Answer:
![\large \boxed{2}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B2%7D)
Explanation:
Often, the best way to start is to balance all atoms other than O and H, then balance O, then balance H.
1. Put a 1 in front of the most complicated-looking formula (KClO₃):
<u>1</u>KClO₃ ⟶ 2KCl + O₂
2. Balance K:
We have fixed 1 K on the left. We need 1 K on the right. Put a 1 in front of KCl.
<u>1</u>KClO₃ ⟶ <u>1</u>KCl + O₂
3. Balance Cl:
Done.
4. Balance O:
We have fixed 3 O on the left. We need 3O on the right. Uh, oh. Fractions.
Double every underlined coefficient
<u>2</u>KClO₃ ⟶ <u>3</u>KCl + O₂
Now, we have 6 O on the left. we can put a 3 in front of O₂.
<u>2</u>KClO₃ ⟶ <u>3</u>KCl + <u>3</u>O₂
Every formula now has a coefficient. The equation should be balanced.
5. Check that atoms balance:
![\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{K} & 2 &2\\\text{Cl} & 2 & 2\\\text{O} & 6 & 6\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bccc%7D%5Ctextbf%7BAtom%7D%20%26%20%5Ctextbf%7BOn%20the%20left%7D%20%26%20%5Ctextbf%7BOn%20the%20right%7D%5C%5C%5Ctext%7BK%7D%20%26%202%20%262%5C%5C%5Ctext%7BCl%7D%20%26%202%20%26%202%5C%5C%5Ctext%7BO%7D%20%26%206%20%26%206%5C%5C%5Cend%7Barray%7D)
The equation is now balanced.