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

1 Answer
Dec 15, 2015

3 Hg(OH)2+2 H3PO4Hg3(PO4)2+6 H2O

Explanation:

Here's your unbalanced equation:

Hg(OH)2+H3PO4Hg3(PO4)2+H2O

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+H3PO4Hg3(PO4)2+H2O

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 H3PO4

3 Hg(OH)2+2 H3PO4Hg3(PO4)2+H2O

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 H2O, so adding a coefficient of 6 will give us our final answer:

3 Hg(OH)2+2 H3PO4Hg3(PO4)2+6 H2O

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