Question

Show the complete ionic equation and net ionic equation for all the equations below, then state whether or not a precipitate (insoluble compound) will form. To receive full credit, you must show ALL your work.
Cacl2(aq) + K2co3(aq) + -------->
Bacl2(aq) + MgSO4(aq) + -------->
AgNO3(aq) + Kl(aq) →
Nacl(aq) + (NH4)2Cro4(aq) →

Answer 1

Answer:

(a): Precipitate of calcium carbonate will form.

(b): Precipitate of barium sulfate will form.

(c): Precipitate of silver iodide will form.

(d): Precipitate of sodium chromate will form.

Explanation:

Complete ionic equation is defined as the equation in which all the substances that are strong electrolytes present in an aqueous state and are represented in the form of ions.

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

(a):

The balanced molecular equation is:

$CaCl_2(aq)+K_2CO_3(aq)\rightarrow 2KCl(aq)+CaCO_3(s)$

The complete ionic equation follows:

$Ca^{2+}(aq)+2Cl^-(aq)+2K^+(aq)+CO_3^{2-}(aq)\rightarrow 2K^+(aq)+2Cl^-(aq)+CaCO_3(s)$

As potassium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)$

Precipitate of calcium carbonate will form.

(b)

The balanced molecular equation is:

$BaCl_2(aq)+MgSO_4(aq)\rightarrow MgCl_2(aq)+BaSO_4(s)$

The complete ionic equation follows:

$Ba^{2+}(aq)+2Cl^-(aq)+Mg^{2+}(aq)+SO_4^{2-}(aq)\rightarrow Mg^{2+}(aq)+2Cl^-(aq)+BaSO_4(s)$

As magnesium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

Precipitate of barium sulfate will form.

(c):

The balanced molecular equation is:

$AgNO_3(aq)+KI(aq)\rightarrow KNO_3(aq)+AgI(s)$

The complete ionic equation follows:

$Ag^{+}(aq)+NO_3^-(aq)+K^+(aq)+I^{-}(aq)\rightarrow K^+(aq)+NO_3^-(aq)+AgI(s)$

As potassium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ag^{+}(aq)+I^{-}(aq)\rightarrow AgI(s)$

Precipitate of silver iodide will form.

(d):

The balanced molecular equation is:

$2NaCl(aq)+(NH_4)_2CrO_4(aq)\rightarrow 2NH_4Cl(aq)+Na_2CrO_4(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+2Cl^-(aq)+2NH_4^+(aq)+CrO_4^{2-}(aq)\rightarrow 2NH_4^+(aq)+2Cl^-(aq)+Na_2CrO_4(s)$

As ammonium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$2Na^{+}(aq)+CrO_4^{2-}(aq)\rightarrow Na_2CrO_4(s)$

Precipitate of sodium chromate will form.