Fusion of chromite ore with sodium carbonate in excess of air gives a compound (X) (yellow solution) X when treated with sulphuric acid gives:

Options

A. `Na_(2)Cr_(2)O_(7)`
B. `Na_(2)Cr_(2)O_(7)*5H_(2)O`
C. `Na_(2)Cr_(2)O_(7)*10H_(2)O`
D. `Na_(2)Cr_(2)O_(7)*2H_(2)O`

Answer

To solve the problem step-by-step, we will analyze the fusion of chromite ore with sodium carbonate and the subsequent reaction with sulfuric acid.

### Step 1: Identify the Chromite Ore
The chromite ore is represented by the chemical formula FeCr2O4, which contains iron (Fe) and chromium (Cr).

### Step 2: Fusion with Sodium Carbonate
When chromite ore is fused with sodium carbonate (Na2CO3) in the presence of excess air (O2), a chemical reaction occurs. The oxygen from the air oxidizes the chromium in the chromite ore.

The reaction can be represented as:
\[ \text{FeCr}_2\text{O}_4 + 4 \text{Na}_2\text{CO}_3 + 3 \text{O}_2 \rightarrow 2 \text{Na}_2\text{Cr}_2\text{O}_7 + \text{Fe}_2\text{O}_3 + 4 \text{CO}_2 \]

### Step 3: Identify the Compound (X)
From the reaction, the compound (X) formed is sodium dichromate (Na2Cr2O7), which is known to produce a yellow solution.

### Step 4: Treatment with Sulfuric Acid
When sodium dichromate (X) is treated with sulfuric acid (H2SO4), it undergoes a reaction to form chromic acid (H2Cr2O7) and sodium sulfate (Na2SO4).

The reaction can be represented as:
\[ \text{Na}_2\text{Cr}_2\text{O}_7 + 2 \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{Cr}_2\text{O}_7 + \text{Na}_2\text{SO}_4 \]

### Step 5: Conclusion
The final product when compound (X) is treated with sulfuric acid is chromic acid (H2Cr2O7).

### Final Answer
When the compound (X) is treated with sulfuric acid, it gives chromic acid (H2Cr2O7).

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