A white substance (A) reacts with dilute `H_(2)SO_(4)` to produce a colourless suffocating gas (B) and a colourless solution (C )/ The reaction of gas (B) with potassium iodate and starch solution produce a blue colour solution. Aqueous Solution of (A) gives a white precipitate with `BaCl_(2)` solution which is soluble in dilute `HCl`. Addition of aqueous `NH_(3)` or `NaOH` to (C ) produced first a precipitate which dissolves in excess of the respective reagent to poduced a clear solution. Similarly addition of excess of potassium ferrocyanide to (C ) produced a precipitate (D) which also dissolves in aqeous `NaOH` giving a clear solution. Identify (A),(B),(C ) and (D ). Write the equation of the reactions involved.

Answer

To solve the given problem step by step, we will identify the substances (A, B, C, and D) and write the corresponding chemical equations for the reactions involved.

### Step 1: Identify Substance A
The white substance (A) is identified as **Zinc Sulfide (ZnS)**. This is based on the information that it reacts with dilute sulfuric acid to produce a colorless gas and a colorless solution.

### Step 2: Reaction of A with Dilute H₂SO₄
When Zinc Sulfide reacts with dilute sulfuric acid, the following reaction occurs:
\[ \text{ZnS} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{SO}_2 + \text{H}_2\text{O} \]
- Here, **B** (the colorless suffocating gas) is **Sulfur Dioxide (SO₂)**.
- **C** (the colorless solution) is **Zinc Sulfate (ZnSO₄)**.

### Step 3: Reaction of Gas B with Potassium Iodate and Starch
The gas (B), which is SO₂, reacts with potassium iodate (KIO₃) and starch to produce a blue color due to the formation of iodine (I₂):
\[ \text{SO}_2 + \text{KIO}_3 + \text{H}_2\text{O} \rightarrow \text{I}_2 + \text{K}_2\text{SO}_4 + \text{H}^+ \]
- The blue color indicates the presence of iodine in solution.

### Step 4: Aqueous Solution of A with BaCl₂
The aqueous solution of (A) gives a white precipitate with BaCl₂:
\[ \text{Zn}^{2+} + \text{BaCl}_2 \rightarrow \text{BaSO}_4 (s) + \text{ZnSO}_4 \]
- The white precipitate formed is **Barium Sulfate (BaSO₄)**, which is soluble in dilute HCl.

### Step 5: Reaction of C with NH₃ or NaOH
When aqueous solution C (ZnSO₄) is treated with NH₃ or NaOH, it first forms a white precipitate:
\[ \text{ZnSO}_4 + 2\text{NaOH} \rightarrow \text{Zn(OH)}_2 (s) + \text{Na}_2\text{SO}_4 \]
- The white precipitate is **Zinc Hydroxide (Zn(OH)₂)**, which dissolves in excess NaOH to form:
\[ \text{Zn(OH)}_2 + 2\text{OH}^- \rightarrow \text{Zn(OH)}_4^{2-} \]

### Step 6: Reaction of C with Potassium Ferrocyanide
When excess potassium ferrocyanide is added to C (ZnSO₄), it produces a precipitate (D):
\[ 3\text{ZnSO}_4 + 2\text{K}_4[\text{Fe(CN)}_6] \rightarrow \text{K}_2[\text{Zn}_3[\text{Fe(CN)}_6]_2] + 6\text{K}_2\text{SO}_4 \]
- The precipitate (D) is **Zinc Ferrocyanide (K₂Zn₃[Fe(CN)₆]₂)**, which also dissolves in excess NaOH.

### Summary of Identifications
- **A**: Zinc Sulfide (ZnS)
- **B**: Sulfur Dioxide (SO₂)
- **C**: Zinc Sulfate (ZnSO₄)
- **D**: Zinc Ferrocyanide (K₂Zn₃[Fe(CN)₆]₂)

### Chemical Equations
1. \( \text{ZnS} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{SO}_2 + \text{H}_2\text{O} \)
2. \( \text{SO}_2 + \text{KIO}_3 + \text{H}_2\text{O} \rightarrow \text{I}_2 + \text{K}_2\text{SO}_4 + \text{H}^+ \)
3. \( \text{ZnSO}_4 + 2\text{NaOH} \rightarrow \text{Zn(OH)}_2 (s) + \text{Na}_2\text{SO}_4 \)
4. \( 3\text{ZnSO}_4 + 2\text{K}_4[\text{Fe(CN)}_6] \rightarrow \text{K}_2[\text{Zn}_3[\text{Fe(CN)}_6]_2] + 6\text{K}_2\text{SO}_4 \)