Question

How would you balance ${\text{HBr}}\,{\text{ + }}\,{\text{Mg(OH}}{{\text{)}}_2}\,\, \to \,{\text{MgB}}{{\text{r}}_{\text{2}}}\, + \,{{\text{H}}_2}{\text{O}}$

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 the 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.
${\text{HBr}}\,{\text{ + }}\,{\text{Mg(OH}}{{\text{)}}_2}\,\, \to \,{\text{MgB}}{{\text{r}}_{\text{2}}}\, + \,{{\text{H}}_2}{\text{O}}$
Magnesium is one on both sides of the arrow so it is balanced. Bromine is one on the reactant side and two on the product side, so1 we will add coefficient two in front of HBr at the reactant side to balance the bromine.
${\text{2HBr}}\,{\text{ + }}\,{\text{Mg(OH}}{{\text{)}}_2}\,\, \to \,{\text{MgB}}{{\text{r}}_{\text{2}}}\, + \,{{\text{H}}_2}{\text{O}}$
Now we have two oxygen atoms on the reactant side and one on the product side so, we will add coefficient $2$ in front of ${{\text{H}}_2}{\text{O}}$ on the reactant side to balance the oxygen.
${\text{2HBr}}\,{\text{ + }}\,{\text{Mg(OH}}{{\text{)}}_2}\,\, \to \,{\text{MgB}}{{\text{r}}_{\text{2}}}\, + \,2\,{{\text{H}}_2}{\text{O}}$
So, the balanced equation is ${\text{2HBr}}\,{\text{ + }}\,{\text{Mg(OH}}{{\text{)}}_2}\,\, \to \,{\text{MgB}}{{\text{r}}_{\text{2}}}\, + \,2\,{{\text{H}}_2}{\text{O}}$.

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 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.