100 mL of 0.05 M Cu^{2+} aqueous solution i | vedclass

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Question

(D) The balanced chemical equation for the reaction is: $2Cu^{2+} + 4I^- ightarrow Cu_2I_2 + I_2$. Moles of $Cu^{2+} = M 	imes V(L) = 0.05 	imes 0

Vedclass · Accepted Answer

$2.5 	imes 10^{-3}, 2.5 	imes 10^{-3}$

Vedclass · Answer

(D) The balanced chemical equation for the reaction is: $2Cu^{2+} + 4I^- ightarrow Cu_2I_2 + I_2$. Moles of $Cu^{2+} = M 	imes V(L) = 0.05 	imes 0.1 = 5 	imes 10^{-3} \ mol$. Moles of $I^- = M 	imes V(L) = 0.1 	imes 1 = 0.1 \ mol$. Since $I^-$ is in excess, the reaction is limited by $Cu^{2+}$. According to the stoichiometry, $2 \ mol$ of $Cu^{2+}$ produces $1 \ mol$ of $I_2$ and $1 \ mol$ of $Cu_2I_2$. Therefore, $5 	imes 10^{-3} \ mol$ of $Cu^{2+}$ will produce: Moles of $I_2 = \frac{5 	imes 10^{-3}}{2} = 2.5 	imes 10^{-3} \ mol$. Moles of $Cu_2I_2 = \frac{5 	imes 10^{-3}}{2} = 2.5 	imes 10^{-3} \ mol$.

$100 \ mL$ of $0.05 \ M$ $Cu^{2+}$ aqueous solution is added to $1 \ L$ of $0.1 \ M$ $KI$ solution. The number of moles of $I_2$ and $Cu_2I_2$ formed are respectively:

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