Chemistry Reference
In-Depth Information
KClO
3
KCl + O
2
Elements
{
# of atoms
{
KClO KClO
113112
Not balanced!
Our tally shows us that we have one more atom of oxygen on the reac-
tant side than on the product side. Balancing this equation is not as easy as
you might assume when you see that we are only off by one atom. If you
look at the product side, there is no way to add only one atom to that side,
because the oxygen has a subscript of “2.” Putting a coefficient of “2” in
front of the oxygen on the product side will leave us 4 atoms of oxygen, and
we will be no closer to solving the equation. This is when some students get
desperate enough to violate our rule, leading them to simply change the
subscript on the reactant side to “2.” Please, remember that changing the
subscripts is never an option!
Our plan of attack is suggested by the subscript of “2” that comes with
the O
2
on the product side. Can you see that it is impossible to get an odd
number of oxygen atoms on the product side? Placing either an even or
odd coefficient in front of the oxygen will still result in an even number of
oxygen atoms on the right-hand side; for example, 3O
2
= 6 atoms of oxy-
gen. This tells us that the real problem is on the left-hand side of the equa-
tion, where we want to change the odd number of oxygen atoms to an even
number, perhaps by adding a coefficient of 2 before the KClO
3
.
2KClO
3
KCl + O
2
Elements
{
# of atoms
{
KClO KClO
226112
Not balanced!
I know that it seems that we just moved further away from the solution,
but some of these equations are akin to puzzles that need to be solved one
step at a time. Now, we have an even number of oxygen atoms on the left-
hand side, which we can match on the right by adding a coefficient of “3” in
front of the oxygen, as I show next.
2KClO
3
KCl + 3O
2
Elements
{
# of atoms
{
KClO KClO
226116
Not balanced!
