• # question_answer Which of the following combination will produce ${{\operatorname{H}}_{2}}gas?$ A) Fe metal and cone. $HN{{O}_{3}}$ B) Cu metal and cone. $HN{{O}_{3}}$ C) Zn metal and $NaOH\text{ }(aq)$ D) Au metal and $NaCN\text{ }(aq)$ in the presence of air

Solution :

 Fe becomes passive on reaction with concentrated$HN{{O}_{3}}$. However, cold relatively cone.${{\operatorname{HNO}}_{3}}$, reacts with Fe as below. $Fe+6HN{{O}_{3}}\to Fe{{\left( N{{O}_{3}} \right)}_{3}}+3N{{O}_{2}}+3{{H}_{2}}O$ $Fe+6HN{{O}_{3}}\to Fe{{\left( N{{O}_{3}} \right)}_{3}}+3N{{O}_{2}}+3{{H}_{2}}O$$+2{{H}_{2}}O$ $4Au+8NaCN+{{O}_{2}}+2{{H}_{2}}O\to$$4Na[Au {{(CN)}_{2}}]+ 4NaOH$ $Zn+2NaOH\to N{{a}_{2}}Zn{{O}_{2}}+{{H}_{2}},$

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