The treatment of an aqueous solution of 3.74 | vedclass

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(D) Step $1$: Calculate moles of $Cu(NO_3)_2$. Molar mass of $Cu(NO_3)_2 = 63 + 2 	imes (14 + 3 	imes 16) = 63 + 2 	imes 62 = 187 \ g/mol$. Moles o

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$0.32$

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(D) Step $1$: Calculate moles of $Cu(NO_3)_2$. Molar mass of $Cu(NO_3)_2 = 63 + 2 	imes (14 + 3 	imes 16) = 63 + 2 	imes 62 = 187 \ g/mol$. Moles of $Cu(NO_3)_2 = \frac{3.74}{187} = 0.02 \ mol$.Step $2$: Reaction with $KI$: $2Cu(NO_3)_2 + 4KI \longrightarrow Cu_2I_2 \downarrow + I_2 + 4KNO_3$. The $I_2$ formed reacts with excess $KI$ to form $KI_3$ (brown solution): $I_2 + KI \longrightarrow KI_3$. Thus,$2 \ mol$ of $Cu(NO_3)_2$ produces $1 \ mol$ of $I_2$,which forms $1 \ mol$ of $KI_3$.Step $3$: Reaction with $H_2S$: $KI_3 + H_2S \longrightarrow S \downarrow + KI + 2HI$. Here,$1 \ mol$ of $KI_3$ produces $1 \ mol$ of $S$ (precipitate $X$).Step $4$: Since $2 \ mol$ of $Cu(NO_3)_2$ gives $1 \ mol$ of $KI_3$,then $0.02 \ mol$ of $Cu(NO_3)_2$ gives $0.01 \ mol$ of $KI_3$.Step $5$: $0.01 \ mol$ of $KI_3$ produces $0.01 \ mol$ of $S$. Mass of $S = 0.01 	imes 32 = 0.32 \ g$.

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