How many moles of oxygen gas are produced by the decomposition of $245$ g of potassium chlorate?

\[\ce{2KClO3(s)} \rightarrow \ce{2KCl(s)} + \ce{3O2(g)}\]

Given:
Molar Mass/ $\text{g} \cdot \text{mol}^{-1}$
$\ce{KClO3}$: $122.6$

$ \textbf{(A)}\hspace{.05in}1.50 \qquad\textbf{(B)}\hspace{.05in}2.00 \qquad\textbf{(C)}\hspace{.05in}2.50 \qquad\textbf{(D)}\hspace{.05in}3.00 \qquad $
1. **Calculate the number of moles of potassium chlorate ($\ce{KClO3}$):**
   \[
   \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{245 \text{ g}}{122.6 \text{ g/mol}}
   \]
   \[
   \text{Number of moles} = 2.00 \text{ mol}
   \]

2. **Use the stoichiometry of the reaction to find the moles of oxygen gas ($\ce{O2}$) produced:**
   The balanced chemical equation is:
   \[
   \ce{2KClO3(s)} \rightarrow \ce{2KCl(s)} + \ce{3O2(g)}
   \]
   According to the equation, 2 moles of $\ce{KClO3}$ produce 3 moles of $\ce{O2}$.

3. **Calculate the moles of $\ce{O2}$ produced from 2.00 moles of $\ce{KClO3}$:**
   \[
   \text{Moles of } \ce{O2} = \left(\frac{3 \text{ moles of } \ce{O2}}{2 \text{ moles of } \ce{KClO3}}\right) \times 2.00 \text{ moles of } \ce{KClO3}
   \]
   \[
   \text{Moles of } \ce{O2} = \frac{3}{2} \times 2.00 = 3.00 \text{ moles}
   \]

The final answer is $\boxed{3.00}$