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

Nov 1, 2015

$2 {H}_{3} P {O}_{4}$ + $3 M g {\left(O H\right)}_{2}$ = $M {g}_{3} {\left(P {O}_{4}\right)}_{2}$ + $6 {H}_{2} O$

#### Explanation:

I can either tally all the atoms one by one or use my knowledge of ionic bonds to make the tally system much simpler.

I'll do the simpler version.

First re-write the chemical equation to avoid confusing myself:

${H}_{3} P {O}_{4}$ + $M g {\left(O H\right)}_{2}$ = $M {g}_{3} {\left(P {O}_{4}\right)}_{2}$ + $H$-$O H$

Notice that I had shown the line structure between the ${H}^{+}$ and $O {H}^{-}$ ions. Now, let's tally based on the subscripts.

left side:

${H}^{+}$ = 3

$P {O}_{4}^{\text{3-}}$ = 1

$M {g}^{\text{2+}}$ = 1

$O {H}^{-}$ = 2

right side:

${H}^{+}$ = 1

$P {O}_{4}^{\text{3-}}$ = 2

$M {g}^{\text{2+}}$ = 3

$O {H}^{-}$ = 1

Let's start balancing the largest ion, $P {O}_{4}^{\text{3-}}$.

$\textcolor{red}{2} {H}_{3} P {O}_{4}$ + $M g {\left(O H\right)}_{2}$ = $M {g}_{3} {\left(P {O}_{4}\right)}_{2}$ + $H$-$O H$

left side:

${H}^{+}$ = 3 x $\textcolor{red}{2}$ = 6

$P {O}_{4}^{\text{3-}}$ = 1 x $\textcolor{red}{2}$ = 2

$M {g}^{\text{2+}}$ = 1

$O {H}^{-}$ = 2

right side:

${H}^{+}$ = 1

$P {O}_{4}^{\text{3-}}$ = 2

$M {g}^{\text{2+}}$ = 3

$O {H}^{-}$ = 1

Do not forget that since ${H}_{3} P {O}_{4}$ is a substance, you need to also multiply the coefficient with the number of $H$ atoms.

Next,

$2 {H}_{3} P {O}_{4}$ + $\textcolor{g r e e n}{3} M g {\left(O H\right)}_{2}$ = $M {g}_{3} {\left(P {O}_{4}\right)}_{2}$ + $H$-$O H$

left side:

${H}^{+}$ = 3 x 2 = 6

$P {O}_{4}^{\text{3-}}$ = 1 x 2 = 2

$M {g}^{\text{2+}}$ = 1 x $\textcolor{g r e e n}{3}$ = 3

$O {H}^{-}$ = 2 x $\textcolor{g r e e n}{3}$ = 6

right side:

${H}^{+}$ = 1

$P {O}_{4}^{\text{3-}}$ = 2

$M {g}^{\text{2+}}$ = 3

$O {H}^{-}$ = 1

Again, since $M g {\left(O H\right)}_{2}$ is a substance, whatever coefficient you apply to $M {g}^{\text{2+}}$ ion will also be applied to $O {H}^{-}$ ion.

$2 {H}_{3} P {O}_{4}$ + $3 M g {\left(O H\right)}_{2}$ = $M {g}_{3} {\left(P {O}_{4}\right)}_{2}$ + $\textcolor{b l u e}{6}$$H$-$O H$

left side:

${H}^{+}$ = 3 x 2 = 6

$P {O}_{4}^{\text{3-}}$ = 1 x 2 = 2

$M {g}^{\text{2+}}$ = 1 x 3 = 3

$O {H}^{-}$ = 2 x 3 = 6

right side:

${H}^{+}$ = 1 x $\textcolor{b l u e}{6}$ = 6

$P {O}_{4}^{\text{3-}}$ = 2

$M {g}^{\text{2+}}$ = 3

$O {H}^{-}$ = 1 x $\textcolor{b l u e}{6}$ = 6

Reverting back to the original equation,

$2 {H}_{3} P {O}_{4}$ + $3 M g {\left(O H\right)}_{2}$ = $M {g}_{3} {\left(P {O}_{4}\right)}_{2}$ + $6 {H}_{2} O$

The equation is now balanced.

Please take note this method is only applicable for IONIC bonding of polyatomic ions.