A) 1 : 1
B) 1 : 2
C) 2 : 1
D) None of these
Correct Answer: B
Solution :
100 mL of 1 M \[AgN{{O}_{3}}\equiv 0.1\,mol\,AgN{{O}_{3}}\] 100 mL of 1M \[CuS{{O}_{4}}=0.1\,\,mol\,Cu\,S{{O}_{4}}\] \[\underset{\begin{smallmatrix} 2\,mol \\ 0.1\,mol \end{smallmatrix}}{\mathop{2AgN{{O}_{3}}}}\,\,\,\,+\,\,\,\underset{\begin{smallmatrix} 1\,mol \\ 0\,\,0.5\,mol \end{smallmatrix}}{\mathop{{{H}_{2}}S}}\,\xrightarrow{{}}\,\,A{{g}_{2}}S+2HN{{O}_{3}}\] \[\underset{\begin{smallmatrix} 1\,\,mol \\ 0.1\,mol \end{smallmatrix}}{\mathop{CuS{{O}_{4}}}}\,\,\,\,+\,\,\,\underset{\begin{smallmatrix} 1\,\,mol \\ 0.1\,\,mol \end{smallmatrix}}{\mathop{{{H}_{2}}S}}\,\xrightarrow{{}}\,\,CuS+{{H}_{2}}S{{O}_{4}}\] \[\therefore \] Ratio of the amounts of \[{{H}_{2}}S\] obtained = 0.05 : 0.1 = 1 : 2
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