The weight of a pure sample of ${\mathrm{KClO}}_3$ to be decomposed in order to get 0.96g of ${\mathrm{O}}_2$ is

$2{\mathrm{KClO}}_3\to 2\mathrm{KCl}+3{\mathrm{O}}_2$

2 × 122.5 $\to$ 3 × 2 × 16g of ${\mathrm{O}}_2$

2 × 122.5 $\to$ 96 g of ${\mathrm{O}}_2$

           ?   $\leftarrow$0.96g of O₂

96 × $\mathrm{X}$ = 2 × 122.5 × 0.96

$\mathrm{x}=\frac{2\times 122.5\times 0.96}{96}$

$\mathrm{x}=\frac{2\times 122.5\times 96}{96\times 100}$

x = 2 × 1.225

x=2.45 g