The ratio of amounts of $H_{2} S$ needed to precipitate all the metal ions from 100 ml of 1 M $A g N O_{3}$ and 100 ml of 1 M $C u S O_{4}$ will be :

1 : 2

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2 : 1

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1 : 1

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None of the above

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Solution

The correct option is A $1 : 2$
$2 A g N O_{3} + H_{2} S \to A g_{2} S + 2 H N O_{3} C u S O_{4} + H_{2} S \to C u S + H_{2} S O_{4} 2 m o l A g N O_{3} r e q u i r e s 1 m o l H_{2} S ∴ 1 m o l A g N O_{3} r e q u i r e s 1 / 2 m o l H_{2} S 1 m o l C u S O_{4} r e q u i r e s 1 m o l H_{2} S ∴ r a t i o o f H_{2} S r e q u i r e s t o p r e c i p i t a t e A g N O_{3} t o C u S O_{4} w i l l b e 1 / 2 : o r 1 : 2$

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The ratio of amounts of H2S needed to precipitate all the metal ions from 100 ml of 1 M AgNO3 and 100 ml of 1 M CuSO4 will be :

The correct option is A 1:22AgNO3+H2S→Ag2S+2HNO3CuSO4+H2S→CuS+H2SO42molAgNO3requires1molH2S∴1molAgNO3requires1/2molH2S1molCuSO4requires1molH2S∴ratioofH2SrequirestoprecipitateAgNO3toCuSO4will be1/2:or1:2

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