A) \[153.12g\] of \[KC1{{O}_{3}}\]
B) \[122.5\text{ }g\]of \[KC1{{O}_{3}}\]
C) \[245\text{ }g\]of \[KC1{{O}_{3}}\]
D) \[98.0\text{ }g\]of \[KC1{{O}_{3}}\]
Correct Answer: A
Solution :
\[\begin{matrix} KCl{{O}_{3}} \\ 1\,mol \\ 122.5 \\ \end{matrix}\begin{matrix} \xrightarrow{{}} \\ {} \\ {} \\ \end{matrix}\begin{matrix} KCl+ \\ {} \\ {} \\ \end{matrix}\begin{matrix} \frac{3}{2} \\ \frac{3}{2}mol \\ \frac{3}{2}\times 32=48g \\ \end{matrix}\begin{matrix} {{O}_{2}} \\ {} \\ {} \\ \end{matrix}\] 48g \[{{O}_{2}}\] is produced from \[122.5g\] of pure \[KCl{{O}_{3}}\] \[\therefore \] 80% \[KCl{{O}_{3}}\] needed \[=\frac{122.5\times 100}{80}=153.12\,g\]
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