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JEE Main & Advanced JEE Main Paper (Held On 9 April 2016)

  • question_answer
    The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess \[{{H}_{2}}S\]in the presence of conc. HCl (assuming 100% conversion)   JEE Main Online Paper (Held On 09 April 2016)

    A) 0.25 mol                                              

    B) 0.125 mol

    C) 0.333 mol                                           

    D) 0.50 mol

    Correct Answer: B

    Solution :

                    \[{{H}_{3}}As{{O}_{4}}\xrightarrow[{}]{{{H}_{2}}S/HCl}A{{s}_{2}}{{S}_{5}}\] Assuming 100% conversion of As, apply POAC rule for 'As' atom \[1\times {{n}_{{{H}_{3}}AsO{{ & }_{4}}}}=2\times {{n}_{A{{s}_{2}}{{O}_{5}}}}\] \[\frac{35.5}{142}=2\times {{n}_{A{{s}_{2}}{{O}_{5}}}}\]\[\therefore \]\[{{N}_{A{{s}_{2}}O{{ & }_{5}}}}=0.125mol\]                


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