Answer:
[tex]\large \boxed{2}[/tex]
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):
1KClO₃ ⟶ 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.
1KClO₃ ⟶ 1KCl + 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
2KClO₃ ⟶ 3KCl + O₂
Now, we have 6 O on the left. we can put a 3 in front of O₂.
2KClO₃ ⟶ 3KCl + 3O₂
Every formula now has a coefficient. The equation should be balanced.
5. Check that atoms balance:
[tex]\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}[/tex]
The equation is now balanced.
[tex]\text{The coefficient of KClO$_{3}$ is $\large \boxed{\mathbf{2}}$}[/tex]
