A) 153.12 g of \[KCl{{O}_{3}}\]
B) 122.5 g of \[KCl{{O}_{3}}\]
C) 245 g of\[KCl{{O}_{3}}\]
D) 98 g of \[KCl{{O}_{3}}\]
Correct Answer: A
Solution :
[a] \[\underset{\begin{smallmatrix} 2\times 122.5g \\ \,\,\,\,\,\,\,245 \end{smallmatrix}}{\mathop{2KCl{{O}_{3}}}}\,\xrightarrow{^{heat}}2KCl+\underset{\begin{smallmatrix} 3\times 32g \\ \,\,\,\,\,96 \end{smallmatrix}}{\mathop{3{{O}_{2}}}}\,\] 48 g of oxygen will be produced from 122.5 g of \[KCl{{O}_{3}}\] \[\therefore \] Amount of 80% \[KCl{{O}_{3}}\] needed \[=\frac{100}{80}\times 122.5=153.12g\]
You need to login to perform this action.
You will be redirected in
3 sec
