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BCECE Engineering BCECE Engineering Solved Paper-2001

  • question_answer
    \[Mn{{O}_{2}}(s)+4HCl(aq)\xrightarrow{Heat}\] \[MnC{{l}_{2}}(aq)+2{{H}_{2}}O+C{{l}_{2}}(g)\] The equivalent weight of \[Mn{{O}_{2}}\]in the above reaction is:

    A) 23.4                       

    B)        33.6

    C) 43.5                       

    D)        53.6

    Correct Answer: C

    Solution :

    \[\underset{1\,\text{mole}\,\text{(M)}}{\mathop{Mn{{O}_{2}}(g)}}\,+4HCl(aq)\xrightarrow{\Delta }\] \[MnC{{l}_{2}}+2{{H}_{2}}O+\underset{(71\,g)}{\mathop{C{{l}_{2}}(g)}}\,\] \[\because \]71 got \[C{{l}_{2}}\]is displaced by M g of \[\text{Mn}{{\text{O}}_{\text{2}}}\] \[\therefore \]\[35.5\,g\]of \[C{{l}_{2}}\]is displaced by \[\frac{M\times 35.5}{71}\] \[=\frac{M}{2}g\]of \[Mn{{O}_{2}}\] \[\therefore \]Eq. wt. of \[Mn{{O}_{2}}=\frac{\text{Molecular}\,\text{weight}\,\text{of}\,\text{Mn}{{\text{O}}_{\text{2}}}}{\text{2}}\]                 \[=\frac{87}{2}=43.5\]


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