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Number of moles of acidified potassium dichromate required to liberate 6 moles of iodine from an aqueous solution of Iodate ions

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?

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