How do you balance `(NH_4)_3PO_4 + Pb(NO_3)_4 -> Pb_3(PO_4)_4 + NH_4NO_3`?

The unbalanced equation is

`("NH"_4)_3"PO"_4 +"Pb"("NO"_3)_4 → "Pb"_3("PO"_4)_4 + "NH"_4"NO"_3`

One way to balance the equation is to recognize that the polyatomic ions stay together.

Then we can make some substitutions.

Let `"A" = "NH"_4`; `"X" = "PO"_4`; `"Y" = "NO"_3`.

Then the equation becomes

`"A"_3"X" + "PbY"_4 → "Pb"_3"X"_4 + "AY"`

We start with the most complicated formula, `"Pb"_3"X"_4`, and put a 1 in front of it.

`"A"_3"X" + "PbY"_4 → color(red)(1)"Pb"_3"X"_4 + "AYX"`

Balance `"Pb"`

We have fixed 3 atoms of `"Pb"` on the right, so we need 3 atoms of `"Pb"` on the left. Put a 3 in front of `"PbY"_4`.

`"A"_3"X" + color(blue)(3)"PbY"_4 → color(red)(1)"Pb"_3"X"_4 + "AY"`

Balance `"X"`

We have fixed 4 atoms of `"X"` on the right, so we need 4 atoms of `"X"` on the left. Put a 4 in front of `"A"_3"X"`.

Balance `"A"`

We have fixed 12 atoms of `"A"` on the left, so we need 12 atoms of `"A"` on the right.

`color(orange)(4)"A"_3"X" + color(blue)(3)"PbY"_4 → color(red)(1)"Pb"_3"X"_4 + color(teal)(12)"AY"`

Every formula has a coefficient, and the equation is balanced.

Now, we replace the original formulas in the equation.

The balanced equation is

`4("NH"_4)_3"PO"_4 +"3Pb"("NO"_3)_4 → "Pb"_3("PO"_4)_4 + "12NH"_4"NO"_3`