# How do you balance Ca_3(PO_4)_2 + H_3PO_4 -> Ca(H_2PO_4)_2?

Apr 24, 2018

$\text{Ca"_3("PO"_4)_2+4"H"_3"PO"_4}$$\rightarrow$$3 \text{Ca(H"_2"PO"_4)_2}$

#### Explanation:

Given equation:

$\text{Ca"_3("PO"_4)_2+"H"_3"PO"_4}$$\rightarrow$$\text{Ca(H"_2"PO"_4)_2}$

There are three Ca atoms on the left-hand side (LHS), and one on the right-hand side (RHS). Place a coefficient of $3$ in front of "Ca(H"_2"PO"_4)_2.

$\text{Ca"_3("PO"_4)_2+"H"_3"PO"_4}$$\rightarrow$$\textcolor{t e a l}{3} \text{Ca(H"_2"PO"_4)_2}$

Now there are twelve H atoms on the RHS $\left(3 \times 2 \times 2\right)$, and three on the LHS. Place a coefficient of $4$ in front of $\text{H"_3"PO"_4}$.

$\text{Ca"_3("PO"_4)_2+color(magenta)4"H"_3"PO"_4}$$\rightarrow$$\textcolor{t e a l}{3} \text{Ca(H"_2"PO"_4)_2}$

Now lets count the number of atoms of each element on each side of the equation.

$\text{LHS:}$ $\text{3 Ca atoms}$, $\text{6 P atoms}$, $\text{12 H atoms}$, $\text{24 O atoms}$

$\text{RHS:}$ $\text{3 Ca atoms}$, $\text{6 P atoms}$, $\text{12 H atoms}$, $\text{24 O atoms}$

Since there are the same number of atoms of each element on both sides of the equation, it is balanced.