A) 1:1
B) 1:2
C) 2:1
D) None of these
Correct Answer: B
Solution :
\[AgN{{O}_{3}}\equiv 2A{{g}^{+}}+\underset{({{H}_{2}}S)}{\mathop{{{S}^{2-}}}}\,\to A{{g}_{2}}S\] \[\because \] 2 mole \[\to \] 1 mole [100×1 =100 millimole] \[\therefore \] 100 miliimole \[\to \]50 millimole \[{{H}_{2}}S\] required \[CuS{{O}_{4}}\equiv C{{u}^{+2}}+\underset{({{H}_{2}}S)}{\mathop{{{S}^{2-}}}}\,\to CuS\] \[\because \] 1 mole \[\to \]1 mole [100×1=100 millimole] \[\therefore \] 100 millimole \[\to \] 100 millimole \[{{H}_{2}}S\] required Ratio \[\frac{50}{100}=\frac{1}{2}\].You need to login to perform this action.
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