1.5 g of oxygen is produced by heating ${\mathrm{KClO}}_3$ . How much KCl is produced in the reaction?

$2{\mathrm{KClO}}_3\left(s\right)⟶2\mathrm{KCl}\left(s\right)+3{\mathrm{O}}_2\left(g\right)$

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

moles of ${\mathrm{O}}_2=\frac{1.5}{32}$

$\Rightarrow 4.68\times {10}^{-2}\mathrm{mol}$

From balanced equation,

3 moles of ${\mathrm{O}}_2=2\text{ moles }$of $\mathrm{KCl}$

$4.68\times {10}^{-2}\mathrm{mol} {\mathrm{O}}_2=\frac{2\times 4.68\times {10}^{-2}}{3}=3.12\times {10}^{-2}\mathrm{mol}$

Mass of $\mathrm{KCl}=$$\text{ moles }\times \text{ molarmass }$

                       $=2.33 \mathrm{g}$.