Question

When a chromite ore (A) is fused with an aqueous solution of sodium carbonate in free excess of air, a yellow solution of compound (B) is obtained. This solution is filtered and acidified with sulphuric acid to form compound (C). Compound (C) on treatment with solution of KCl gives orange crystals of compound (D). Write the chemical formulae of compounds A to D.

Answer 1

A=FeCr2O4,

B=Na2CrO4,

C=Na2Cr2O7.2H2O,

D= K2Cr2O7

• When a chromite ore is fused with sodium carbonate, a yellow solution of sodium chromate is formed

           4FeCr2O4 (A) + 8Na2CO3 + 7O2 → 8Na2CrO4 (B) + 2Fe2O3 + 8CO2

• When this yellow solution of sodium chromate is treated with acid, it changes to sodium dichromate

             2NaCrO4 + H2+ → Na2Cr2O7 (C) + 2Na2SO4 + H2O

• Sodium dichromate on treatment with KCl gives orange crystals of potassium dichromate

             Na2Cr2O7 + 2KCl → K2Cr2O7 (D) + 2NaCl

Answer 2

Chemical formulae of compounds A to D:

A = $\mathrm{FeCr}_{2} \mathrm{O}_{4}$

B = $\mathrm{Na}_{2} \mathrm{CrO}_{4}$

C = $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$

D = $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$

Explanation:

• When chromite reacts with sodium carbonate to form sodium chromate (yellow solution).

$4 \mathrm{FeCr}_{2} \mathrm{O}_{4}+8 \mathrm{Na}_{2} \mathrm{CO}_{3}+7 \mathrm{O}_{2} \longrightarrow 8 \mathrm{Na}_{2} \mathrm{CrO}_{4}+2 \mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{BCO}_{2}$

• When sodium chromate reacts with sulphuric acid to form sodium dichromate (orange crystalline).

$2 \mathrm{NaCrO}_{4}+2 \mathrm{H}^{+} \longrightarrow \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{Na}^{+}+\mathrm{H}_{2} \mathrm{O}$

• When sodium dichromate reacts with potassium chloride to form potassium dichromate.

$\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{KCl} \longrightarrow \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{NaCl}$