JEE Main & Advanced JEE Main Online Paper (Held on 9 April 2013)

  • question_answer
                    Given: \[\operatorname{X}\operatorname{N}{{\operatorname{a}}_{2}}{{\operatorname{HAsO}}_{3}}+\operatorname{Y}{{\operatorname{NaBrO}}_{3}}+\operatorname{ZHCI}\to \operatorname{NaBr}\]\[\text{+}{{\text{H}}_{\text{2}}}\text{As}{{\text{O}}_{\text{4}}}\text{+NaCl}\]                 The values of X, Y and Z in above reaction are respectively:           JEE Main Online Paper (Held On 09 April 2013)        

    A)                 2, 1, 2                

    B)                 2, 1, 3                

    C)                 3, 1, 6                

    D)                 3, 1, 4                

    Correct Answer: C

    Solution :

                    \[3N{{a}_{2}}HAs{{O}_{3}}+NaBr{{O}_{3}}+6HCl\]\[\xrightarrow{\,}\,NaBr+{{H}_{3}}As{{O}_{4}}+NaCl\]                


You need to login to perform this action.
You will be redirected in 3 sec spinner