(C{{O}_{3}})}_{3}}\downarrow +6{{H}^{+}}(aq)+6B{{r}^{-}}(aq)$
The net ionic equation is, $2A{{u}^{3+}}(aq)+C{{O}_{3}}^{2-}(aq)\to A{{u}_{2}}{{(C{{O}_{3}})}_{3}}\downarrow $
For the reaction$Bi{{(N{{O}_{3}})}_{3}}+BaS$ ,
Molecular equation is, $2Bi{{(N{{O}_{3}})}_{3}}+3BaS\to B{{i}_{2}}{{S}_{3}}\downarrow +3Ba{{(N{{O}_{3}})}_{2}}(aq)$
Total ionic equation is, $2Bi(aq)+6N{{O}_{3}}^{-}(aq)+3B{{a}^{2+}}(aq)+3{{S}^{2-}}(aq)\to B{{i}_{2}}{{S}_{3}}\downarrow +3B{{a}^{2+}}(aq)+6N{{O}_{3}}^{-}(aq)$
The net ionic equation is, $2Bi(aq)+3{{S}^{2-}}(aq)\to B{{i}_{2}}{{S}_{3}}\downarrow $
Hence, the molecular, total ionic, and net ionic equations of the reactions are expressed above.

Note: Any displacement reaction takes place with the evolution of gases, or formation of a precipitate (solid). The sign $\downarrow $in the above reactions indicate a solid precipitate is formed. While the sign $\uparrow $ indicates that a gas is evolved.