I. 4FeCr_2O_4 + 16NaOH + 7O_2 rightarrow 8Na_ | vedclass

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(A) The extraction of chromium from chromite ore $(FeCr_2O_4)$ involves the following steps:Step $I$: The chromite ore is fused with sodium hydroxide

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$Na_2CrO_4, H_2SO_4$

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(A) The extraction of chromium from chromite ore $(FeCr_2O_4)$ involves the following steps:Step $I$: The chromite ore is fused with sodium hydroxide in the presence of air to form sodium chromate $(Na_2CrO_4)$.$4FeCr_2O_4 + 16NaOH + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8H_2O$Here,$(A) = Na_2CrO_4$.Step $II$: The yellow solution of sodium chromate is acidified with concentrated sulfuric acid $(H_2SO_4)$ to convert it into orange sodium dichromate $(Na_2Cr_2O_7)$.$2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O$Here,$(B) = H_2SO_4$.Thus,the compounds $(A)$ and $(B)$ are $Na_2CrO_4$ and $H_2SO_4$ respectively.

$I$. $4FeCr_2O_4 + 16NaOH + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8H_2O$ (where $A = Na_2CrO_4$)
$II$. $2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O$ (where $B = H_2SO_4$)
Compounds $(A)$ and $(B)$ are :

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$I$. $4FeCr_2O_4 + 16NaOH + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8H_2O$ (where $A = Na_2CrO_4$)$II$. $2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O$ (where $B = H_2SO_4$)Compounds $(A)$ and $(B)$ are :