One last example:

Na₃PO₄ + BaCl₂→ Ba₃(PO₄)₂ + NaCl

Counting the elements and ions we get:

Left side	Right side
Na = 3	Na = 1
PO4 = 1	PO4 = 2
Ba = 1	Ba = 3
Cl = 2	Cl = 1

If we start at the left with Na...

Na₃PO₄ + BaCl₂→ Ba₃(PO₄)₂ + 3 NaCl

Left side	Right side
Na = 3	Na = 3
PO4 = 1	PO4 = 2
Ba = 1	Ba = 3
Cl = 2	Cl = 3

But now, if we move on to the phosphates...we end up UNbalancing the sodium's

2 Na₃PO₄ + BaCl₂→ Ba₃(PO₄)₂ + 3 NaCl

Left side	Right side
Na = 6	Na = 3
PO4 = 2	PO4 = 2
Ba = 1	Ba = 3
Cl = 2	Cl = 3

At this point, you may see the solution, but if not, it might be worth going back to the beginning and starting with a different element or ion.

Na₃PO₄ + BaCl₂→ Ba₃(PO₄)₂ + NaCl

Counting the elements and ions we get:

Left side	Right side
Na = 3	Na = 1
PO4 = 1	PO4 = 2
Ba = 1	Ba = 3
Cl = 2	Cl = 1

If we take a quick look at this reaction, we see that the barium phosphate on the right side is the most complex formula in the problem – that is, this is the compound with the most subscripts in it. Let's start there, with the barium (although it would work just as well to start with the phosphates.

Na₃PO₄ + 3 BaCl₂→ Ba₃(PO₄)₂ + NaCl

Left side	Right side
Na = 3	Na = 1
PO4 = 1	PO4 = 2
Ba = 3	Ba = 3
Cl = 6	Cl = 1

Since this changed the value for chloride, lets look at that next...

Na₃PO₄ + 3 BaCl₂→ Ba₃(PO₄)₂ + 6 NaCl

Left side	Right side
Na = 3	Na = 6
PO4 = 1	PO4 = 2
Ba = 3	Ba = 3
Cl = 6	Cl = 6

We can now look at the phosphates or the sodium's to finish the equation

2 Na₃PO₄ + 3 BaCl₂→ Ba₃(PO₄)₂ + 6 NaCl

...and everything is balanced.

A suggestion:

When balancing reactions, it is often best to start with the compound that has the most subscripts. For instance, it worked best in this example to start with the Ba₃(PO₄)₂.

Now, try these on your own

Pb(C₂H₃O2)₂ + KBr → KC₂H₃O₂ + PbBr₂

AlCl₃ + LiOH → LiCl + Al(OH)₃

Cu(NO₂)₂ + FeCl₃ → CuCl₂ + Fe(NO₂)₃