Question

Complete and balance the following reaction:
(i). ${\text{NaCl}}\,{\text{ + }}\,{\text{Mn}}{{\text{O}}_2}{\text{ + }}\,{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\,\, \to$
(ii). ${\text{KMn}}{{\text{O}}_{\text{4}}}\,{\text{ + }}\,{\text{S}}{{\text{O}}_2}{\text{ + }}\,{{\text{H}}_2}{\text{O}} \to$

Answer 1

To determine the answer we should know what balancing the equation means. Balancing the equation means we have to determine the stoichiometry coefficients of each compound or molecule. For this, we have to count the number of atoms on both side of the arrow. On the side, where less number of atoms are present, we will add a suitable coefficient at that side in the front of the atom which is less in number. Similarly, we add the coefficient wherever required. At last, the total number of atoms on the left side will be equal to the total number of atoms on the right side.

Complete step-by-step answer:
We will write the products on the right side of the arrow. First, we will balance the atoms which are less in number. We will balance the oxygen and hydrogen at last.

(i) ${\text{NaCl}}\,{\text{ + }}\,{\text{Mn}}{{\text{O}}_2}{\text{ + }}\,{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\, \to {\text{NaHS}}{{\text{O}}_4}\, + \,{\text{MnS}}{{\text{O}}_{\text{4}}}\, + \,{\text{C}}{{\text{l}}_{\text{2}}}\, + \,{{\text{H}}_{\text{2}}}{\text{O}}$
Sodium, sulphur and manganese are one on both sides of the reaction so they are balanced. Chlorine is one on the reactant side and two on the product side, so we will add coefficient two in front of NaCl to balance the chlorine.
${\text{2}}\,{\text{NaCl}}\,{\text{ + }}\,{\text{Mn}}{{\text{O}}_2}{\text{ + }}\,{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\, \to {\text{NaHS}}{{\text{O}}_4}\, + \,{\text{MnS}}{{\text{O}}_{\text{4}}}\, + \,{\text{C}}{{\text{l}}_{\text{2}}}\, + \,{{\text{H}}_{\text{2}}}{\text{O}}$
Now we have two sodium on the reactant side and one on the product side so, we will add coefficient two in front of ${\text{NaHS}}{{\text{O}}_4}$ on the product side to balance the sodium.
${\text{2}}\,{\text{NaCl}}\,{\text{ + }}\,{\text{Mn}}{{\text{O}}_2}{\text{ + }}\,{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\, \to 2\,{\text{NaHS}}{{\text{O}}_4}\, + \,{\text{MnS}}{{\text{O}}_{\text{4}}}\, + \,{\text{C}}{{\text{l}}_{\text{2}}}\, + \,{{\text{H}}_{\text{2}}}{\text{O}}$
Now, sodium, manganese, sulphur, and chlorine are balanced. So, we have two hydrogens on the reactant side and four hydrogens on the product. We have six oxygen on the reactant side and thirteen oxygen on the product.
So, we will add the coefficient two in front of $\,{{\text{H}}_{\text{2}}}{\text{O}}$ and three in front of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$.
${\text{2}}\,{\text{NaCl}}\,{\text{ + }}\,{\text{Mn}}{{\text{O}}_2}{\text{ + }}\,3{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\, \to 2\,{\text{NaHS}}{{\text{O}}_4}\, + \,{\text{MnS}}{{\text{O}}_{\text{4}}}\, + \,{\text{C}}{{\text{l}}_{\text{2}}}\, + \,2\,{{\text{H}}_{\text{2}}}{\text{O}}$

(ii). ${\text{KMn}}{{\text{O}}_{\text{4}}}\,{\text{ + }}\,{\text{S}}{{\text{O}}_2}{\text{ + }}\,{{\text{H}}_2}{\text{O}} \to \,{\text{MnS}}{{\text{O}}_{\text{4}}}\, + \,{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\, + \,{{\text{H}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$
Manganese is one on both sides of the reaction, so it is balanced. Potassium is one on the reactant side and two on the product side so, we will add coefficient two in front of ${\text{KMn}}{{\text{O}}_{\text{4}}}$NaCl to balance the potassium.
${\text{2}}\,{\text{KMn}}{{\text{O}}_{\text{4}}}\,{\text{ + }}\,{\text{S}}{{\text{O}}_2}{\text{ + }}\,{{\text{H}}_2}{\text{O}} \to \,{\text{MnS}}{{\text{O}}_{\text{4}}}\, + \,{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\, + \,{{\text{H}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$
Now we have two manganese on the reactant side and one on the product side so, we will add coefficient two in front of ${\text{MnS}}{{\text{O}}_{\text{4}}}$ on the product side to balance the manganese.
${\text{2}}\,{\text{KMn}}{{\text{O}}_{\text{4}}}\,{\text{ + }}\,{\text{S}}{{\text{O}}_2}{\text{ + }}\,{{\text{H}}_2}{\text{O}} \to \,2\,{\text{MnS}}{{\text{O}}_{\text{4}}}\, + \,{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\, + \,{{\text{H}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$
Now we have one sulphur on the reactant side and three on the product side so, we will add coefficient five in front of ${\text{S}}{{\text{O}}_2}$ on the reactant side and two in front of ${{\text{H}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$ on the product side to balance the sulphur.
${\text{2}}\,{\text{KMn}}{{\text{O}}_{\text{4}}}\,{\text{ + }}\,5\,{\text{S}}{{\text{O}}_2}{\text{ + }}\,{{\text{H}}_2}{\text{O}} \to \,2\,{\text{MnS}}{{\text{O}}_{\text{4}}}\, + \,{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\, + \,2\,{{\text{H}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$
We have two hydrogens on the reactant side and four hydrogens on the product so, we will add coefficient two in front of ${{\text{H}}_2}{\text{O}}$ on the reactant side to balance the hydrogen.
${\text{2}}\,{\text{KMn}}{{\text{O}}_{\text{4}}}\,{\text{ + }}\,5\,{\text{S}}{{\text{O}}_2}{\text{ + }}\,2\,{{\text{H}}_2}{\text{O}} \to \,2\,{\text{MnS}}{{\text{O}}_{\text{4}}}\, + \,{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\, + \,2\,{{\text{H}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$

Note: We add the coefficients in front of the atoms or molecules, the coefficients are not added as subscript or superscript. We add the coefficient at the side where fewer atoms are present. Multiplication of the complete equation with the same number does not cause the change in the number of atoms. To count the total number of an atom, we multiply the subscript of that atom with the coefficient present in front of that atom or the molecule containing that atom.