500mL of 0.25M `Na_(2)SO_(4)` solution is added to an aquesous solution is 15g of `BaCl_(2)` resulting in the formation of a white precipatate of insoluble `BaSO_(4)`. How many moles and how many grams of `BaSO_(4)` are formed.

Answer

To solve the problem, we need to follow these steps:

### Step 1: Calculate the moles of `Na2SO4`
We have a solution of `Na2SO4` with a concentration of 0.25 M and a volume of 500 mL.

1. Convert the volume from mL to L:
\[ 500 \, \text{mL} = 0.500 \, \text{L} \]

2. Use the molarity formula to find the number of moles:
\[ \text{Moles of } Na2SO4 = \text{Molarity} \times \text{Volume (L)} = 0.25 \, \text{mol/L} \times 0.500 \, \text{L} = 0.125 \, \text{mol} \]

### Step 2: Calculate the moles of `BaCl2`
Next, we need to find the number of moles of `BaCl2` in 15 g.

1. First, calculate the molar mass of `BaCl2`:
- Molar mass of Ba = 137.33 g/mol
- Molar mass of Cl = 35.45 g/mol
- Molar mass of `BaCl2` = 137.33 + (2 × 35.45) = 137.33 + 70.90 = 208.23 g/mol

2. Now, calculate the moles of `BaCl2`:
\[ \text{Moles of } BaCl2 = \frac{\text{mass}}{\text{molar mass}} = \frac{15 \, \text{g}}{208.23 \, \text{g/mol}} \approx 0.072 \, \text{mol} \]

### Step 3: Determine the limiting reagent
The reaction between `Na2SO4` and `BaCl2` can be represented as:
\[ Na2SO4 + BaCl2 \rightarrow BaSO4 \downarrow + 2NaCl \]

From the balanced equation, we see that 1 mole of `Na2SO4` reacts with 1 mole of `BaCl2` to produce 1 mole of `BaSO4`.

- We have 0.125 moles of `Na2SO4` and 0.072 moles of `BaCl2`. Since `BaCl2` has fewer moles, it is the limiting reagent.

### Step 4: Calculate the moles of `BaSO4` formed
Since `BaCl2` is the limiting reagent, the moles of `BaSO4` produced will be equal to the moles of `BaCl2` reacted:
\[ \text{Moles of } BaSO4 = \text{Moles of } BaCl2 = 0.072 \, \text{mol} \]

### Step 5: Calculate the mass of `BaSO4` formed
Now, we need to calculate the mass of `BaSO4` produced.

1. First, calculate the molar mass of `BaSO4`:
- Molar mass of `BaSO4` = 137.33 (Ba) + 32.07 (S) + (4 × 16.00 (O)) = 137.33 + 32.07 + 64.00 = 233.40 g/mol

2. Now, calculate the mass of `BaSO4`:
\[ \text{Mass of } BaSO4 = \text{moles} \times \text{molar mass} = 0.072 \, \text{mol} \times 233.40 \, \text{g/mol} \approx 16.83 \, \text{g} \]

### Final Answer
- Moles of `BaSO4` formed: **0.072 mol**
- Mass of `BaSO4` formed: **16.83 g**

---