How many moles of acidified K_{2}Cr_{2}O_{7} | vedclass

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Question

(A) The balanced chemical equation for the reaction is:$Cr_{2}O_{7}^{2-} + 6I^{-} + 14H^{+} \longrightarrow 2Cr^{3+} + 3I_{2} + 7H_{2}O$From the stoic

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$2$

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(A) The balanced chemical equation for the reaction is:$Cr_{2}O_{7}^{2-} + 6I^{-} + 14H^{+} \longrightarrow 2Cr^{3+} + 3I_{2} + 7H_{2}O$From the stoichiometry of the reaction,$1$ mole of $K_{2}Cr_{2}O_{7}$ produces $3$ moles of $I_{2}$.Therefore,to liberate $6$ moles of $I_{2}$,the required moles of $K_{2}Cr_{2}O_{7}$ is:$\frac{1 	ext{ mol } K_{2}Cr_{2}O_{7}}{3 	ext{ mol } I_{2}} 	imes 6 	ext{ mol } I_{2} = 2 	ext{ moles of } K_{2}Cr_{2}O_{7}$.

How many moles of acidified $K_{2}Cr_{2}O_{7}$ are required to liberate $6$ moles of $I_{2}$ from an aqueous solution of $I^{-}$?

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