How many moles of K_2Cr_2O_7 can be reduced b | vedclass

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(A) The balanced half-reactions are:$Cr_2O_7^{2-} + 14H^{+} + 6e^- 	o 2Cr^{3+} + 7H_2O$$(Sn^{2+} 	o Sn^{4+} + 2e^-) 	imes 3$Adding these gives the

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$1/3$

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(A) The balanced half-reactions are:$Cr_2O_7^{2-} + 14H^{+} + 6e^- 	o 2Cr^{3+} + 7H_2O$$(Sn^{2+} 	o Sn^{4+} + 2e^-) 	imes 3$Adding these gives the overall balanced equation:$Cr_2O_7^{2-} + 14H^{+} + 3Sn^{2+} 	o 3Sn^{4+} + 2Cr^{3+} + 7H_2O$From the stoichiometry of the balanced equation,$3 \ mol$ of $Sn^{2+}$ are required to reduce $1 \ mol$ of $Cr_2O_7^{2-}$.Therefore,$1 \ mol$ of $Sn^{2+}$ will reduce $\frac{1}{3} \ mol$ of $K_2Cr_2O_7$.

How many moles of $K_2Cr_2O_7$ can be reduced by $1 \ mol$ of $Sn^{2+}$?

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