BCECE Engineering BCECE Engineering Solved Paper-2015

  • question_answer
     Consider the following reaction,\[\text{ZnC}{{\text{l}}_{\text{2}}}\text{+NaOH}\to \underset{\text{White ppt}\text{.}}{\mathop{B}}\,\to \xrightarrow{NaOH}\underset{\text{soluble}}{\mathop{C}}\,\] Then, the final product C is

    A) \[\text{N}{{\text{a}}_{\text{2}}}\text{Zn}{{\text{O}}_{\text{2}}}\]   

    B) \[\text{Zn}{{\left( \text{OH} \right)}_{\text{2}}}\]

    C) \[\text{ZnO}\]                                  

    D) \[\text{ZnS}{{\text{O}}_{\text{4}}}\]

    Correct Answer: A

    Solution :

    Salts like \[\text{ZnC}{{\text{l}}_{\text{2}}}\text{,SnC}{{\text{l}}_{\text{2}}},\text{AlC}{{\text{l}}_{\text{3}}},\], etc, give white precipitate with \[\text{NaOH}\], however precipitate dissolves in excess of it. \[\text{ZnC}{{\text{l}}_{\text{2}}}\text{+2NaOH}\to \underset{\text{B}}{\mathop{\underset{\text{(whiteppt}\text{.)}}{\mathop{\text{Zn(OH}{{\text{)}}_{\text{2}}}}}\,}}\,\xrightarrow{\text{NaOH}}\underset{\text{C}}{\mathop{\underset{\text{Soluble}}{\mathop{\text{N}{{\text{a}}_{\text{2}}}\text{Zn}{{\text{O}}_{\text{2}}}}}\,}}\,\]                  


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