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

1 Answer
Dec 15, 2015

Answer:

#"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.