A) passing\[C{{O}_{2}}\]into\[Ba{{O}_{2}}\]
B) adding\[Mn{{O}_{2}}\]to dil.\[{{H}_{2}}S{{O}_{4}}\]
C) adding\[N{{a}_{2}}{{O}_{2}}\]to cold water
D) adding\[Pb{{O}_{2}}\]into\[KMn{{O}_{4}}\].
Correct Answer: A
Solution :
Hydrogen peroxide is prepared by the action of\[C{{O}_{2}}\]on barium peroxide\[(Ba{{O}_{2}}).\] \[\underset{\begin{smallmatrix} (Barium \\ Peroxide) \end{smallmatrix}}{\mathop{Ba{{O}_{2}}}}\,+C{{O}_{2}}+{{H}_{2}}O\xrightarrow[{}]{{}}\underset{\begin{smallmatrix} (Barium \\ carbonate) \end{smallmatrix}}{\mathop{BaC{{O}_{3}}}}\,+\underset{\begin{smallmatrix} (Hydrogen \\ peroxide) \end{smallmatrix}}{\mathop{{{H}_{2}}{{O}_{2}}}}\,\]You need to login to perform this action.
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