`0.5` mole of `H_(2)SO_(4)` is mixed with `0.2` mole of `Ca(OH)_(2)`. The maximum number of mole of `CaSO_(4)` formed is:

Options

A. 0.2
B. 0.5
C. 0.4
D. 0.1

Explanation

To solve the problem of how many moles of \( CaSO_4 \) are formed when \( 0.5 \) moles of \( H_2SO_4 \) are mixed with \( 0.2 \) moles of \( Ca(OH)_2 \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction between sulfuric acid and calcium hydroxide is:

\[ H_2SO_4 + Ca(OH)_2 \rightarrow CaSO_4 + 2H_2O \]

### Step 2: Identify the stoichiometry
From the balanced equation, we can see that:
- \( 1 \) mole of \( H_2SO_4 \) reacts with \( 1 \) mole of \( Ca(OH)_2 \) to produce \( 1 \) mole of \( CaSO_4 \).

### Step 3: Determine the limiting reagent
We have:
- \( 0.5 \) moles of \( H_2SO_4 \)
- \( 0.2 \) moles of \( Ca(OH)_2 \)

To find the limiting reagent, we compare the mole ratio needed for the reaction:
- For \( 0.5 \) moles of \( H_2SO_4 \), we would need \( 0.5 \) moles of \( Ca(OH)_2 \).
- However, we only have \( 0.2 \) moles of \( Ca(OH)_2 \).

Since \( Ca(OH)_2 \) is present in a smaller amount than required, it is the limiting reagent.

### Step 4: Calculate the moles of \( CaSO_4 \) produced
Since \( Ca(OH)_2 \) is the limiting reagent, we can determine the amount of \( CaSO_4 \) produced:
- From the balanced equation, \( 1 \) mole of \( Ca(OH)_2 \) produces \( 1 \) mole of \( CaSO_4 \).
- Therefore, \( 0.2 \) moles of \( Ca(OH)_2 \) will produce \( 0.2 \) moles of \( CaSO_4 \).

### Step 5: Conclusion
The maximum number of moles of \( CaSO_4 \) formed is \( 0.2 \) moles.

### Final Answer
The maximum number of moles of \( CaSO_4 \) formed is \( 0.2 \).

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