`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. `1.5`

Explanation

To solve the problem of how many moles of \( \text{CaSO}_4 \) can be formed when \( 0.5 \) moles of \( \text{H}_2\text{SO}_4 \) are mixed with \( 0.2 \) moles of \( \text{Ca(OH)}_2 \), we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between sulfuric acid and calcium hydroxide can be represented by the following balanced equation:

\[ \text{H}_2\text{SO}_4 + \text{Ca(OH)}_2 \rightarrow \text{CaSO}_4 + 2\text{H}_2\text{O} \]

From the equation, we can see that 1 mole of \( \text{H}_2\text{SO}_4 \) reacts with 1 mole of \( \text{Ca(OH)}_2 \) to produce 1 mole of \( \text{CaSO}_4 \).

### Step 2: Determine the limiting reactant
We have:
- \( 0.5 \) moles of \( \text{H}_2\text{SO}_4 \)
- \( 0.2 \) moles of \( \text{Ca(OH)}_2 \)

From the balanced equation, we need equal moles of \( \text{H}_2\text{SO}_4 \) and \( \text{Ca(OH)}_2 \) for the reaction to proceed completely.

Since we have \( 0.5 \) moles of \( \text{H}_2\text{SO}_4 \), we would need \( 0.5 \) moles of \( \text{Ca(OH)}_2 \) to fully react with it. However, we only have \( 0.2 \) moles of \( \text{Ca(OH)}_2 \). Therefore, \( \text{Ca(OH)}_2 \) is the limiting reactant.

### Step 3: Calculate the moles of \( \text{CaSO}_4 \) produced
Since \( \text{Ca(OH)}_2 \) is the limiting reactant, we can use its amount to find out how much \( \text{CaSO}_4 \) can be formed. According to the balanced equation, 1 mole of \( \text{Ca(OH)}_2 \) produces 1 mole of \( \text{CaSO}_4 \).

Thus, \( 0.2 \) moles of \( \text{Ca(OH)}_2 \) will produce:

\[ 0.2 \text{ moles of } \text{CaSO}_4 \]

### Final Answer
The maximum number of moles of \( \text{CaSO}_4 \) formed is \( 0.2 \) moles.

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