Question

Write balanced complete ionic equation for K2SO4(aq)+CaI2(aq)→CaSO4(s)+KI(aq)

Answer 1

The balanced complete ionic equation is as follows:$\boxed{{\text{2}}{{\text{K}}^ + }\left( {aq} \right) + {\text{SO}}_4^{2 - }\left( {aq} \right) + {\text{C}}{{\text{a}}^{2 + }}\left( {aq} \right) + {\text{2}}{{\text{I}}^ - }\left( {aq} \right) \to {\text{CaS}}{{\text{O}}_4}\left( s \right) + {{\text{K}}^ + }\left( {aq} \right) + {{\text{I}}^ - }\left( {aq} \right)}$

Further Explanation:

The three types of equations that are used to represent the chemical reaction are as follows:

1. Molecular equation

2. Total ionic equation

3. Net ionic equation

The reactants and products remain in undissociated form in the molecular equation. In the case of total ionic equation, all the ions that are dissociated and present in the reaction mixture are represented while in the case of net or overall ionic equation only the useful ions that participate in the reaction are represented.

The steps to write the complete or total ionic reaction are as follows:

Step 1: Write the molecular equation for the reaction with the phases in the bracket.

In the reaction, ${{\text{K}}_2}{\text{S}}{{\text{O}}_4}$reacts with ${\text{Ca}}{{\text{I}}_{\text{2}}}$to form ${\text{CaS}}{{\text{O}}_{\text{4}}}$and KI. The balanced molecular equation of the reaction is as follows:

${{\text{K}}_2}{\text{S}}{{\text{O}}_4}\left( {aq}\right)+{\text{Ca}}{{\text{I}}_2}\left({aq} \right)\to{\text{CaS}}{{\text{O}}_4}\left(s\right)+{\text{KI}}\left( {aq}\right)$

Step 2: Dissociate all the compounds with the aqueous phase to write the total ionic equation. The compounds with solid and liquid phases remain the same. The total ionic equation is as follows:

${\text{2}}{{\text{K}}^+}\left( {aq}\right)+{\text{SO}}_4^{2-}\left( {aq}\right)+{\text{C}}{{\text{a}}^{2+}}\left({aq}\right)+{\text{2}}{{\text{I}}^-}\left({aq}\right) \to {\text{CaS}}{{\text{O}}_4}\left(s\right)+{{\text{K}}^+}\left( {aq}\right)+{{\text{I}}^-}\left( {aq} \right)$

Therefore, the complete ionic equation obtained  is as follows:

${\text{2}}{{\text{K}}^ + }\left( {aq} \right) + {\text{SO}}_4^{2 - }\left( {aq} \right) + {\text{C}}{{\text{a}}^{2 + }}\left( {aq} \right) + {\text{2}}{{\text{I}}^ - }\left( {aq} \right) \to {\text{CaS}}{{\text{O}}_4}\left( s \right) + {{\text{K}}^ + }\left( {aq} \right) + {{\text{I}}^ - }\left( {aq} \right)$

Learn more:

1. Balanced chemical equation: brainly.com/question/1405182

2. Oxidation and reduction reaction: brainly.com/question/2973661

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: net ionic equation, complete ionic equation, K+, Cl-, CaSO4, KI, CaI2, K2SO4, solid phase, liquid phase, aqueous phase.

Answer 2

First let us write the equation. By the law of conservation of mass, the reactant side must have the same mass as the product side. 

K₂SO₄ + CaI₂ → CaSO₄ + KI 
If you visually check there are surpluses of elements in both sides. To correct this, correct coefficients must be added to each reactant. Here are the correct coefficients.  

K₂SO₄ + CaI₂ → CaSO₄ + 2KI
Inspecting, there are now 2 potassium, 1 sulfur, 2 oxygen,  1 calcium, 2 iodine. the equation now is balanced.