# How would you balance this equation: ?H_3PO_4 + ?Mg(OH)_2 -> ? Mg_3(PO_4)_2 + ? H_2O?

Nov 1, 2016

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

#### Explanation:

To do it, I took the following steps :

• Balance Mg:

I have 3 Mg in $M {g}_{3} {\left(P {O}_{4}\right)}_{2}$ so I added 3 on the other side $3 M g {\left(O H\right)}_{2}$

• Balance P :

Again I have 2 P in $M {g}_{3} {\left(P {O}_{4}\right)}_{2}$ so I added 2 on the other side $2 {H}_{3} P {O}_{4}$

• Balance H :

I have 6 H in $2 {H}_{3} P {O}_{4}$ $\left(2 \times 3\right)$ and 6 H in $3 M g {\left(O H\right)}_{2}$ a total of 12 H so I added 6 next to ${H}_{2} O$
$6 {H}_{2} O$

Notice that the O is automatically balanced : 8 O in $2 {H}_{3} P {O}_{4}$ + 6 O in $3 M g {\left(O H\right)}_{2}$ = 8 O in $M {g}_{3} {\left(P {O}_{4}\right)}_{2}$ + 6 O in $6 {H}_{2} O$

I hope that was clear