Question

How‌ ‌would‌ ‌you‌ ‌balance‌ ‌${\text{Ag}}\,{\text{‌ ‌+‌ ‌}}\,{{\text{O}}_2}\,\,‌ ‌\to‌ ‌ \,{\text{A}}{{\text{g}}_{\text{2}}}{\text{O}}$‌ ‌

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 sides 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: Right side of the arrow represents the products and left side of the arrow represents the reactants.First, we will balance the atoms which are less in number. We will balance the oxygen and hydrogen at last.
${\text{Ag}}\,{\text{ + }}\,{{\text{O}}_2}\,\, \to \,{\text{A}}{{\text{g}}_{\text{2}}}{\text{O}}$
Silver is one on the reactant side and two on the product side, so we will add coefficient two in front of Ag at the reactant side to balance the silver.
${\text{2}}\,{\text{Ag}}\,{\text{ + }}\,{{\text{O}}_2}\,\, \to \,{\text{A}}{{\text{g}}_{\text{2}}}{\text{O}}$
Now we have two oxygen atoms on the reactant side and one on the product side, so we will add a coefficient $1/2$ in front of ${{\text{O}}_2}$ on the reactant side to balance the oxygen.
${\text{2}}\,{\text{Ag}}\,{\text{ + }}\,1/2\,{{\text{O}}_2}\,\, \to \,{\text{A}}{{\text{g}}_{\text{2}}}{\text{O}}$
Now, we will multiply the whole equation with coefficient $2$ to convert the integer into whole numbers.
${\text{2}}\,{\text{Ag}}\,{\text{ + }}\,{{\text{O}}_2}\,\, \to \,2\,{\text{A}}{{\text{g}}_{\text{2}}}{\text{O}}$
So, the balanced equation is ${\text{2}}\,{\text{Ag}}\,{\text{ + }}\,1/2\,{{\text{O}}_2}\,\, \to \,{\text{A}}{{\text{g}}_{\text{2}}}{\text{O}}$.

Note: We do not add the coefficients as subscript or superscript. We add the coefficients in front of the atoms or molecules. 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 the ratio of atoms of products and reactants. 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.