Explanation: Hints: \(PV = nRT\)
Solve:
\[
2KClO_3 \rightarrow 2KCl + 3O_2
\]
\(122.5 \, \text{g}\)
\(PV = nRT\)
\(n_{O_2} = \frac{PV}{RT} = \frac{1 \times 1.8}{0.0821 \times 295} = 0.075 \, \text{mole}\)
\(3 \, \text{mole } O_2 \equiv 2 \, \text{mole } KClO_3\)
\(0.075 \, \text{mole } O_2 = \frac{2 \times 0.075}{3} \, \text{mole } KClO_3 = 0.05 \, \text{mole } KClO_3\)
\[
0.05 \times 122.5 = 6.125 \, \text{g } KClO_3
\]
Ans. (D)
