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Calcination

FeCr2O4 Chrom...

Question

$F e C r_{2} O_{4}$ (Chromite) is a good source of chromium and its compound like $N a_{2} C r O_{4}$ and $N a_{2} C r_{2} O_{7}$

I. $F e C r_{2} O_{4} + N a O H + a i r ⟶ (A) + F e_{2} O_{3}$

II. $(A) + (B) ⟶ N a_{2} C r_{2} O_{7}$

III. $N a_{2} C r_{2} O_{7} + X △ ⟶ C r_{2} O_{3}$

IV. $C r_{2} O_{3} + Y △ ⟶ C r$

Compounds (A) and (B) are:

N a_{2} C r O_{4}, H_{2} S O_{4}

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N a_{2} C r_{2} O_{7}, H C l

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N a_{2} C r O_{6}, H_{2} S O_{4}

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N a_{4} [F e (O H)_{6}], H_{2} S O_{4}

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Solution

The correct option is D $N a_{2} C r O_{4}, H_{2} S O_{4}$
Compounds (A) and (B) are $N a_{2} C r O_{4} a n d H_{2} S O_{4}$ .

$F e C r_{2} O_{4} + N a O H + a i r \to (A) N a_{2} C r O_{4} + F e_{2} O_{3}$

$N a_{2} C r O_{4} (A) + H_{2} S O_{4} (B) \to N a_{2} C r_{2} O_{7}$

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Similar questions

F e C r_{2} O_{4} + N a O H + a i r \to (A) + F e_{2} O_{3}

(A) + (B) \to N a_{2} C r_{2} 07

N a_{2} C r_{2} O_{7} + X Δ - \to C r_{2} O_{3}

C r_{2} O_{3} + Y Δ - \to C r

F e C r_{2} O_{4}

N a_{2} C O_{3}

7 O_{2}

\to

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H_{2} S O_{4}

\to

N a_{2} C r_{2} O_{7}

2 K C l

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N H_{4} C l / Z n

H_{2} S O_{4} / N a_{2} C r_{2} O_{7}

F e C r_{2} O_{4}

C r

I \to

N a C r O_{4}

I I - \to

C r_{2} O_{3}

I I I - - \to

C r

K_{2} C r_{2} O_{7}

2 F e C r O_{4} + 2 N a_{2} C O_{3} + O \to F e_{2} O_{3} + 2 N a_{2} C r O_{4} + 2 C O_{2}

2 N a_{2} C r O_{4} + H_{2} S O_{4} \to N a_{2} C r_{2} O_{7} + H_{2} O + N a_{2} S O_{4}

N a_{2} C r_{2} O_{7} + 2 K C l \to K_{2} C r_{2} O_{7} + 2 N a C l

0.25

K_{2} C r_{2} O_{7}

50 %

F e C r O_{4}

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