Question

Balance the following chemical equation:
a. ${\text{Al}}\,{\text{ + }}\,{\text{HCl}}\,\, \to {\text{AlC}}{{\text{l}}_{\text{3}}}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}$
b. ${\text{C}}{{\text{S}}_2}{\text{ + }}\,{{\text{O}}_2}\,\, \to {\text{C}}{{\text{O}}_2}\,{\text{ + }}\,\,{\text{S}}{{\text{O}}_{\text{2}}}$
c. ${{\text{P}}_4}{\text{ + }}\,{{\text{O}}_2}\,\, \to {{\text{P}}_2}{{\text{O}}_5}$
d. ${\text{N}}{{\text{H}}_3}\,{\text{ + }}\,{{\text{O}}_2}\, \to {{\text{N}}_2}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}{\text{O}}$
e. ${\text{Mg}}\,{\text{ + }}\,{{\text{N}}_2}\, \to {\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_2}$

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. For this, we have to count the number of atoms on the 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 solution:
First, we will balance the atoms which are less in number. We will balance the oxygen and hydrogen at last.
a. ${\text{Al}}\,{\text{ + }}\,{\text{HCl}}\,\, \to {\text{AlC}}{{\text{l}}_{\text{3}}}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}$
Aluminium is one on both sides of the reaction so it is balanced. Chlorine is one on the reactant side and three on the product side, so we will add coefficient three in front of HCl to balance the chlorine.
${\text{Al}}\,{\text{ + }}\,3\,{\text{HCl}}\,\, \to {\text{AlC}}{{\text{l}}_{\text{3}}}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}$
Now we have three hydrogens on the reactant side and two on the product side, so we will add a coefficient $3/2$ in front of hydrogen on the product side to balance the hydrogen.
${\text{Al}}\,{\text{ + }}\,3\,{\text{HCl}}\,\, \to {\text{AlC}}{{\text{l}}_{\text{3}}}\,{\text{ + }}\,3/2\,{{\text{H}}_{\text{2}}}$
Now we will multiply the whole reaction with coefficient two to convert the integer into a whole number. So, the balanced equation is,
${\text{2}}\,{\text{Al}}\,{\text{ + }}\,6\,{\text{HCl}}\,\, \to 2\,{\text{AlC}}{{\text{l}}_{\text{3}}}\,{\text{ + }}\,3\,{{\text{H}}_{\text{2}}}$
b. ${\text{C}}{{\text{S}}_2}{\text{ + }}\,{{\text{O}}_2}\,\, \to {\text{C}}{{\text{O}}_2}\,{\text{ + }}\,\,{\text{S}}{{\text{O}}_{\text{2}}}$
Carbon is one on both sides of the reaction so it is balanced. Sulphur is two on the reactant side and one on the product side so, we will add coefficient two in front of ${\text{S}}{{\text{O}}_{\text{2}}}$ to balance the sulphur.
${\text{C}}{{\text{S}}_2}{\text{ + }}\,{{\text{O}}_2}\,\, \to {\text{C}}{{\text{O}}_2}\,{\text{ + }}\,\,2\,{\text{S}}{{\text{O}}_{\text{2}}}$
Now we have two oxygen on the reactant side and six on the product side, so we will add coefficient three in front of oxygen on the reactant side to balance the oxygen.
${\text{C}}{{\text{S}}_2}{\text{ + }}\,3\,{{\text{O}}_2}\,\, \to {\text{C}}{{\text{O}}_2}\,{\text{ + }}\,\,2\,{\text{S}}{{\text{O}}_{\text{2}}}$
So, the balanced equation is, ${\text{C}}{{\text{S}}_2}{\text{ + }}\,3\,{{\text{O}}_2}\,\, \to {\text{C}}{{\text{O}}_2}\,{\text{ + }}\,\,2\,{\text{S}}{{\text{O}}_{\text{2}}}$.
c. ${{\text{P}}_4}{\text{ + }}\,{{\text{O}}_2}\,\, \to {{\text{P}}_2}{{\text{O}}_5}$
Phosphorous is four on the reactant side and two on the product side so, we will add coefficient two in front of ${{\text{P}}_2}{{\text{O}}_5}$ to balance the phosphorous.
${{\text{P}}_4}{\text{ + }}\,{{\text{O}}_2}\,\, \to 2\,{{\text{P}}_2}{{\text{O}}_5}$
Now we have two oxygen on the reactant side and ten on the product side, so we will add coefficient five in front of oxygen on the reactant side to balance the oxygen.
${{\text{P}}_4}{\text{ + }}\,5\,{{\text{O}}_2}\,\, \to 2\,{{\text{P}}_2}{{\text{O}}_5}$
So, the balanced equation is, ${{\text{P}}_4}{\text{ + }}\,5\,{{\text{O}}_2}\,\, \to 2\,{{\text{P}}_2}{{\text{O}}_5}$.
d. ${\text{N}}{{\text{H}}_3}\,{\text{ + }}\,{{\text{O}}_2}\, \to {{\text{N}}_2}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}{\text{O}}$
Nitrogen is one on the reactant side and two on the product side, so we will add coefficient two in front of ${\text{N}}{{\text{H}}_3}$ to balance the nitrogen.
${\text{2}}\,{\text{N}}{{\text{H}}_3}\,{\text{ + }}\,{{\text{O}}_2}\, \to {{\text{N}}_2}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}{\text{O}}$
Now we have six hydrogens on the reactant side and two on the product side so, we will add coefficient three in front of ${{\text{H}}_{\text{2}}}{\text{O}}$on product side to balance the hydrogen.
${\text{2}}\,{\text{N}}{{\text{H}}_3}\,{\text{ + }}\,{{\text{O}}_2}\, \to {{\text{N}}_2}\,{\text{ + }}\,\,{\text{3}}\,{{\text{H}}_{\text{2}}}{\text{O}}$
Now we have two oxygen on the reactant side and three on the product side so, we will add coefficient $3/2$ in front of ${{\text{O}}_{\text{2}}}$on the product side to balance the oxygen.
${\text{2}}\,{\text{N}}{{\text{H}}_3}\,{\text{ + }}\,3/2{{\text{O}}_2}\, \to {{\text{N}}_2}\,{\text{ + }}\,\,{\text{3}}\,{{\text{H}}_{\text{2}}}{\text{O}}$
Now we will multiply the whole reaction with coefficient two to convert the integer into a whole number. So, the balanced equation is,
$4\,{\text{N}}{{\text{H}}_3}\,{\text{ + }}\,3\,{{\text{O}}_2}\, \to 2\,{{\text{N}}_2}\,{\text{ + }}\,\,6\,{{\text{H}}_{\text{2}}}{\text{O}}$
i.e. ${\text{Mg}}\,{\text{ + }}\,{{\text{N}}_2}\, \to {\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_2}$
Magnesium is one on the reactant side and three on the product side, so, we will add coefficient three in front of ${\text{Mg}}$ to balance the magnesium.
${\text{3}}\,{\text{Mg}}\,{\text{ + }}\,{{\text{N}}_2}\, \to {\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_2}$
So, the balanced equation is, ${\text{3}}\,{\text{Mg}}\,{\text{ + }}\,{{\text{N}}_2}\, \to {\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_2}$.

Note: 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. The coefficient will be added in front of the atoms or molecule, not as subscript or superscript. The coefficient will be added at that side where fewer atoms are present. Multiplication of the whole equation with the same number does not alter the number of atoms.