FeCr_2O_4 (chromite) is converted to Cr by fo | vedclass

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(B) Step $I$: Chromite ore $(FeCr_2O_4)$ is fused with $NaOH$ in the presence of air to form sodium chromate $(Na_2CrO_4)$.$4FeCr_2O_4 + 8NaOH + 7O_2

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$NaOH/air, \Delta$ $-$ $C, \Delta$ $-$ $Al, \Delta$

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(B) Step $I$: Chromite ore $(FeCr_2O_4)$ is fused with $NaOH$ in the presence of air to form sodium chromate $(Na_2CrO_4)$.$4FeCr_2O_4 + 8NaOH + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8H_2O$.Step $II$: Sodium chromate is treated with carbon to reduce it to chromium$(III)$ oxide $(Cr_2O_3)$.$2Na_2CrO_4 + 2C \rightarrow Cr_2O_3 + Na_2CO_3 + CO$.Step $III$: Chromium$(III)$ oxide is reduced to metallic chromium using the aluminothermic process ($Al$ as a reducing agent).$Cr_2O_3 + 2Al \rightarrow 2Cr + Al_2O_3$.

$FeCr_2O_4$ (chromite) is converted to $Cr$ by following steps:
Chromite $\xrightarrow{I} Na_2CrO_4$ $\xrightarrow{II} Cr_2O_3$ $\xrightarrow{III} Cr$
Reagents in $I, II$ and $III$ steps might be:
$I$ Step $-$ $II$ Step $-$ $III$ Step

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$FeCr_2O_4$ (chromite) is converted to $Cr$ by following steps:Chromite $\xrightarrow{I} Na_2CrO_4$ $\xrightarrow{II} Cr_2O_3$ $\xrightarrow{III} Cr$Reagents in $I, II$ and $III$ steps might be:$I$ Step $-$ $II$ Step $-$ $III$ Step