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

Apr 19, 2017

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

#### Explanation:

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 ${\text{A" = "NH}}_{4}$; ${\text{X" = "PO}}_{4}$; ${\text{Y" = "NO}}_{3}$.

Then the equation becomes

$\text{A"_3"X" + "PbY"_4 → "Pb"_3"X"_4 + "AY}$

We start with the most complicated formula, ${\text{Pb"_3"X}}_{4}$, and put a 1 in front of it.

$\text{A"_3"X" + "PbY"_4 → color(red)(1)"Pb"_3"X"_4 + "AYX}$

Balance $\text{Pb}$

We have fixed 3 atoms of $\text{Pb}$ on the right, so we need 3 atoms of $\text{Pb}$ on the left. Put a 3 in front of ${\text{PbY}}_{4}$.

$\text{A"_3"X" + color(blue)(3)"PbY"_4 → color(red)(1)"Pb"_3"X"_4 + "AY}$

Balance $\text{X}$

We have fixed 4 atoms of $\text{X}$ on the right, so we need 4 atoms of $\text{X}$ on the left. Put a 4 in front of $\text{A"_3"X}$.

Balance $\text{A}$

We have fixed 12 atoms of $\text{A}$ on the left, so we need 12 atoms of $\text{A}$ on the right.

$\textcolor{\mathmr{and} a n \ge}{4} \text{A"_3"X" + color(blue)(3)"PbY"_4 → color(red)(1)"Pb"_3"X"_4 + color(teal)(12)"AY}$

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