Consider the following reactions:K_2Cr_2O_7 x | vedclass

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(C) The reaction of potassium dichromate $(K_2Cr_2O_7)$ with potassium hydroxide $(KOH)$ yields potassium chromate $(K_2CrO_4)$ and water:$K_2Cr_2O_7

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$K_2CrO_4$ and $K_2Cr_2O_7$

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(C) The reaction of potassium dichromate $(K_2Cr_2O_7)$ with potassium hydroxide $(KOH)$ yields potassium chromate $(K_2CrO_4)$ and water:$K_2Cr_2O_7 + 2KOH \rightarrow 2K_2CrO_4 + H_2O$Thus,$[A] = K_2CrO_4$.Next,the reaction of potassium chromate $(K_2CrO_4)$ with sulfuric acid $(H_2SO_4)$ yields potassium dichromate $(K_2Cr_2O_7)$,potassium sulfate $(K_2SO_4)$,and water:$2K_2CrO_4 + H_2SO_4 \rightarrow K_2Cr_2O_7 + K_2SO_4 + H_2O$Thus,$[B] = K_2Cr_2O_7$.

Consider the following reactions:
$K_2Cr_2O_7$ $\xrightarrow[-H_2O]{KOH} [A]$ $\xrightarrow[-H_2O]{H_2SO_4} [B] + K_2SO_4$
The products $[A]$ and $[B]$,respectively,are:

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Consider the following reactions:$K_2Cr_2O_7$ $\xrightarrow[-H_2O]{KOH} [A]$ $\xrightarrow[-H_2O]{H_2SO_4} [B] + K_2SO_4$The products $[A]$ and $[B]$,respectively,are: