Identify A, B, C, D and E.
A white substance `(A)` reacts with dilute `H_(2)SO_(4)` to produce a colourless gas `(B)` and a colourless solution `( C)`.The reaction between `(B)` and acidified `K_(2)Cr_(2)O_(7)` solution produces a green solution and a slightly coloured precipitate `(D)`.The substance `(D)` burns in air to produce a gas `(E)` which reacts with `(B)` to yield `(D)` and a colourless liquid. Anhydrous copper sulphate is turned blue on addition of this colouless liquid.Addition of aqueous `NH_(3)` or `NaOH` to `( C)` produces first a precipitate

Explanation

To solve the problem step by step, we will identify the substances A, B, C, D, and E based on the reactions described in the question.

### Step 1: Identify Substance A
- **Given**: A white substance (A) reacts with dilute H₂SO₄.
- **Reaction**: When zinc sulfide (ZnS) reacts with dilute sulfuric acid (H₂SO₄), it produces hydrogen sulfide (H₂S) and zinc sulfate (ZnSO₄).
- **Conclusion**:
- **A = Zinc sulfide (ZnS)**

### Step 2: Identify Substance B
- **Given**: The reaction produces a colorless gas (B).
- **From Step 1**: The gas produced from the reaction of ZnS with H₂SO₄ is hydrogen sulfide (H₂S).
- **Conclusion**:
- **B = Hydrogen sulfide (H₂S)**

### Step 3: Identify Substance C
- **Given**: The reaction also produces a colorless solution (C).
- **From Step 1**: The solution produced is zinc sulfate (ZnSO₄).
- **Conclusion**:
- **C = Zinc sulfate (ZnSO₄)**

### Step 4: Identify Substance D
- **Given**: The reaction between B (H₂S) and acidified K₂Cr₂O₇ produces a green solution and a slightly colored precipitate (D).
- **Reaction**: H₂S reacts with acidified potassium dichromate (K₂Cr₂O₇) to produce chromium (III) ions (Cr³⁺), which are green, and sulfur (S) as a precipitate.
- **Conclusion**:
- **D = Sulfur (S)**

### Step 5: Identify Substance E
- **Given**: Substance D (sulfur) burns in air to produce a gas (E).
- **Reaction**: Sulfur burns in air to produce sulfur dioxide (SO₂).
- **Conclusion**:
- **E = Sulfur dioxide (SO₂)**

### Step 6: Confirm the Reaction of E with B
- **Given**: Gas E reacts with B (H₂S) to yield D (sulfur) and a colorless liquid.
- **Reaction**: The reaction of SO₂ with H₂S produces sulfur and water (H₂O).
- **Conclusion**: The colorless liquid is water (H₂O).

### Step 7: Identify the Color Change with Anhydrous Copper Sulfate
- **Given**: Anhydrous copper sulfate turns blue on addition of the colorless liquid.
- **Conclusion**: This is due to the addition of water (H₂O), which hydrates the copper sulfate to form blue copper(II) sulfate (CuSO₄·5H₂O).

### Step 8: Identify the Precipitate Formation
- **Given**: Addition of aqueous NH₃ or NaOH to C (zinc sulfate) produces a precipitate.
- **Reaction**: ZnSO₄ reacts with NaOH or NH₃ to form zinc hydroxide (Zn(OH)₂), which is a white precipitate.
- **Conclusion**: The precipitate formed is zinc hydroxide (Zn(OH)₂).

### Summary of Identifications
- **A = Zinc sulfide (ZnS)**
- **B = Hydrogen sulfide (H₂S)**
- **C = Zinc sulfate (ZnSO₄)**
- **D = Sulfur (S)**
- **E = Sulfur dioxide (SO₂)**