For the reaction
`Ba(OH)_(2)+2HClO_(3)rarr Ba(ClO_(3))_(2)+2H_(2)O` calculate the no. of moles of `H_(2)O` formed when 0.1 mole of `Ba(OH)_(2)` is treated with 0.0250 moles of `HClO_(3)`.

Options

A. 0.1
B. 0.125
C. 0.25
D. 0.025

Answer

To solve the problem of calculating the number of moles of water (H₂O) formed when 0.1 mole of barium hydroxide (Ba(OH)₂) is treated with 0.0250 moles of hydrochloric acid (HClO₃), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ \text{Ba(OH)}_2 + 2 \text{HClO}_3 \rightarrow \text{Ba(ClO}_3)_2 + 2 \text{H}_2\text{O} \]

### Step 2: Identify the stoichiometry of the reaction
From the balanced equation, we can see that:
- 1 mole of Ba(OH)₂ reacts with 2 moles of HClO₃ to produce 2 moles of H₂O.

### Step 3: Determine the amount of HClO₃ needed for the reaction
For 0.1 moles of Ba(OH)₂, the amount of HClO₃ required can be calculated as follows:
\[ \text{Required moles of HClO}_3 = 2 \times \text{moles of Ba(OH)}_2 = 2 \times 0.1 = 0.2 \text{ moles} \]

### Step 4: Compare the required moles of HClO₃ with the available moles
We have only 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 H₂O produced based on the limiting reagent
From the stoichiometry of the reaction:
- 2 moles of HClO₃ produce 2 moles of H₂O.
This means that 1 mole of HClO₃ produces 1 mole of H₂O.

Using the available moles of HClO₃:
\[ \text{Moles of H}_2\text{O produced} = \text{moles of HClO}_3 = 0.0250 \text{ moles} \]

### Conclusion
The number of moles of H₂O formed when 0.1 mole of Ba(OH)₂ is treated with 0.0250 moles of HClO₃ is **0.0250 moles**.

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