To determine the mass of \( \text{KClO}_3 \) needed to produce 32.0 g of \( \text{O}_2 \) given a yield of 65.0%, we'll follow these steps:
1. **Calculate the moles of \( \text{O}_2 \) produced:**
The molar mass of \( \text{O}_2 \) is:
\[
M_{\text{O}_2} = 32.0 \, \text{g/mol}
\]
Moles of \( \text{O}_2 \):
\[
\text{moles of } \text{O}_2 = \frac{32.0 \, \text{g}}{32.0 \, \text{g/mol}} = 1.0 \, \text{mol}
\]
2. **Determine the moles of \( \text{KClO}_3 \) needed:**
According to the balanced chemical equation:
\[
2 \, \text{KClO}_3 \rightarrow 2 \, \text{KCl} + 3 \, \text{O}_2
\]
This means 2 moles of \( \text{KClO}_3 \) produce 3 moles of \( \text{O}_2 \), or:
\[
2 \, \text{moles of } \text{KClO}_3 \text{ produce 3 moles of } \text{O}_2
\]
So, the moles of \( \text{KClO}_3 \) required for 1.0 mol of \( \text{O}_2 \) is:
\[
\text{moles of } \text{KClO}_3 = \frac{2}{3} \times 1.0 \, \text{mol} = \frac{2}{3} \, \text{mol} = 0.667 \, \text{mol}
\]
3. **Calculate the theoretical mass of \( \text{KClO}_3 \) needed:**
The molar mass of \( \text{KClO}_3 \) is:
\[
M_{\text{KClO}_3} = 39.1 \, (\text{K}) + 35.5 \, (\text{Cl}) + 3 \times 16.0 \, (\text{O}) = 122.6 \, \text{g/mol}
\]
Therefore, the theoretical mass of \( \text{KClO}_3 \) required is:
\[
\text{mass of } \text{KClO}_3 = 0.667 \, \text{mol} \times 122.6 \, \text{g/mol} = 81.8 \, \text{g}
\]
4. **Account for the yield of the reaction:**
Given a yield of 65.0%, the actual mass of \( \text{KClO}_3 \) needed is:
\[
\text{actual mass of } \text{KClO}_3 = \frac{81.8 \, \text{g}}{0.65} = 125.8 \, \text{g}
\]
Rounding to three significant figures, this is approximately:
\[
126 \, \text{g}
\]
Therefore, the correct answer is (D) 126 g.
