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

| May 27, 2018

You follow a systematic procedure to balance the 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.
##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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