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QUESTION

# How can I balance this equation? ____ KClO_3 -&gt; ____ KCl + ____ O_2

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