Question

What is the correct representation of reaction occurring when ${\text{HCl}}$ is heated with ${\text{Mn}}{{\text{O}}_{\text{2}}}$?
A. ${\text{MnO}}_4^ - \, + {\text{5C}}{{\text{l}}^ - }\, + \,8{{\text{H}}^{\text{ + }}}\, \to \,{\text{M}}{{\text{n}}^{2 + }}\, + {\text{5C}}{{\text{l}}^ - } + \,{\text{5}}{{\text{H}}_2}{\text{O}}$
B. ${\text{Mn}}{{\text{O}}_2}\, + {\text{2C}}{{\text{l}}^ - }\, + \,4\,{{\text{H}}^{\text{ + }}}\, \to \,{\text{M}}{{\text{n}}^{2 + }}\, + {\text{C}}{{\text{l}}_2} + \,{\text{2}}\,{{\text{H}}_2}{\text{O}}$
C. ${\text{2}}\,{\text{Mn}}{{\text{O}}_2}\, + {\text{4C}}{{\text{l}}^ - }\, + \,8\,{{\text{H}}^{\text{ + }}}\, \to \,2\,{\text{M}}{{\text{n}}^{2 + }}\, + 2\,{\text{C}}{{\text{l}}_2} + \,{\text{4}}\,{{\text{H}}_2}{\text{O}}$
D. ${\text{Mn}}{{\text{O}}_2}\, + {\text{4}}\,{\text{HCl}}\,\, \to \,{\text{MnC}}{{\text{l}}_{\text{4}}}\,{\text{ + }}\,{\text{C}}{{\text{l}}_2} + \,{{\text{H}}_2}{\text{O}}$

Answer 1

Hint:We will write the reaction of hydrochloric acid with manganese oxide. Then balanced the reaction. Then write the reaction in ionic form and cancel the spectator ions to get the correct representation.

Complete step by step solution:
 The reaction of the manganese oxide with hydrochloric acid is as follows:
${\text{Mn}}{{\text{O}}_2}{\text{(s)}} + {\text{HCl(aq)}}\,\, \to \,{\text{MnC}}{{\text{l}}_2}{\text{(aq)}}\,{\text{ + }}\,{\text{C}}{{\text{l}}_2}{\text{(g)}} + \,{{\text{H}}_2}{\text{O(l)}}$
We will assign the oxidation state of each ion as follows:
$\mathop {{\text{Mn}}}\limits^{ + 4} \mathop {{{\text{O}}_2}}\limits^{ - 2} {\text{(s)}}\, + \mathop {\text{H}}\limits^{ + 1} \mathop {{\text{Cl}}}\limits^{ - 1} \,\,({\text{aq)}} \to \,\mathop {{\text{Mn}}}\limits^{ + 2} \mathop {{\text{C}}{{\text{l}}_2}}\limits^{ - 1} {\text{(aq)}}\,\,\,{\text{ + }}\,\mathop {{\text{C}}{{\text{l}}_2}{\text{(g)}}}\limits^0 + \,\mathop {{{\text{H}}_2}}\limits^{ + 1} \mathop {\text{O}}\limits^{ - 2} {\text{(l)}}$
Manganese atoms are given two electrons and chlorine gets two electrons, so the electrons are balanced in the reaction.
Now we will balance the manganese and chlorine after that oxygen and hydrogen.
Manganese is already balanced so, balance chlorine as,
On the left side, four chlorines are present, so add $4$ in front of ${\text{HCl}}$.
${\text{Mn}}{{\text{O}}_2}{\text{(s)}} + 4\,{\text{HCl(aq)}}\,\, \to \,{\text{MnC}}{{\text{l}}_2}{\text{(aq)}}\,{\text{ + }}\,{\text{C}}{{\text{l}}_2}{\text{(g)}} + \,{{\text{H}}_2}{\text{O(l)}}$
Balance oxygen, on the right side two oxygen atoms are present, so add $2$ in front of ${{\text{H}}_2}{\text{O}}$.
${\text{Mn}}{{\text{O}}_2}{\text{(s)}} + 4\,{\text{HCl(aq)}}\,\, \to \,{\text{MnC}}{{\text{l}}_2}{\text{(aq)}}\,{\text{ + }}\,{\text{C}}{{\text{l}}_2}{\text{(g)}} + \,2\,{{\text{H}}_2}{\text{O(l)}}$
Hydrogen atoms are already balanced.
The reaction in ionic form is as follows:
${\text{Mn}}{{\text{O}}_2}{\text{(s)}} + 4\,{{\text{H}}^ + }{\text{(aq)}}\,{\text{ + }}\,{\text{4C}}{{\text{l}}^ - }{\text{(aq)}}\,\, \to \,{\text{M}}{{\text{n}}^{2 + }}{\text{(aq)}} + \,2\,{\text{C}}{{\text{l}}^ - }{\text{(aq)}}\,{\text{ + }}\,{\text{C}}{{\text{l}}_2}{\text{(g)}} + \,{{\text{H}}_2}{\text{O(l)}}$
Cancel the sector ions as and write the reaction as follows:
Spectator ions are the ions which occur on both sides of the reactions.
${\text{Mn}}{{\text{O}}_2}{\text{(s)}} + 4\,{{\text{H}}^ + }{\text{(aq)}}\,{\text{ + }}\,2{\text{C}}{{\text{l}}^ - }{\text{(aq)}}\,\, \to \,{\text{M}}{{\text{n}}^{2 + }}{\text{(aq)}}\,{\text{ + }}\,{\text{C}}{{\text{l}}_2}{\text{(g)}} + \,{{\text{H}}_2}{\text{O(l)}}$
So, the correct representation of reaction occurring when ${\text{HCl}}$ is heated with ${\text{Mn}}{{\text{O}}_{\text{2}}}$is, ${\text{Mn}}{{\text{O}}_2}\, + {\text{2C}}{{\text{l}}^ - }\, + \,4\,{{\text{H}}^{\text{ + }}}\, \to \,{\text{M}}{{\text{n}}^{2 + }}\, + {\text{C}}{{\text{l}}_2} + \,{\text{2}}\,{{\text{H}}_2}{\text{O}}$

Therefore, option (B) ${\text{Mn}}{{\text{O}}_2}\, + {\text{2C}}{{\text{l}}^ - }\, + \,4\,{{\text{H}}^{\text{ + }}}\, \to \,{\text{M}}{{\text{n}}^{2 + }}\, + {\text{C}}{{\text{l}}_2} + \,{\text{2}}\,{{\text{H}}_2}{\text{O}}$ is correct.

Note:
During the reaction electrons get exchanged between two species. One gets reduced and another gets oxidized. To write the correct representation, electrons should be balanced. Solid and gas do not remain in ionic form in solution. Only the aqueous species remains in ionic form. The representation obtained after the cancelation of Spectator ions is known as a net ionic reaction.