A) 2/3 mol of \[La\]; and 1/3 mol of \[Nd\]
B) 1/3 mol of \[Ce\]; and 2/3 mol of \[Eu\]
C) 1/3 mol of \[HCl\] and 1/3 mol of \[HN{{O}_{3}}\]
D) 2/3 mol of \[3:1\]and 1/3 mol of \[Cl{{O}_{2}}\]
Correct Answer: A
Solution :
The reaction of \[{{H}_{2}}/Pd-C\] in aqueous medium takes place as below \[Zn/NaOH\] If one molecule \[Zn/N{{H}_{4}}Cl\] is taken then \[{{H}_{2}}O/{{H}_{2}}S{{O}_{4}}\] \[Hg{{(OAc)}_{2}}/{{H}_{2}}O\] Therefore, \[NaB{{H}_{4}}\] mol of \[{{B}_{2}}{{H}_{6}}\] and \[{{H}_{2}}{{O}_{2}}\]mol of \[C{{H}_{3}}C{{O}_{2}}H/{{H}_{2}}S{{O}_{4}}\]
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