Question

How would you balance the following chemical equation: mercury (II) hydroxide + phosphoric acid --> mercury (II) phosphate + water?

Answer 1

`"3 Hg(OH)"_2 + "2 H"_3"PO"_4 -> "Hg"_3("PO"_4)_2 + "6 H"_2"O"`

Explanation:

Here's your unbalanced equation:

`"Hg(OH)"_2 + "H"_3"PO"_4 -> "Hg"_3("PO"_4)_2 + "H"_2"O"`

Begin by fixing the obvious. We need 3 `"Hg"` on the left side and we only have 1, so add a coefficient of 3 to `"Hg(OH)"_2`

`"3 Hg(OH)"_2 + "H"_3"PO"_4 -> "Hg"_3("PO"_4)_2 + "H"_2"O"`

This solves our mercury problem, now on to phosphate. There is one phosphate ion on the left and two on right, so add a coefficient of two to `"H"_3"PO"_4`

`"3 Hg(OH)"_2 + "2 H"_3"PO"_4 -> "Hg"_3("PO"_4)_2 + "H"_2"O"`

The only thing left to the balance is the water. On the left side, we have a total of 12 `"H"` and 6 `"O"` (not counting the phosphate). This gives us exactly 6 `"H"_2"O"`, so adding a coefficient of 6 will give us our final answer:

`"3 Hg(OH)"_2 + "2 H"_3"PO"_4 -> "Hg"_3("PO"_4)_2 + "6 H"_2"O"`

Always double check your answer to make sure all the elements balance. In this case, they do, so we are done.