When an orange coloured crystalline compound (A) was heated with common salt and concentrated sulphuric acid an orange-yellow coloured gas (B) was evolved. The gas (B) when passed through caustic soda solution gave a yellow solution (C) which in turn gave the following reactions.
(i) Addition of silver nitrate solution to (C) gave first a white precipitate which then turns red. Quantitatively, 0.155 g of the gas (B) required 2.0 m moles of `AgNO_(3)` to produce the first trace of red colour.
(ii) Acidification of the solution (C) with dil. `H_(2)SO_(4)` gave an orange solution which contained chromium in +6 oxidation state. The solution liberated iodine from aqueous potassium iodide, leaving a green solution containing chromium in +3. oxidation state. Quantitatively, 0.155 g of the gas (B) liberated 1.5 m moles of iodine.
Deduce the formula of A, B and C and explain the reactions.

Explanation

To solve the problem, we need to identify the compounds A, B, and C based on the given reactions and observations. Let's break it down step by step:

### Step 1: Identify Compound A
The question states that an orange-colored crystalline compound (A) is heated with common salt (NaCl) and concentrated sulfuric acid (H₂SO₄) to produce an orange-yellow gas (B). The only compound that fits this description is **Potassium Dichromate (K₂Cr₂O₇)**.

**Conclusion:**
- Compound A = K₂Cr₂O₇ (Potassium Dichromate)

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