(i) A white solid mixture of two salts containing a common cations in insoluble in water. It dissolves in dilute HCl producing some gases (with effervescence) that turn an acidified dichromate solution gren. After the gases are passed through the acidified dichromate solution, the emerging gas turns baryta water milky.
(ii) On treatment with dilute `HNO_(3)`, the white solid gives a solution which does not directly give a precipitate with a `BaCl_(2)` solution but gives a white precipitate when warmed with `H_(2)O_(2)` and then treated with a `BaCl_(2)` solution.
(iii) The solution of the mixture in dilute HCl, when treated with `NH_(4)Cl,NH_(4)OH and ` an `Na_(2)HPO_(4)` solution, gives a white precipitate.
Q. The white precipitate obtained in (iii) consists of:

Options

A. `Ba_(3)(PO_(4))_(2)`
B. `Sr_(3)(PO_(4))_(2)`
C. `Ca_(3)(PO_(4))_(2)`
D. `MgNH_(4)PO_(4)*6H_(2)O`

Answer

To solve the problem step by step, we will analyze each part of the question carefully.

### Step 1: Identify the salts in the mixture
The problem states that we have a white solid mixture of two salts containing a common cation that is insoluble in water. The fact that it dissolves in dilute HCl and produces gases indicates that one of the salts likely contains a carbonate or bicarbonate, which would release carbon dioxide (CO₂) upon reaction with HCl.

**Hint:** Consider common salts that contain cations which are insoluble in water and can react with acids.

### Step 2: Analyze the gas produced
The gases produced during the reaction with dilute HCl turn an acidified dichromate solution green. This suggests that one of the gases is likely hydrogen sulfide (H₂S), which can reduce dichromate ions (Cr₂O₇²⁻) to chromium ions (Cr³⁺), causing the solution to change color.

**Hint:** Think about gases that can reduce dichromate ions and their common sources in inorganic salts.

### Step 3: Identify the emerging gas
After passing the gases through the acidified dichromate solution, the emerging gas turns baryta water milky. This indicates the presence of carbon dioxide (CO₂), which reacts with barium hydroxide to form barium carbonate (BaCO₃), a white precipitate.

**Hint:** Remember that CO₂ is a common product of carbonate reactions and can turn lime water or baryta water milky.

### Step 4: Examine the reaction with dilute HNO₃
When the white solid is treated with dilute HNO₃, it produces a solution that does not give a precipitate with BaCl₂ initially but does upon warming with H₂O₂. This suggests the presence of a substance that can form barium sulfate (BaSO₄) only after oxidation, indicating the presence of sulfite ions (SO₃²⁻) in the original salts.

**Hint:** Consider the behavior of sulfites and their oxidation products when treated with HNO₃.

### Step 5: Analyze the final precipitation step
The final part of the problem states that the solution of the mixture in dilute HCl, when treated with NH₄Cl, NH₄OH, and Na₂HPO₄, gives a white precipitate. The presence of ammonium ions (NH₄⁺) and phosphate ions (HPO₄²⁻) suggests the formation of magnesium ammonium phosphate (MAP), which is a white crystalline precipitate.

**Hint:** Recall the reaction between magnesium ions, ammonium ions, and phosphate ions to form magnesium ammonium phosphate.

### Conclusion
The white precipitate obtained in the third step consists of magnesium ammonium phosphate (MgNH₄PO₄·6H₂O).

**Final Answer:** The white precipitate obtained in (iii) consists of magnesium ammonium phosphate.