To determine how many moles of oxygen ([tex]$O_2$[/tex]) are produced when 2 moles of potassium chlorate ([tex]$KClO_3$[/tex]) decompose, we can use stoichiometry based on the given balanced chemical equation:
[tex]\[
2 KClO_3 \rightarrow 2 KCl + 3 O_2
\][/tex]
1. Identify the molar ratios in the balanced equation:
- The coefficients indicate that 2 moles of [tex]$KClO_3$[/tex] produce 3 moles of [tex]$O_2$[/tex].
2. Set up the ratio based on the balanced equation:
- 2 moles of [tex]$KClO_3$[/tex] corresponds to 3 moles of [tex]$O_2$[/tex].
- Therefore, the ratio of [tex]$KClO_3$[/tex] to [tex]$O_2$[/tex] is [tex]\( \frac{3 \text{ moles } O_2}{2 \text{ moles } KClO_3} \)[/tex].
3. Calculate the moles of [tex]$O_2$[/tex] produced:
- For 2 moles of [tex]$KClO_3$[/tex], apply the ratio:
[tex]\[
\text{Moles of } O_2 = 2 \text{ moles } KClO_3 \times \frac{3 \text{ moles } O_2}{2 \text{ moles } KClO_3}
\][/tex]
4. Simplify the calculation:
[tex]\[
\text{Moles of } O_2 = \frac{2 \times 3}{2} = 3 \text{ moles } O_2
\][/tex]
Therefore, when 2 moles of potassium chlorate ([tex]$KClO_3$[/tex]) decompose, 3 moles of oxygen ([tex]$O_2$[/tex]) are produced.
Thus, the correct answer is:
[tex]\[
\boxed{3}
\][/tex]
