Answer :
We begin by writing the balanced chemical equation for the decomposition of potassium chlorate:
$$
2\,KClO_3 \rightarrow 2\,KCl \ +\ 3\,O_2
$$
**Step 1. Calculate the moles of $KClO_3$:**
The molar mass of potassium chlorate is approximately
$$
M_{KClO_3} = 39.1 + 35.45 + 3(16) = 122.55\ \text{g/mol}.
$$
Thus, the number of moles of $KClO_3$ is
$$
\text{moles of } KClO_3 = \frac{35.8\ \text{g}}{122.55\ \text{g/mol}} \approx 0.2921\ \text{mol}.
$$
**Step 2. Determine the theoretical moles of $O_2$:**
From the balanced equation, we see that
$$
2\ \text{mol}\ KClO_3 \rightarrow 3\ \text{mol}\ O_2.
$$
This means that one mole of $KClO_3$ yields $\frac{3}{2}$ moles of oxygen gas. Therefore, the theoretical moles of $O_2$ produced are
$$
\text{moles of } O_2 = 0.2921\ \text{mol} \times \frac{3}{2} \approx 0.4382\ \text{mol}.
$$
**Step 3. Calculate the theoretical mass of $O_2$:**
The molar mass of oxygen gas ($O_2$) is
$$
M_{O_2} = 2 \times 16.00 = 32.00\ \text{g/mol}.
$$
Hence, the theoretical mass of oxygen is
$$
\text{mass of } O_2 = 0.4382\ \text{mol} \times 32.00\ \text{g/mol} \approx 14.0220\ \text{g}.
$$
**Step 4. Calculate the percent yield:**
The percent yield is given by
$$
\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%.
$$
Given that the actual yield of oxygen is $11.0\ \text{g}$, we have
$$
\text{Percent Yield} = \left(\frac{11.0\ \text{g}}{14.0220\ \text{g}}\right) \times 100\% \approx 78.45\%.
$$
Thus, the percent yield of the oxygen gas is approximately $78.45\%$.
$$
2\,KClO_3 \rightarrow 2\,KCl \ +\ 3\,O_2
$$
**Step 1. Calculate the moles of $KClO_3$:**
The molar mass of potassium chlorate is approximately
$$
M_{KClO_3} = 39.1 + 35.45 + 3(16) = 122.55\ \text{g/mol}.
$$
Thus, the number of moles of $KClO_3$ is
$$
\text{moles of } KClO_3 = \frac{35.8\ \text{g}}{122.55\ \text{g/mol}} \approx 0.2921\ \text{mol}.
$$
**Step 2. Determine the theoretical moles of $O_2$:**
From the balanced equation, we see that
$$
2\ \text{mol}\ KClO_3 \rightarrow 3\ \text{mol}\ O_2.
$$
This means that one mole of $KClO_3$ yields $\frac{3}{2}$ moles of oxygen gas. Therefore, the theoretical moles of $O_2$ produced are
$$
\text{moles of } O_2 = 0.2921\ \text{mol} \times \frac{3}{2} \approx 0.4382\ \text{mol}.
$$
**Step 3. Calculate the theoretical mass of $O_2$:**
The molar mass of oxygen gas ($O_2$) is
$$
M_{O_2} = 2 \times 16.00 = 32.00\ \text{g/mol}.
$$
Hence, the theoretical mass of oxygen is
$$
\text{mass of } O_2 = 0.4382\ \text{mol} \times 32.00\ \text{g/mol} \approx 14.0220\ \text{g}.
$$
**Step 4. Calculate the percent yield:**
The percent yield is given by
$$
\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%.
$$
Given that the actual yield of oxygen is $11.0\ \text{g}$, we have
$$
\text{Percent Yield} = \left(\frac{11.0\ \text{g}}{14.0220\ \text{g}}\right) \times 100\% \approx 78.45\%.
$$
Thus, the percent yield of the oxygen gas is approximately $78.45\%$.
