Give the Reaction of ethylalcohol with Boric Acid.

Explanation

### Step-by-Step Text Solution:

1. **Identify the Reactants**:
- Ethyl alcohol (C2H5OH) and boric acid (H3BO3) are the two reactants involved in this reaction.

2. **Write the Chemical Reaction**:
- When ethyl alcohol reacts with boric acid, the products formed are ethyl borate and water. The reaction can be represented as:
\[ C_2H_5OH + H_3BO_3 \rightarrow C_2H_5O(B(OH)_2) + H_2O \]

3. **Formation of Ethyl Borate**:
- Ethyl borate is formed from the reaction of ethyl alcohol with boric acid. The formula for ethyl borate can be represented as:
\[ C_2H_5O(B(OH)_2) \text{ or } C_2H_5O \cdot B(OH)_2 \]

4. **Balancing the Reaction**:
- To balance the reaction, we need to ensure that the number of atoms of each element is the same on both sides of the equation. The balanced equation is:
\[ 3 C_2H_5OH + H_3BO_3 \rightarrow C_2H_5O(B(OH)_2) + 3 H_2O \]

5. **Final Balanced Equation**:
- The final balanced equation can be summarized as:
\[ 3 C_2H_5OH + H_3BO_3 \rightarrow 3 C_2H_5OBO_3 + 3 H_2O \]