I. Na_2Cr_2O_7 + X xrightarrow{Delta} Cr_2O_3 | vedclass

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(A) In reaction $I$,sodium dichromate $(Na_2Cr_2O_7)$ is reduced to chromium$(III)$ oxide $(Cr_2O_3)$ using carbon $(C)$ as a reducing agent: $Na_2Cr_

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$C$ and $Al$

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(A) In reaction $I$,sodium dichromate $(Na_2Cr_2O_7)$ is reduced to chromium$(III)$ oxide $(Cr_2O_3)$ using carbon $(C)$ as a reducing agent: $Na_2Cr_2O_7 + 2C \xrightarrow{\Delta} Na_2CO_3 + Cr_2O_3 + CO$. In reaction $II$,chromium$(III)$ oxide $(Cr_2O_3)$ is reduced to metallic chromium $(Cr)$ using aluminum $(Al)$ in the thermite process: $Cr_2O_3 + 2Al \xrightarrow{\Delta} Al_2O_3 + 2Cr$. Therefore,$X = C$ and $Y = Al$.

$I$. $Na_2Cr_2O_7 + X \xrightarrow{\Delta} Cr_2O_3$
$II$. $Cr_2O_3 + Y \xrightarrow{\Delta} Cr$
$X$ and $Y$ are :

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$I$. $Na_2Cr_2O_7 + X \xrightarrow{\Delta} Cr_2O_3$$II$. $Cr_2O_3 + Y \xrightarrow{\Delta} Cr$$X$ and $Y$ are :