Question

How would you balance the following chemical equation
Mercury(II) hydroxide + phosphoric acid mercury(II) phosphate + water

Answer 1

The stoichiometry of the reaction gives us an idea about the amount of reactant reacts in a particular ratio of mole or mass or number. The stoichiometry ratio of each and every reaction is fixed and it follows the Law of definite proportion or constant proportion.

Complete step by step answer:
Here’s our unbalanced equation:
$\text{Hg(OH}{{\text{)}}_{\text{2}}}+{{\text{H}}_{3}}\text{P}{{\text{O}}_{4}}\to \text{H}{{\text{g}}_{3}}{{(\text{P}{{\text{O}}_{4}})}_{2}}+6{{\text{H}}_{2}}\text{O}$
By looking at the above equation we can clearly see that the reaction is not balanced.
We will use a hit and trial method to balance this equation in which coefficients before the formula or symbols of the reactants and products are adjusted, not the subscripts after the formula.
When balancing a reaction look at the ions, rather than atoms.
Begin by fixing the obvious. We need $3$ Hg on the left side and we only have $1$ so we will add a coefficient of $3$ on $\text{Hg(OH}{{\text{)}}_{\text{2}}}$
$\text{3Hg(OH}{{\text{)}}_{\text{2}}}+{{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}\to \text{H}{{\text{g}}_{\text{3}}}{{\text{(P}{{\text{O}}_{\text{4}}}\text{)}}_{\text{2}}}+6{{\text{H}}_{\text{2}}}\text{O}$
Now we will move the phosphate ion. There is one phosphate ion on left and two on right so add a coefficient of two to ${{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}$
$\text{3Hg(OH}{{\text{)}}_{2}}+2{{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}\to \text{H}{{\text{g}}_{\text{3}}}{{\text{(P}{{\text{O}}_{\text{4}}}\text{)}}_{\text{2}}}+{{\text{H}}_{\text{2}}}\text{O}$
The only thing left is water there are $12\text{H}$ and $60$ on left side this means $\text{6}{{\text{H}}_{\text{2}}}\text{O}$ , so adding a coefficient of $6$ will balance our equation.

Final Answer :
$\text{3Hg(OH}{{\text{)}}_{\text{2}}}+2{{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}\to \text{H}{{\text{g}}_{\text{3}}}{{\text{(P}{{\text{O}}_{\text{4}}}\text{)}}_{\text{2}}}+6{{\text{H}}_{\text{2}}}\text{O}$

Note: In every reaction, atom conservation is very important. Balancing a chemical reaction ensures that the same number of atoms of each element appear on both sides of the reaction.In any chemical reaction every time conservation of atom is there, by the help of atom conservation we can balance any chemical reaction.