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 (H₂O) are produced when a specific amount of barium hydroxide (Ba(OH)₂) reacts with chloric acid (HClO₃).

The balanced chemical equation is:
\[ \text{Ba(OH)}_2 + 2 \text{HClO}_3 \rightarrow \text{Ba(ClO}_3)_2 + 2 \text{H}_2\text{O} \]

### Step 1: Identify the Stoichiometry
From the balanced equation, we can see that:
- 1 mole of barium hydroxide reacts with 2 moles of chloric acid to produce 2 moles of water.

### Step 2: Determine the Moles of Reactants
We are given:
- 0.1 moles of Ba(OH)₂
- 0.0250 moles of HClO₃

### Step 3: Calculate the Required Moles of HClO₃
To react completely with 0.1 moles of Ba(OH)₂, we need:
\[ 2 \text{ moles of HClO}_3 \text{ for every 1 mole of Ba(OH)}_2 \]
Thus, for 0.1 moles of Ba(OH)₂:
\[ 0.1 \text{ moles Ba(OH)}_2 \times 2 = 0.2 \text{ moles of HClO}_3 \]

### Step 4: Identify the Limiting Reagent
We only have 0.0250 moles of HClO₃ available, which is less than the 0.2 moles required. Therefore, HClO₃ is the limiting reagent.

### Step 5: Calculate the Moles of Water Produced
According to the balanced equation, 2 moles of HClO₃ produce 2 moles of H₂O. Therefore, the moles of water produced from 0.0250 moles of HClO₃ can be calculated as follows:
\[ \text{Moles of H}_2\text{O} = \left(\frac{2 \text{ moles H}_2\text{O}}{2 \text{ moles HClO}_3}\right) \times 0.0250 \text{ moles HClO}_3 \]
\[ \text{Moles of H}_2\text{O} = 0.0250 \text{ moles H}_2\text{O} \]

### Final Answer
Thus, the number of moles of water (H₂O) formed is **0.0250 moles**.

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