The amount of `BaSO_(4)` formed upon mixing 100mL of 20.8% `BaCl_(2)` solution with 50mL of 9.8% `H_(2)SO_(4)` solution will be :
(Ba = 137, Cl = 35.5, S = 32, H = 1 and O =16)

Options

A. 23.3g
B. 11.65 g
C. 30.6 g
D. None of these

Explanation

To solve the problem of determining the amount of barium sulfate (BaSO₄) formed when mixing barium chloride (BaCl₂) and sulfuric acid (H₂SO₄), we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between barium chloride and sulfuric acid can be represented as:
\[ \text{BaCl}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{HCl} \]

### Step 2: Calculate the moles of BaCl₂
1. **Find the mass of BaCl₂ in the solution:**
- The concentration of BaCl₂ is 20.8% w/v, which means there are 20.8 grams of BaCl₂ in 100 mL of solution.
- Therefore, in 100 mL, the mass of BaCl₂ is:
\[ \text{Mass of BaCl}_2 = 20.8 \text{ grams} \]

2. **Calculate the molar mass of BaCl₂:**
- Molar mass of Ba = 137 g/mol
- Molar mass of Cl = 35.5 g/mol (2 Cl atoms)
- Molar mass of BaCl₂ = 137 + (35.5 × 2) = 137 + 71 = 208 g/mol

3. **Calculate the number of moles of BaCl₂:**
\[ \text{Moles of BaCl}_2 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{20.8 \text{ g}}{208 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 3: Calculate the moles of H₂SO₄
1. **Find the mass of H₂SO₄ in the solution:**
- The concentration of H₂SO₄ is 9.8% w/v, which means there are 9.8 grams of H₂SO₄ in 100 mL of solution.
- Therefore, in 50 mL, the mass of H₂SO₄ is:
\[ \text{Mass of H}_2\text{SO}_4 = \frac{9.8 \text{ grams}}{2} = 4.9 \text{ grams} \]

2. **Calculate the molar mass of H₂SO₄:**
- Molar mass of H = 1 g/mol (2 H atoms)
- Molar mass of S = 32 g/mol
- Molar mass of O = 16 g/mol (4 O atoms)
- Molar mass of H₂SO₄ = (1 × 2) + 32 + (16 × 4) = 2 + 32 + 64 = 98 g/mol

3. **Calculate the number of moles of H₂SO₄:**
\[ \text{Moles of H}_2\text{SO}_4 = \frac{4.9 \text{ g}}{98 \text{ g/mol}} = 0.05 \text{ moles} \]

### Step 4: Determine the limiting reagent
From the balanced equation, we see that the stoichiometry is 1:1 between BaCl₂ and H₂SO₄. We have:
- 0.1 moles of BaCl₂
- 0.05 moles of H₂SO₄

Since H₂SO₄ is present in lesser moles, it is the limiting reagent.

### Step 5: Calculate the moles of BaSO₄ produced
According to the reaction stoichiometry:
- 1 mole of H₂SO₄ produces 1 mole of BaSO₄.
- Therefore, 0.05 moles of H₂SO₄ will produce 0.05 moles of BaSO₄.

### Step 6: Calculate the mass of BaSO₄ produced
1. **Calculate the molar mass of BaSO₄:**
- Molar mass of Ba = 137 g/mol
- Molar mass of S = 32 g/mol
- Molar mass of O = 16 g/mol (4 O atoms)
- Molar mass of BaSO₄ = 137 + 32 + (16 × 4) = 137 + 32 + 64 = 233 g/mol

2. **Calculate the mass of BaSO₄ produced:**
\[ \text{Mass of BaSO}_4 = \text{Moles} \times \text{Molar Mass} = 0.05 \text{ moles} \times 233 \text{ g/mol} = 11.65 \text{ grams} \]

### Final Answer
The amount of BaSO₄ formed is **11.65 grams**.

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