Question

Complete and balance the following chemical equations
(A)- ${\text{Zn}}\left( {\text{s}} \right){\text{ + HCl}}\left( {{\text{aq}}} \right) \to $
(B)- ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}\left( {\text{s}} \right){\text{ + 2HCl}}\left( {{\text{aq}}} \right) \to $
(C)- ${\text{NaHC}}{{\text{O}}_{\text{3}}}\left( {\text{s}} \right){\text{ + HCl}}\left( {{\text{aq}}} \right) \to $
(D)- ${\text{NaOH}}\left( {{\text{aq}}} \right){\text{ + HCl}}\left( {{\text{aq}}} \right) \to $
(E)- ${\text{CuO}}\left( {\text{s}} \right){\text{ + HCl}}\left( {{\text{aq}}} \right) \to $

Answer 1

Hint: For the completion of any reaction we have to know about the reactivity of each species with another species. And we balance the chemical reaction by managing the molecularity of each atom to the reactant side as well as to the product side of the chemical reaction.

Complete answer:
Completion and balancing of above given reactions are as follow:
-Zinc (${\text{Zn}}$) on reacting with Hydrochloric acid produces Zinc chloride (${\text{ZnC}}{{\text{l}}_{\text{2}}}$) and hydrogen gas${\text{(}}{{\text{H}}_{\text{2}}})$and chemical reaction is display as follow:
${\text{Zn}}\left( {\text{s}} \right){\text{ + HCl}}\left( {{\text{aq}}} \right) \to {\text{ZnC}}{{\text{l}}_{\text{2}}}\left( {{\text{aq}}} \right){\text{ + }}{{\text{H}}_{\text{2}}}\left( {\text{g}} \right)$
And balanced form of the above reaction is as follow:
${\text{Zn}}\left( {\text{s}} \right){\text{ + 2HCl}}\left( {{\text{aq}}} \right) \to {\text{ZnC}}{{\text{l}}_{\text{2}}}\left( {{\text{aq}}} \right){\text{ + }}{{\text{H}}_{\text{2}}}\left( {\text{g}} \right)$
In which all species show the same molecularity on the left hand side as well as on the right hand side also.
-Sodium carbonate reacting with Hydrochloric acid produces sodium chloride solution, water and carbon dioxide gas. And chemical reaction is display as follow:
${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}\left( {\text{s}} \right){\text{ + 2HCl}}\left( {{\text{aq}}} \right) \to {\text{NaCl}}\left( {{\text{aq}}} \right){\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\left( {{\text{aq}}} \right){\text{ + C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$
And balanced form of the above reaction is as follow:
${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}\left( {\text{s}} \right){\text{ + 2HCl}}\left( {{\text{aq}}} \right) \to 2{\text{NaCl}}\left( {{\text{aq}}} \right){\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\left( {{\text{aq}}} \right){\text{ + C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$
In which all species show the same molecularity on the left hand side as well as on the right hand side also.
-Sodium bicarbonate reacting with Hydrochloric acid produces sodium chloride solution, water and carbon dioxide gas. And chemical reaction is display as follow:
${\text{NaHC}}{{\text{O}}_{\text{3}}}\left( {\text{s}} \right){\text{ + HCl}}\left( {{\text{aq}}} \right) \to {\text{NaCl}}\left( {{\text{aq}}} \right){\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\left( {{\text{aq}}} \right){\text{ + C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$
And the above reaction is present in the balanced form because all the species show the same molecularity on the left hand side as well as on the right hand side also.
-Sodium hydroxide on reacting with Hydrochloric acid produces sodium chloride solution and water and chemical reaction is display as follow:
${\text{NaOH}}\left( {{\text{aq}}} \right){\text{ + HCl}}\left( {{\text{aq}}} \right) \to {\text{NaCl}}\left( {{\text{aq}}} \right){\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\left( {{\text{aq}}} \right)$
And the above reaction is present in the balanced form because all the species show the same molecularity on the left hand side as well as on the right hand side also.
-Copper oxide on reacting with hydrochloric acid produces copper ii chloride and water, and chemical reaction is display as follow:
${\text{CuO}}\left( {\text{s}} \right){\text{ + HCl}}\left( {{\text{aq}}} \right) \to {\text{CuC}}{{\text{l}}_{\text{2}}}\left( {\text{s}} \right){\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\left( {{\text{aq}}} \right)$
And balanced form of the above reaction is as follow:
${\text{CuO}}\left( {\text{s}} \right){\text{ + 2HCl}}\left( {{\text{aq}}} \right) \to {\text{CuC}}{{\text{l}}_{\text{2}}}\left( {\text{s}} \right){\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\left( {{\text{aq}}} \right)$
In which all species show the same molecularity on the left hand side as well as on the right hand side also.

Note:
Here some of you may write the wrong balanced equation, if you think that molecularity and oxidation state of any species is the same thing. So, always balanced the equation by keeping in mind that oxidation value and molecularity are different.