Sodium borate on reaction with conc. `H_2 SO_4` and `C_2 H_5 OH` gives a compound A which burns with a green edged flame. The compound A is

Options

A. `H_(2)B_(4)O_(7)`
B. `(C_(2)H_(5))_(2)B_(4)O_(7)`
C. `H_(3)BO_(3)`
D. `(C_(2)H_(5))_(3)BO_(3)`

Answer

To solve the problem step by step, we will analyze the reaction of sodium borate with concentrated sulfuric acid and ethanol, leading to the formation of compound A that burns with a green-edged flame.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are sodium borate (Na3BO3), concentrated sulfuric acid (H2SO4), and ethanol (C2H5OH).

2. **Reaction with Concentrated H2SO4**:
- When sodium borate reacts with concentrated sulfuric acid, it produces sodium sulfate (Na2SO4) and boric acid (H3BO3).
- The balanced reaction can be written as:
\[ \text{Na}_3\text{BO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_3\text{BO}_3 \]

3. **Balancing the Reaction**:
- To balance the reaction, we can multiply the reactants and products accordingly:
\[ 2 \text{Na}_3\text{BO}_3 + 3 \text{H}_2\text{SO}_4 \rightarrow 3 \text{Na}_2\text{SO}_4 + 2 \text{H}_3\text{BO}_3 \]

4. **Reaction of Boric Acid with Ethanol**:
- Next, boric acid (H3BO3) reacts with ethanol (C2H5OH) to form triethyl borate (B(OEt)3) and water (H2O).
- The reaction can be represented as:
\[ \text{H}_3\text{BO}_3 + 3 \text{C}_2\text{H}_5\text{OH} \rightarrow \text{B(OEt)}_3 + 3 \text{H}_2\text{O} \]

5. **Identify Compound A**:
- The product formed from the reaction of boric acid with ethanol is triethyl borate (B(OEt)3), which can also be written as C2H5O3B.
- This compound is the one that burns with a green-edged flame.

6. **Conclusion**:
- Therefore, the compound A is triethyl borate (C2H5)3BO3.

### Final Answer:
The compound A is **triethyl borate (C2H5)3BO3**.