`10 mL` of `1 M BaCl_(2)` solution & `5 mL 0.5 M K_(2)SO_(4)` are mixed together to precipitate out `BaSO_(4)`. The amount of `BaSO_(4)` precipated will be

Options

A. 0.005 mol
B. 0.00025 mol`
C. 0.025 mol
D. 0.0025 mol

Explanation

To solve the problem of determining the amount of barium sulfate (BaSO₄) precipitated when mixing 10 mL of 1 M BaCl₂ and 5 mL of 0.5 M K₂SO₄, we can follow these steps:

### Step 1: Calculate the number of moles of BaCl₂
To find the number of moles of BaCl₂, we use the formula:
\[ \text{Number of moles} = \text{Volume (L)} \times \text{Molarity (M)} \]
Given:
- Volume of BaCl₂ = 10 mL = 0.010 L
- Molarity of BaCl₂ = 1 M

Calculating the moles:
\[ \text{Moles of BaCl₂} = 0.010 \, \text{L} \times 1 \, \text{M} = 0.010 \, \text{moles} = 10 \, \text{mmoles} \]

### Step 2: Calculate the number of moles of K₂SO₄
Using the same formula:
\[ \text{Number of moles} = \text{Volume (L)} \times \text{Molarity (M)} \]
Given:
- Volume of K₂SO₄ = 5 mL = 0.005 L
- Molarity of K₂SO₄ = 0.5 M

Calculating the moles:
\[ \text{Moles of K₂SO₄} = 0.005 \, \text{L} \times 0.5 \, \text{M} = 0.0025 \, \text{moles} = 2.5 \, \text{mmoles} \]

### Step 3: Determine the limiting reagent
The balanced chemical equation for the reaction is:
\[ \text{BaCl}_2 + \text{K}_2\text{SO}_4 \rightarrow \text{BaSO}_4 \downarrow + 2 \text{KCl} \]
From the equation, we see that 1 mole of BaCl₂ reacts with 1 mole of K₂SO₄ to produce 1 mole of BaSO₄.

- Moles of BaCl₂ available = 10 mmoles
- Moles of K₂SO₄ available = 2.5 mmoles

Since K₂SO₄ is present in a smaller amount, it is the limiting reagent.

### Step 4: Calculate the amount of BaSO₄ produced
According to the stoichiometry of the reaction, the moles of BaSO₄ produced will be equal to the moles of K₂SO₄ reacted:
\[ \text{Moles of BaSO₄} = \text{Moles of K₂SO₄} = 2.5 \, \text{mmoles} = 0.0025 \, \text{moles} \]

### Final Answer
The amount of BaSO₄ precipitated will be **0.0025 moles**.

---