5.7 g of KCl is produced.
Explanation:
First we have to write the balanced equation as,
[tex]2 K C l O_{3} \rightarrow 2 K C l+3 O_{2} \uparrow[/tex]
Then we have to find the moles of potassium chlorate as,
[tex]Moles of $K C l O_{3}=9.3 g$ of $K C l O_{3} \times \frac{1 \text { mol of } K C l O_{3}}{122.55 g K C l O_{3}}=0.076$ moles[/tex]
Then we have to find the moles of KCl as,
[tex]Moles of $\mathrm{KCl}=0.076$ moles of $\mathrm{KClO}_{3} \times \frac{2 \text { mol of } \mathrm{KCl}}{2 \text { mol } \mathrm{KClo}_{3}}=0.076$ moles[/tex]
From the moles of KCl, we have to find the mass using its molar mass by multiplying them as,
[tex]\text { Mass of } \mathrm{KCl}=0.076 \text { mol } \times 74.55 \frac{g}{\mathrm{mol}}=5.7 \mathrm{g} \text { of } \mathrm{KCl}[/tex]
