The ratio of amounts of H_2S needed to precip | vedclass

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(B) For $AgNO_3$: The reaction is $2AgNO_3 + H_2S \rightarrow Ag_2S + 2HNO_3$.$2 \ mol$ of $AgNO_3$ react with $1 \ mol$ of $H_2S$.Amount of $AgNO_3 =

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$1:2$

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(B) For $AgNO_3$: The reaction is $2AgNO_3 + H_2S ightarrow Ag_2S + 2HNO_3$.$2 \ mol$ of $AgNO_3$ react with $1 \ mol$ of $H_2S$.Amount of $AgNO_3 = 100 \ mL 	imes 1 \ M = 100 \ mmol$.$H_2S$ required $= \frac{100}{2} = 50 \ mmol$.For $CuSO_4$: The reaction is $CuSO_4 + H_2S ightarrow CuS + H_2SO_4$.$1 \ mol$ of $CuSO_4$ reacts with $1 \ mol$ of $H_2S$.Amount of $CuSO_4 = 100 \ mL 	imes 1 \ M = 100 \ mmol$.$H_2S$ required $= 100 \ mmol$.Ratio of $H_2S$ required $= \frac{50}{100} = \frac{1}{2}$ or $1:2$.

The ratio of amounts of $H_2S$ 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

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