A) 1 : 1
B) 1 : 2
C) 2 : 1
D) None of these
Correct Answer: B
Solution :
100 mL of \[1\,M\,AgN{{O}_{3}}\equiv 0.1\,mol\,AgN{{O}_{3}}\] \[100\,mL\,of\,1M\,CuS{{O}_{4}}=0.1\,mol\,CuS{{O}_{4}}\] \[\underset{\begin{smallmatrix} \text{2}\,\text{mol} \\ \text{0}\text{.1}\,\text{mol} \end{smallmatrix}}{\mathop{2\,AgN{{O}_{3}}}}\,+\underset{\begin{smallmatrix} \text{1}\,\text{mol} \\ \text{0}\text{.05}\,\text{mol} \end{smallmatrix}}{\mathop{{{H}_{2}}S}}\,\xrightarrow{{}}A{{g}_{2}}S+2HN{{O}_{3}}\] \[\underset{\begin{smallmatrix} \text{1}\,\text{mol} \\ \text{0}\text{.1}\,\text{mol} \end{smallmatrix}}{\mathop{CuS{{O}_{4}}}}\,+\underset{\begin{smallmatrix} \text{1}\,\text{mol} \\ \text{0}\text{.1}\,\text{mol} \end{smallmatrix}}{\mathop{{{H}_{2}}S}}\,\xrightarrow{{}}CuS+{{H}_{2}}S{{O}_{4}}\] \[\therefore \] Ratio of the amounts of \[{{\text{H}}_{\text{2}}}\text{S}\]obtained \[=0.05:0.1=1:2\]
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