# How can I balance this equation? ____ ##KClO_3 ->## ____ ##KCl +## ____ ##O_2##

You follow a systematic procedure to balance the equation.

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 ##color(red)(1)## in front of it and mark the ##color(red)(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

##color(red)(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 ##color(blue)(1)## in front of the ##"KCl"##.

##color(red)(1)"KClO"_3 → color(blue)(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.

##color(red)(2)"KClO"_3 → color(blue)(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##.

##color(red)(2)"KClO"_3 → color(blue)(2)"KCl" + color(orange)(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

##"2KClO"_3 → "2KCl" + "3O"_2##

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