I came across this question in a recent exam I had taken:
How many moles of acidified $\ce{K2Cr2O7}$ is required to liberate 6 moles of $\ce{I2}$ from an aqueous solution of $\ce{I-}$ ?
So this is how I approached it (by writing the corresponding equations):
$$ \require{cancel} \begin{align} \ce{ Cr2O7^2- + 14 H+ + \cancel{6e^-} &-> 2Cr^3+ + 7H2O} \\ \ce{6 I- &-> 3I2 + \cancel{6e^-}}\\ \hline \ce{Cr2O7^2- + 6I- + 14H+&-> 2Cr^3+ + 3 I2 + 7H2O} \end{align} $$
Clearly, every mole of $\ce{K2Cr2O7}$ liberates $3$ moles of $\ce{I2}$ from an aqueous solution containing $\ce{I-}$.
So, I concluded that $2$ moles of $\ce{K2Cr2O7}$ are required to liberate 6 moles of $\ce{I2}$.
But the answer key given to me says that the correct answer for this question is $1$. Am I going wrong somewhere?