Question

The equation which is balanced and represents the correct product is:
(A) ${{\left[ Mg{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}+{{\left( EDTA \right)}^{4-}}\xrightarrow{excess\text{ }NaOH}{{\left[ Mg\left( EDTA \right) \right]}^{2+}}+6{{H}_{2}}O$
(B) $CuS{{O}_{4}}+4KCN\to {{K}_{2}}\left[ Cu{{\left( CN \right)}_{4}} \right]+{{K}_{2}}S{{O}_{4}}$
(C) $L{{i}_{2}}O+2KCl\to 2LiCl+{{K}_{2}}O$
(D) ${{\left[ CoCl{{\left( N{{H}_{3}} \right)}_{5}} \right]}^{+}}+5{{H}^{+}}\to C{{o}^{2+}}+5N{{H}_{4}}^{+}+C{{l}^{-}}$

Answer 1

Hint: As we know, a balanced chemical equation is an equation for a reaction in which the total charge and number of atoms for each element in the reaction is the same for both the reactants and the products. Or in other words, both the charge and mass are balanced on both sides of the chemical reaction. Let’s take each of the given options and check whether how many of them obeys the rules of balanced chemical equation.

Complete step by step answer:
 - The balancing of charge in a balanced chemical equation means that the total charge will be zero on both sides of the chemical equation and balancing for mass produces the same kinds and numbers of atoms on both sides of the equation.

(i) \[{{\left[ Mg{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}+{{\left( EDTA \right)}^{4-}}\xrightarrow{excess\text{ }NaOH}{{\left[ Mg\left( EDTA \right) \right]}^{2+}}+6{{H}_{2}}O\]
 The charge on the reactants can be found as, −4+2=−2 whereas the charge on the products side is +2. That is the net charge on reactants and products are different and hence this reaction is unbalanced. If the charge on the product was −2, then the equation will be balanced.

(ii) \[CuS{{O}_{4}}+4KCN\to {{K}_{2}}\left[ Cu{{\left( CN \right)}_{4}} \right]+{{K}_{2}}S{{O}_{4}}\]
The product formed in the reaction is incorrect. Instead of the product ${{K}_{2}}\left[ Cu{{\left( CN \right)}_{4}} \right]$,the real product formed will be${{K}_{3}}\left[ Cu{{\left( CN \right)}_{4}} \right]$. Hence this reaction is incorrect.

(iii) \[L{{i}_{2}}O+2KCl\to 2LiCl+{{K}_{2}}O\]
 This reaction is not favorable in the forward direction since the reactant $L{{i}_{2}}O$ is comparatively stable than the product ${{K}_{2}}O$. Also, a stronger base ${{K}_{2}}O$ cannot be generated by a weaker base $L{{i}_{2}}O$. Hence this reaction is also incorrect.

(iv) \[{{\left[ CoCl{{\left( N{{H}_{3}} \right)}_{5}} \right]}^{+}}+5{{H}^{+}}\to C{{o}^{2+}}+5N{{H}_{4}}^{+}+C{{l}^{-}}\]
 This reaction is correct since all ammine complexes can be destroyed by adding hydrogen ions. Also the charge on both the reactant side and product side are balanced too (+2).
So, the correct answer is “Option C”.

Note: Keep in mind that the law of conservation of mass is applied in all the chemical equations. This means that the number of atoms of reactants present is conserved in the number of atoms of products. For any chemical equation in a sealed system, the mass of the reactants must be equivalent to the mass of the products.