Answer :
Answer:
[tex]\rm K[/tex]: approximately [tex]22.0\%[/tex].
[tex]\rm Cl[/tex]: approximately [tex]19.9\%[/tex].
[tex]\rm O[/tex]: approximately [tex]27.0\%[/tex].
[tex]\rm Al[/tex]: [tex]31.0\%[/tex].
Explanation:
Consider a [tex]100\; \rm g[/tex] sample. There would be [tex]69.0\% \times 100\; \rm g = 69.0\; \rm g[/tex] of [tex]\rm KClO_{3}[/tex] and [tex]100.0\; \rm g - 69.0\; \rm g = 31.0\; \rm g[/tex] of [tex]\rm Al[/tex].
Out of that [tex]69.0\; \rm g[/tex] of [tex]\rm KClO_{3}[/tex]:
[tex]31.9\%[/tex] (by mass) would be [tex]\rm K[/tex]: [tex]31.9\% \times 69.0\; \rm g \approx 22.0\; \rm g[/tex].
[tex]28.9\%[/tex] (by mass) would be [tex]\rm Cl[/tex]: [tex]28.9\% \times 69.0\; \rm g \approx 19.9\; \rm g[/tex].
[tex]39.2\%[/tex] (by mass) would be [tex]\rm O[/tex]: [tex]39.2\% \times 69.0\; \rm g \approx 27.0\; \rm g[/tex].
Overall, the composition (by mass) of each element in this mixture would be:
[tex]\rm K[/tex]:
[tex]\begin{aligned} & \frac{31.9\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 31.9\% \times 69.0\% \\ \approx \; & 22.0\%\end{aligned}[/tex].
[tex]\rm Cl[/tex]:
[tex]\begin{aligned} & \frac{28.9\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 28.9\% \times 69.0\% \\ \approx \; & 19.9\%\end{aligned}[/tex].
[tex]\rm O[/tex]:
[tex]\begin{aligned} & \frac{39.2\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 39.2\% \times 69.0\% \\ \approx \; & 27.0\%\end{aligned}[/tex].
[tex]\rm Al[/tex]:
[tex]\begin{aligned}& \frac{100\; \rm g - 69.0\; \rm g}{100\; \rm g} \\ =\; & 100\% - 69.0\% \\ =\; & 31.0\%\end{aligned}[/tex].