One mole of `H_3PO_3` on reaction with excess of `NaOH` gives :

Options

A. One mole of `Na_2HPO_3`
B. Two moles of `Na_2H_2PO_3`
C. Two moles of `Na_2HPO_3`
D. One mole of `Na_3PO_3`

Explanation

To solve the question regarding the reaction of one mole of \( H_3PO_3 \) (phosphorous acid) with excess \( NaOH \) (sodium hydroxide), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants:**
- We have one mole of \( H_3PO_3 \) and excess \( NaOH \).

2. **Understand the Reaction:**
- \( H_3PO_3 \) is a diprotic acid, meaning it can donate two protons (H⁺ ions) when reacting with a base.
- The reaction involves the neutralization of the acidic protons by the hydroxide ions (OH⁻) from \( NaOH \).

3. **Write the Balanced Reaction:**
- The balanced equation for the reaction can be represented as:
\[ H_3PO_3 + 2 NaOH \rightarrow Na_2HPO_3 + 2 H_2O \]
- Here, \( H_3PO_3 \) donates two protons to two \( NaOH \) molecules, forming \( Na_2HPO_3 \) (sodium hydrogen phosphite) and water.

4. **Determine the Products:**
- From the balanced equation, we see that one mole of \( H_3PO_3 \) reacts with two moles of \( NaOH \) to produce one mole of \( Na_2HPO_3 \) and two moles of water.

5. **Conclusion:**
- Therefore, the reaction of one mole of \( H_3PO_3 \) with excess \( NaOH \) yields one mole of \( Na_2HPO_3 \).

### Final Answer:
One mole of \( H_3PO_3 \) on reaction with excess \( NaOH \) gives **one mole of \( Na_2HPO_3 \)**.

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