The ratio of amounts of `H_(2)S` needed to precipitate all the metal ions from `100 ml` of `1 M AgNO_(3)` and `100 ml` of `1 M CuSO_(4)` will be

Options

A. `1:2`
B. `2:1`
C. Zero
D. Infinity

Answer

A

Explanation

The reaction of `AgNO_(3)` with `H_(2)S` : can be represented as
`2AgNO_(2) + H_(2)S to Ag_(2)S + 2HNO_(3)`
Here, 1 mole of `H_(2)S` reacts with 2 moles of `AgNO_(3)` .
`therefore ` The ratio of amount of `H_(2)S` required to precipitate all the metal ions from 100 mL 1M `AgNO_(3)` is
`1H_(2)S : 2AgNO_(3)`
The reaction of `CuSO_(4)` with `H_(2)S` : can be represented as
`CuSO_(4) + H_(2)S to CuS + H_(2)SO_(4)`
Here 1 mole of `H_(2)S` react with 1 mol of `CuSO_(4)` .
`therefore` The ration of amount of `H_(2)S` required to precipitate all the metal ions from 100 mL `1 M CuSO_(4)` is .
`1H_(2)S, 1CuSO_(4)`