The ratio of the molar amounts of `H_(2)S` needed to precipitate the metal ions from 20 mL each 1 M Cd `(NO_(3))_(2)` and 0.5 M `CuSO_(4)` is
The ratio of the molar amounts of `H_(2)S` needed to precipitate the metal ions from 20 mL each 1 M Cd `(NO_(3))_(2)` and 0.5 M `CuSO_(4)` is
A. `1:1`
B. `2:1`
C. `1:2`
D. indefinite
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Correct Answer - B
`underset(1mol)(Cd(NH_(3))_(2))+underset(1mol)(H_(2)S)to CdS + H_(2)SO_(4)`
`underset(1mol)(CuSO_(4))+underset(1mol)(H_(2)S)to CuS + H_(2)SO_(4)`
20 mL of 1 M `Cd(NH_(3))_(2)`
`= ((20)/(1000)L)xx(1molL^(-1))`
`=0.02 mol`
`:. H_(2)S` required = 0.02 mol
20 mL of `0.5 M CuSO_(4)`
`= ((20)/(1000)L)xx(0.5molL^(-1))`
`=0.01 mol`
`:. H_(2)S` required = 0.01 mol
`:.` Ratio of `H_(2)S` required is 2:1
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