For the reaction `Ba(OH)_(2)+2HClO_(3)rarr Ba(ClO_(3))_(2)+2H_(2)O`. Calculate the number of moles of `H_(2)O` formed when 0.1 mole of `Na(OH)_(2)` is treated with 0.0250 mole `HClO_(3)`.

Options

A. 0.1
B. 0.125
C. 0.025
D. 3.75

Explanation

To solve the problem, we need to analyze the given chemical reaction and determine how many moles of water (\(H_2O\)) are produced when \(0.1\) mole of barium hydroxide (\(Ba(OH)_2\)) reacts with \(0.0250\) mole of chloric acid (\(HClO_3\)).

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ Ba(OH)_2 + 2HClO_3 \rightarrow Ba(ClO_3)_2 + 2H_2O \]

### Step 2: Identify the stoichiometric ratios
From the balanced equation, we can see that:
- \(1\) mole of \(Ba(OH)_2\) reacts with \(2\) moles of \(HClO_3\) to produce \(2\) moles of \(H_2O\).

### Step 3: Determine the limiting reagent
We have:
- \(0.1\) mole of \(Ba(OH)_2\)
- \(0.0250\) mole of \(HClO_3\)

To find out how much \(HClO_3\) is required to react with \(0.1\) mole of \(Ba(OH)_2\):
\[ \text{Required moles of } HClO_3 = 0.1 \text{ moles of } Ba(OH)_2 \times 2 = 0.2 \text{ moles of } HClO_3 \]

Since we only have \(0.0250\) moles of \(HClO_3\), which is less than the \(0.2\) moles required, \(HClO_3\) is the limiting reagent.

### Step 4: Calculate the moles of \(H_2O\) produced
From the balanced equation, we know that:
- \(2\) moles of \(HClO_3\) produce \(2\) moles of \(H_2O\).

Using the stoichiometry of the reaction:
\[ \text{If } 2 \text{ moles of } HClO_3 \text{ produce } 2 \text{ moles of } H_2O, \]
\[ \text{Then } 0.0250 \text{ moles of } HClO_3 \text{ will produce } 0.0250 \text{ moles of } H_2O. \]

### Final Answer
Thus, the number of moles of \(H_2O\) formed is:
\[ \boxed{0.0250} \]