The ratio of the molar amounts of H_2S needed | vedclass

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(B) The precipitation reactions are:$Cd^{2+} + H_2S \rightarrow CdS + 2H^+$$Cu^{2+} + H_2S \rightarrow CuS + 2H^+$For $Cd(NO_3)_2$:$Moles = Molarity \

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$2:1$

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(B) The precipitation reactions are:$Cd^{2+} + H_2S ightarrow CdS + 2H^+$$Cu^{2+} + H_2S ightarrow CuS + 2H^+$For $Cd(NO_3)_2$:$Moles = Molarity 	imes Volume = 1 \ M 	imes 0.020 \ L = 0.020 \ mol = 20 \ mmol$.Since the stoichiometry is $1:1$,$20 \ mmol$ of $H_2S$ is required.For $CuSO_4$:$Moles = Molarity 	imes Volume = 0.5 \ M 	imes 0.020 \ L = 0.010 \ mol = 10 \ mmol$.Since the stoichiometry is $1:1$,$10 \ mmol$ of $H_2S$ is required.The ratio of $H_2S$ required is $20:10 = 2:1$.

The ratio of the molar amounts of $H_2S$ needed to precipitate the metal ions from $20 \ mL$ each of $1 \ M \ Cd(NO_3)_2$ and $0.5 \ M \ CuSO_4$ is

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