Question

When FeCr204 is fused with Na2CO3 in the presence of air it gives a
yellow solution of compound (A). Compound (A) on acidification gives
compound (B). Compound (B) on reaction with KCl forms an orange
coloured compound (C). An acidified solution of compound (C) oxidises
Na2SO3 to (D). Identify (A), (B), (C) and (D).

Answer 1

Answer:

A = FeCr2O4    B = Na2CrO4    C = Na2Cr2O7.2H2O D = K2Cr2O7  

hope..it..helps..

Answer 2

A is Na₂CrO₄.

B is Na₂Cr₂O₇

C is K₂Cr₂O₇

D is  Na₂SO₄

When fused chromite (FeCr₂O₄) is reacted with sodium carbonate (Na₂CO₃), sodium chromate (Na₂CrO₄) is formed,which is yellow in colour.

4FeCr₂O₄ +8 Na₂CO₃+ 7O₂ →8 Na₂CrO₄ + 2Fe₂O₃ + 8CO₂

So, A is Na₂CrO₄.

When Na₂CrO₄ is acidified in presence of H₂SO₄, sodium dichromate is formed (Na₂Cr₂O₇) along with sodium sulphate (Na₂SO₄) and H₂O.

2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O

Na₂Cr₂O₇ on reaction with KCl forms potassium dichromate (K₂Cr₂O₇).

Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl

K₂Cr₂O₇ formed is purple in colour.

So, B is Na₂Cr₂O₇ and C is K₂Cr₂O₇.

Acidic solution of K₂Cr₂O₇ oxidises Na₂SO₃ to Na₂SO₄.

So, D is  Na₂SO₄