An aqueous solution of `FeSO_(4)*Al_(2)(SO_(4))_(3)` and chromium alum is heated with excess of `Na_(2)O_(2)` and filtered. The material obtained are:

Options

A. a colourless filtrate and green residue
B. a yellow filtrate and brown residue
C. a yellow filtrate annd a green residue
D. a green filtrate and a brown residue

Answer

To solve the problem, we need to analyze the reaction that occurs when an aqueous solution of `FeSO4·Al2(SO4)3` and chromium alum is heated with excess `Na2O2`.

### Step-by-Step Solution:

1. **Identify the Components**:
- The solution contains ferrous sulfate (`FeSO4`), aluminum sulfate (`Al2(SO4)3`), and chromium alum (which can be represented as `KCr(SO4)2·12H2O`).

2. **Heating with Sodium Peroxide**:
- When the mixture is heated with excess sodium peroxide (`Na2O2`), it acts as an oxidizing agent.
- Sodium peroxide can oxidize ferrous ions (`Fe^2+`) to ferric ions (`Fe^3+`), and it can also oxidize chromic ions (`Cr^3+`) to chromate ions (`CrO4^2-`).

3. **Formation of Precipitates**:
- The ferrous ions will convert to ferric hydroxide (`Fe(OH)3`) as a green precipitate when they react with hydroxide ions produced in the solution.
- The chromate ions formed from the oxidation of chromium will give a yellow color to the solution.

4. **Filtration**:
- After heating and allowing the reaction to proceed, the mixture is filtered.
- The green precipitate of ferric hydroxide will be retained on the filter paper, while the yellow solution containing chromate ions will pass through as the filtrate.

5. **Conclusion**:
- The material obtained after filtration consists of:
- A **yellow filtrate** (due to the presence of `CrO4^2-` ions).
- A **green residue** (the precipitate of `Fe(OH)3`).

### Final Answer:
The material obtained are a yellow filtrate and a green residue.