Question

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

Answer 1

`"Ca"_3("PO"_4)_2+4"H"_3"PO"_4"` `rarr` `3"Ca(H"_2"PO"_4)_2"`

Explanation:

Given equation:

`"Ca"_3("PO"_4)_2+"H"_3"PO"_4"` `rarr` `"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`.

`"Ca"_3("PO"_4)_2+"H"_3"PO"_4"` `rarr` `color(teal)3"Ca(H"_2"PO"_4)_2"`

Now there are twelve H atoms on the RHS `(3xx2xx2)`, and three on the LHS. Place a coefficient of `4` in front of `"H"_3"PO"_4"`.

`"Ca"_3("PO"_4)_2+color(magenta)4"H"_3"PO"_4"` `rarr` `color(teal)3"Ca(H"_2"PO"_4)_2"`

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

`"LHS:"` `"3 Ca atoms"`, `"6 P atoms"`, `"12 H atoms"`, `"24 O atoms"`

`"RHS:"` `"3 Ca atoms"`, `"6 P atoms"`, `"12 H atoms"`, `"24 O atoms"`

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