500 mL of 0.250 M `Na_(2)SO_(4)` solution is treated with 15.00 g of `BaCl_(2)`. Moles of `BaSO_(4)` formed are

Options

A. `0.72`
B. `0.072`
C. `0.168`
D. `0.0168`

Explanation

To solve the problem of how many moles of `BaSO4` are formed when 500 mL of 0.250 M `Na2SO4` solution is treated with 15.00 g of `BaCl2`, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between `Na2SO4` and `BaCl2` can be represented as:
\[ \text{Na}_2\text{SO}_4 + \text{BaCl}_2 \rightarrow \text{BaSO}_4 + 2 \text{NaCl} \]

### Step 2: Calculate the moles of `BaCl2`
To find the moles of `BaCl2`, we first need to calculate its molar mass:
- Molar mass of `Ba` (Barium) = 137.3 g/mol
- Molar mass of `Cl` (Chlorine) = 35.5 g/mol

Molar mass of `BaCl2`:
\[ \text{Molar mass of BaCl}_2 = 137.3 + 2 \times 35.5 = 137.3 + 71.0 = 208.3 \, \text{g/mol} \]

Now, we can calculate the moles of `BaCl2` using the formula:
\[ \text{Moles of BaCl}_2 = \frac{\text{mass}}{\text{molar mass}} \]
\[ \text{Moles of BaCl}_2 = \frac{15.00 \, \text{g}}{208.3 \, \text{g/mol}} \approx 0.072 \, \text{moles} \]

### Step 3: Calculate the moles of `Na2SO4`
We can find the moles of `Na2SO4` using its molarity and volume:
\[ \text{Moles of Na}_2\text{SO}_4 = \text{Molarity} \times \text{Volume (in L)} \]
Convert volume from mL to L:
\[ 500 \, \text{mL} = 0.500 \, \text{L} \]

Now, calculate the moles:
\[ \text{Moles of Na}_2\text{SO}_4 = 0.250 \, \text{M} \times 0.500 \, \text{L} = 0.125 \, \text{moles} \]

### Step 4: Determine the limiting reagent
From the balanced equation, we see that 1 mole of `Na2SO4` reacts with 1 mole of `BaCl2`. We have:
- Moles of `Na2SO4` = 0.125 moles
- Moles of `BaCl2` = 0.072 moles

Since `BaCl2` has fewer moles than `Na2SO4`, it is the limiting reagent.

### Step 5: Calculate the moles of `BaSO4` formed
According to the balanced equation, 1 mole of `BaCl2` produces 1 mole of `BaSO4`. Therefore, the moles of `BaSO4` formed will be equal to the moles of `BaCl2` used:
\[ \text{Moles of BaSO}_4 = \text{Moles of BaCl}_2 = 0.072 \, \text{moles} \]

### Final Answer
The moles of `BaSO4` formed are approximately **0.072 moles**.

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