# How would you balance the following equation: H3PO4 +Mg(OH)2 --> Mg3(PO4)2 +H20?

Nov 5, 2015

Balanced Equation

${\text{2H"_3"PO"_4" + 3Mg(OH)}}_{2}$$\rightarrow$$\text{Mg"_3"(PO"_4)_2 + "6H"_2"O}$

#### Explanation:

${\text{H"_3"PO"_4" + Mg(OH)}}_{2}$$\rightarrow$$\text{Mg"_3"(PO"_4)_2 + "H"_2"O}$

Balance the magnesium first. There are three magnesium ions on the right side and only one on the left side. Add a coefficient of $3$ in front of $\text{Mg(OH)"_2}$..

${\text{H"_3"PO"_4" + 3Mg(OH)}}_{2}$$\rightarrow$$\text{Mg"_3"(PO"_4)_2 + "H"_2"O}$

Next balance the $\text{PO"_4}$.

There are two $\text{PO"_4}$ ions on the right side and one on the left. Place a coefficient of $2$ in front of $\text{H"_3"PO"_4}$.

${\text{2H"_3"PO"_4" + 3Mg(OH)}}_{2}$$\rightarrow$$\text{Mg"_3"(PO"_4)_2 + "H"_2"O}$

Next balance the $\text{H}$. There are twelve hydrogens on the left side and two on the right. Place a coefficient of $6$ in front of $\text{H"_2"O}$ on the right side.

${\text{2H"_3"PO"_4" + 3Mg(OH)}}_{2}$$\rightarrow$$\text{Mg"_3"(PO"_4)_2 + "6H"_2"O}$

Balance the $\text{O}$. There are fourteen oxygen atoms on both sides of the equation, so it is already balanced.

Balanced Equation

${\text{2H"_3"PO"_4" + 3Mg(OH)}}_{2}$$\rightarrow$$\text{Mg"_3"(PO"_4)_2 + "6H"_2"O}$