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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?