Explanation:
Start with the unbalanced equation:
KClO
3
→
KCl
+
O
2
A method that often works is to balance everything other than
H
and
O
first, then balance
O
, and finally balance
H
.
Another rule is to start with what looks like the most complicated formula.
The most complicated formula looks like
KClO
3
.
We put a
1
in front of it and mark the
1
to remind ourselves that the number is now fixed.
We can't change a coefficient unless we run into a roadblock (like having to use fractions).
My teacher never let me use fractions.
My solution when I hit a roadblock was to erase all the numbers and then start over again with a 2 as the starting coefficient.
We start with
1
KClO
3
→
KCl
+
O
2
Balance
K
:
We have 1
K
on the left, so we need 1
K
on the right. We put a
1
in front of the
KCl
.
1
KClO
3
→
1
KCl
+
O
2
Balance
Cl
:
Cl
is already balanced, with 1
Cl
on each side.
Balance
O
:
We have 3
O
atoms on the left and only
2
on the right. We need 1½
O
2
molecules on the right. Uh, oh! Fractions!
We start over with a 2 as the coefficient.
2
KClO
3
→
2
KCl
+
O
2
Now we have 6
O
atoms on the left. To get 6
O
atoms on the right, we put a 3 in front of the
O
2
.
2
KClO
3
→
2
KCl
+
3
O
2
Every formula now has a fixed coefficient.
We should have a balanced equation.
Let’s check:
Left hand side: 2
K
, 2
Cl
, 6
O
Right hand side: 2
K
, 2
Cl
, 6
O
All atoms balance.
The balanced equation is
